chapter 10 conics, parametric equations, and polar coordinates · 362 chapter 10 conics, parametric...

109
CHAPTER 10 Conics, Parametric Equations, and Polar Coordinates Section 10.1 Conics and Calculus . . . . . . . . . . . . . . . . . . . . 360 Section 10.2 Plane Curves and Parametric Equations . . . . . . . . . . 383 Section 10.3 Parametric Equations and Calculus . . . . . . . . . . . . 392 Section 10.4 Polar Coordinates and Polar Graphs . . . . . . . . . . . . 407 Section 10.5 Area and Arc Length in Polar Coordinates . . . . . . . . 423 Section 10.6 Polar Equations of Conics and Kepler’s Laws . . . . . . . 436 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461

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C H A P T E R 1 0Conics, Parametric Equations,

and Polar Coordinates

Section 10.1 Conics and Calculus . . . . . . . . . . . . . . . . . . . . 360

Section 10.2 Plane Curves and Parametric Equations . . . . . . . . . . 383

Section 10.3 Parametric Equations and Calculus . . . . . . . . . . . . 392

Section 10.4 Polar Coordinates and Polar Graphs . . . . . . . . . . . . 407

Section 10.5 Area and Arc Length in Polar Coordinates . . . . . . . . 423

Section 10.6 Polar Equations of Conics and Kepler’s Laws . . . . . . . 436

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461

C H A P T E R 1 0Conics, Parametric Equations, and Polar Coordinates

Section 10.1 Conics and Calculus

360

10.

Vertex:

Focus:

Directrix: y � 2

�0, �2��0, 0� x

4 8−4−8

−4

−8

−12

(0, 0)

y x2 � 4��2�y

x2 � 8y � 0

12.

Vertex:

Focus:

Directrix: y � 0

�1, �4�x

4 8−4−8

−4

−8

−12

(1, 2)−

y�1, �2�

�x � 1�2 � 4��2�� y � 2�

�x � 1�2 � 8�y � 2� � 0

1.

Vertex:

Opens to the rightMatches graph (h).

p � 1 > 0

�0, 0�

y2 � 4x 2.

Vertex:

Opens upwardMatches graph (a).

p � 2 > 0

�0, 0�

x2 � 8y 3.

Vertex:

Opens downwardMatches graph (e).

p � �12 < 0

��3, 2�

�x � 3�2 � �2�y � 2�

4.

Center:EllipseMatches (b)

�2, �1�

�x � 2�2

16�

�y � 1�2

4� 1 5.

Center:EllipseMatches (f)

�0, 0�

x2

9�

y2

4� 1 6.

Circle radius 3. Matches (g)

x2

9�

y2

9� 1

7.

Hyperbola

Center:

Vertical transverse axisMatches (c)

�0, 0�

y2

16�

x2

1� 1 8.

Hyperbola

Center:

Horizontal transverse axisMatches (d)

��2, 0�

�x � 2�2

9�

y2

4� 1

9.

Vertex:

Focus:

Directrix: x �32

��32, 0� 8

4

8

4

4812x

(0, 0)

y�0, 0�

y2 � �6x � 4��32�x

11.

Vertex:

Focus:

Directrix: x � �2.75

��3.25, 2�

x

2

4

−2

−2−4−6−8

−4

( 3, 2)−

y��3, 2�

�y � 2�2 � 4��14��x � 3�

�x � 3� � � y � 2�2 � 0

Section 10.1 Conics and Calculus 361

14.

Vertex:

Focus:

Directrix: x � 0

��4, �3�

x

4

8

−4−8−12−16−20

−8

−12

( 2, 3)− −

y��2, �3�

�y � 3�2 � 4��2��x � 2�

y2 � 6y � 9 � �8x � 25 � 9

y2 � 6y � 8x � 25 � 0

16.

Vertex:

Focus:

Directrix: x � 4

�0, �2�

x

4

4

2

6−4−6 −2

−8

−6

−4

(2, 2)−

y�2, �2�

�y � 2�2 � 4��2��x � 2�

y2 � 4y � 4 � �8x � 12 � 4

y2 � 4y � 8x � 12 � 0

13.

Vertex:

Focus:

Directrix: x � �2

�0, 2�

6

4

6

42

2

2x

(−1, 2)

y��1, 2�

�y � 2�2 � 4�1��x � 1�

y2 � 4y � 4 � 4x � 4

y2 � 4y � 4x � 0

15.

Vertex:

Focus:

Directrix: y � 3

��2, 1�

x2

4

−2

−4

(−2, 2)

−2−4−6

y��2, 2�

�x � 2�2 � 4��1��y � 2�

x2 � 4x � 4 � �4y � 4 � 4

x2 � 4x � 4y � 4 � 0

17.

Vertex:

Focus:

Directrix:

y2 � �12 � �1

4 � x

y1 � �12 ± �1

4 � x

x �12

�0, �12�

−5 2

−3

2�14, �1

2� � y �

12�2

� 4��14��x �

14�

y2 � y �14 � �x �

14

y2 � x � y � 0 18.

Vertex:

Focus:

Directrix: y �196

�4, 16�−2

−4

10

4�4, 53� �x � 4�2 � 4��3

2��y �53�

�x � 4�2 � �6�y �53�

�6y � 10 � �x � 4�2

�6y � �x � 4�2 � 10

y � �16�x2 � 8x � 6� � �

16�x2 � 8x � 16 � 10�

20.

Vertex:

Focus:

Directrix: y � 1

�1, �3�

−10

10−8

2�1, �1�

�x � 1�2 � 4��2��y � 1�

x2 � 2x � 1 � �8y � 9 � 1

x2 � 2x � 8y � 9 � 019.

Vertex:

Focus:

Directrix:

y2 � �2�x � 1

y1 � 2�x � 1

x � �2

�0, 0�−6

−4

4

6

��1, 0�

� 4�1��x � 1�

y2 � 4x � 4

y2 � 4x � 4 � 0

362 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

21.

y2 � 4y � 8x � 20 � 0

�y � 2�2 � 4��2��x � 3�

23.

x2 � 24y � 96 � 0

x2 � 4�6��y � 4�

�x � h�2 � 4p�y � k�

25.

x2 � y � 4 � 0

y � 4 � x2

27. Since the axis of the parabola is vertical, the form of theequation is Now, substituting the val-ues of the given coordinates into this equation, we obtain

Solving this system, we have Therefore,

or 5x2 � 14x � 3y � 9 � 0.y �53x2 �

143 x � 3

c � 3.b � �143 ,a �

53,

11 � 16a � 4b � c.4 � 9a � 3b � c, 3 � c,

y � ax2 � bx � c.

29.

Center:

Foci:

Vertices:

e ��32

�±2, 0��±�3, 0�

�0, 0�

c2 � 3b2 � 1,a2 � 4,

x2

4�

y2

1� 1

x−1 1

2

−2

(0, 0)

yx2 � 4y2 � 4

31.

Center:

Foci:

Vertices:

e �45

�1, 10�, �1, 0��1, 9�, �1, 1�

�1, 5�

c2 � 16b2 � 9,a2 � 25,

12

8448

4

x

(1, 5)

y�x � 1�2

9�

�y � 5�2

25� 1

22.

x2 � 2x � 8y � 15 � 0

�x � 1�2 � 4��2��y � 2�

24. Vertex:

y2 � 8x � 4y � 4 � 0

�y � 2�2 � 4�2��x � 0�

�0, 2�

26.

x2 � 4x � y � 0

y � 4 � �x � 2�2 � 4x � x2

28. From Example 2: or

Vertex:

x2 � 8x � 8y � 16 � 0

�x � 4�2 � 8�y � 0�

�4, 0�

p � 24p � 8

30.

,

Center:

Foci:

Vertices:

e �2

�14�

�147

�±�14, 0��±2, 0�

�0, 0�

c2 � 4b2 � 10a2 � 14,

x2

14�

y2

10� 1

x2

2

4

4

6

6

−6 −2

−6

−4

y5x2 � 7y2 � 70

32.

Center:

Foci:

Vertices:

e ��32

��1, �4�, ��3, �4�

��2 ±�32

, �4���2, �4�

c2 �34

b2 �14

,a2 � 1,

x−1−2−3−4−5

−5

−4

−3

−2

−1

( 2, 4)− −

y�x � 2�2

1�

�y � 4�2

1�4� 1

Section 10.1 Conics and Calculus 363

33.

Center:

Foci:

Vertices:

e ��53

��2, 6�, ��2, 0���2, 3 ± �5 � 6

2

2246x

(−2, 3)

y

��2, 3�

c2 � 5b2 � 4,a2 � 9,

�x � 2�2

4�

�y � 3�2

9� 1

� 36

9�x2 � 4x � 4� � 4�y2 � 6y � 9� � �36 � 36 � 36

9x2 � 4y2 � 36x � 24y � 36 � 0

34.

Center:

Foci:

Vertices:

(2, −3)

1

1

2 3 4−1

−1

−2

−3

−4

y

x

e �ca

�35

�2 ±�10

4, �3�

�2 ±3�10

20, �3�

�2, �3�

a2, 58

, b2 �25

, c2 � a2 � b2 �9

40

�x � 2�2

�5�8� ��y � 3�2

�2�5� � 1

� 10

16�x2 � 4x � 4� � 25�y2 � 6y � 0� � �279 � 64 � 225

16x2 � 25y2 � 64x � 150y � 279 � 0 35.

Center:

Foci:

Vertices:

Solve for y:

(Graph each of these separately.)

−3 3

−3

1

y � �1 ± �57 � 12x � 12x2

20

�y � 1�2 �57 � 12x � 12x2

20

20�y2 � 2y � 1� � �12x2 � 12x � 37 � 20

�12

± �5, �1��1

2± �2, �1�

�12

, �1�c2 � 2b2 � 3,a2 � 5,

�x � �1�2��2

5�

�y � 1�2

3� 1

� 60

12�x2 � x �14� � 20�y2 � 2y � 1� � 37 � 3 � 20

12x2 � 20y2 � 12x � 40y � 37 � 0

364 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

36.

Center:

Foci:

Vertices:

Solve for y:

(Graph each of these separately.)

−1

2−4

3

y � 2 ±13���36x2 � 48x � 7�

�y � 2�2 ���36x2 � 48x � 7�

9

9�y2 � 4y � 4� � �36x2 � 48x � 43 � 36

��23

, 3�, ��23

, 1���

23

, 2 ±�32 �

��23

, 2�

c2 �34

b2 �14

,a2 � 1,

�x � �2�3��2

1�4�

�y � 2�2

1� 1

� 9

36�x2 �43

x �49� � 9�y2 � 4y � 4� � �43 � 16 � 36

36x2 � 9y2 � 48x � 36y � 43 � 0 37.

Center:

Foci:

Vertices:

Solve for y:

(Graph each of these separately.)

−2

−3

4

1

y � �1 ± �7 � 12x � 4x2

8

�y � 1�2 �12�

74

� 3x � x2�

2�y2 � 2y � 1� � �x2 � 3x �14

� 2

��12

, �1�, �72

, �1��3

2± �2, �1�

�32

, �1�c2 � 2b2 � 2,a2 � 4,

�x � �3�2��2

4�

�y � 1�2

2� 1

�x2 � 3x �94� � 2�y2 � 2y � 1� � �

14

�94

� 2 � 4

x2 � 2y2 � 3x � 4y � 0.25 � 0

38.

Center:

Foci:

Vertices:

Solve for y:

(Graph each of these separately.) y � 3.2 ± �7.12 � 4x � 2x2

�y � 3.2�2 � 7.12 � 4x � 2x2

�y2 � 6.4y � 10.24� � �2x2 � 4.8x � 3.12 � 10.24

��65

, 165

± �10�−1

5−7

7��65

, 165

± �5���

65

, 165 �

a2 � 10, b2 � 5, c2 � 5

�x � �6�5��2

5�

�y � �16�5��2

10� 1

50�x2 �125

x �3625� � 25�y2 �

325

y �25625 � � �78 � 72 � 256 � 250

50x2 � 25y2 � 120x � 160y � 78 � 0

2x2 � y2 � 4.8x � 6.4y � 3.12 � 0

Section 10.1 Conics and Calculus 365

39. Center:

Focus:

Vertex:

Horizontal major axis

x2

9�

y2

5� 1

c � 2 ⇒ b � �5a � 3,

�3, 0��2, 0��0, 0�

41. Vertices:

Minor axis length: 6

Vertical major axis

Center:

�x � 3�2

9�

�y � 5�2

16� 1

a � 4, b � 3

�3, 5�

�3, 1�, �3, 9�

43. Center:

Horizontal major axis

Points on ellipse:

Since the major axis is horizontal,

Substituting the values of the coordinates of the givenpoints into this equation, we have

and

The solution to this system is

Therefore,

x2

16�

y2

16�7� 1,

x2

16�

7y2

16� 1.

b2 � 16�7.a2 � 16,

16a2 � 1.� 9

a2� � � 1b2� � 1,

�x2

a2� � �y2

b2� � 1.

�3, 1�, �4, 0�

�0, 0�

40. Vertices:

Eccentricity:

Horizontal major axis

Center:

�x � 2�2

4�

�y � 2�2

3� 1

c � 1 ⇒ b � �3a � 2,

�2, 2�

12

�0, 2�, �4, 2�

42 Foci:

Major axis length: 14

Vertical major axis

Center:

x2

24�

y2

49� 1

a � 7 ⇒ b � �24c � 5,

�0, 0�

�0, ±5�

44. Center:

Vertical major axis

Points on ellipse:

From the sketch, we can see that

�x � 1�2

4�

�y � 2�2

16� 1.

b � 2a � 4,k � 2,h � 1,

x−2−4

−2

6

4

(1, 6)

(3, 2)(1, 2)

y

�1, 6�, �3, 2�

�1, 2�

46.

Center:

Vertices:

Foci:

Asymptotes: y � ±35

x

�±�34, 0��±5, 0�

−8 −4

−4−2

2468

10

−6−8

−10

4 8 10

y

x

�0, 0�

a � 5, b � 3, c � �a2 � b2 � �34

x2

25�

y2

9� 145.

Center:

Vertices:

Foci:

Asymptotes: y � ±12

x

�0, ±�5 ��0, ±1�

4

4

2

224

2

4

x

y�0, 0�

c � �5b � 2,a � 1,

y2

1�

x2

4� 1

366 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

47.

Center:

Vertices:

Foci:

Asymptotes:

32

1

11

2

4

5

x

y

y � �2 ±12

�x � 1�

�1 ± �5, �2���1, �2�, �3, �2�

�1, �2�

c � �5b � 1,a � 2,

�x � 1�2

4�

�y � 2�2

1� 1

49.

Center:

Vertices:

Foci:

Asymptotes:

6422

2

4

6

x

y

y � �3 ± 3�x � 2��2 ± �10, �3�

�1, �3�, �3, �3��2, �3�

c � �10b � 3,a � 1,

�x � 2�2

1�

�y � 3�2

9� 1

9�x2 � 4x � 4� � �y2 � 6y � 9� � �18 � 36 � 9

9x2 � y2 � 36x � 6y � 18 � 0

51.

Degenerate hyperbola is two lines intersecting at

2

4

2

2

6

4x

y

��1, �3�.

y � 3 � ±13

�x � 1�

�x � 1�2 � 9�y � 3�2 � 0

�x2 � 2x � 1� � 9�y2 � 6y � 9� � 80 � 1 � 81 � 0

x2 � 9y2 � 2x � 54y � 80 � 0

48.

Center:

Vertices:

Foci:

Asymptotes:

1

5

−52

20

−20

6 7 8x

y

y � �1 ±125

�x � 4�

�4, �14�, �4, 12��4, 11�, �4, �13�

�4, �1�

c � �a2 � b2 � 13b � 5,a � 12,

�y � 1�2

122 ��x � 4�2

52 � 1

50.

Center:

Vertices:

Foci:

Asymptotes:

−5

−2 −1 5

5

4x

y

y � ±3�x � 2��2, 2�10�, �2, �2�10�

�2, 6�, �2, �6��2, 0�

c � �a2 � b2 � 2�10b � 2,a � 6,

y2

36�

�x � 2�2

4� 1

y2 � 9�x2 � 4x � 4� � 72 � 36 � 36

y2 � 9x2 � 36x � 72 � 0

52.

Center:

Vertices:

Foci:

Asymptotes: y � 1 ±32

�x � 3�

��3, 1 ±16�13�

��3, 12�, ��3,

32�

��3, 1�

x

1

2

3

−1

−1−3

yc ��13

6b �

13

,a �12

,

�y � 1�2

1�4�

�x � 3�2

1�9� 1

9�x � 3�2 � 4�y � 1�2 � �1

9�x2 � 6x � 9� � 4�y2 � 2y � 1� � �78 � 81 � 4 � �1

Section 10.1 Conics and Calculus 367

54.

Center:

Vertices:

Foci:

Solve for y:

(Graph each curve separately.)

y � 5 ± �9x2 � 54x � 80

�y � 5�2 � 9x2 � 54x � 80

y2 � 10y � 25 � 9x2 � 54x � 55 � 25

��3 ±�10

3, 5�

��3 ±13

, 5���3, 5�

−80

2

10c ��10

3b � 1,a �

13

,

�x � 3�2

1�9�

�y � 5�2

1� 1

� 1

9�x2 � 6x � 9� � �y2 � 10y � 25� � �55 � 81 � 25

9x2 � y2 � 54x � 10y � 55 � 053.

Center:

Vertices:

Foci:

Solve for y:

(Graph each curve separately.)

y � �3 ±13�x2 � 2x � 19

�y � 3�2 �x2 � 2x � 19

9

9�y2 � 6y � 9� � x2 � 2x � 62 � 81

�1, �3 ± 2�5 ��1, �3 ± �2 �

�1, �3�−5

−7

7

1c � 2�5b � 3�2,a � �2,

�y � 3�2

2�

�x � 1�2

18� 1

9�y2 � 6y � 9� � �x2 � 2x � 1� � �62 � 1 � 81 � 18

9y2 � x2 � 2x � 54y � 62 � 0

55.

Center:

Vertices:

Foci:

Solve for y:

(Graph each curve separately.)

y � �3 ± �3�x2 � 2x � 3�2

�y � 3�2 �3x2 � 6x � 9

2

2�y2 � 6y � 9� � 3x2 � 6x � 27 � 18

�1 ± �10, �3���1, �3�, �3, �3�

�1, �3�−5

−7

7

1c � �10b � �6,a � 2,

�x � 1�2

4�

�y � 3�2

6� 1

3�x2 � 2x � 1� � 2�y2 � 6y � 9� � 27 � 3 � 18 � 12

3x2 � 2y2 � 6x � 12y � 27 � 0 56.

Center:

Vertices:

Foci:

Solve for y:

(Graph each curve separately.)

y � 2 ± �x2 � 6x � 123

�y � 2�2 �x2 � 6x � 12

3

3�y2 � 4y � 4� � x2 � 6x � 12

�3, 0�, �3, 4��3, 1�, �3, 3�

�3, 2�

−4

−4

10

6c � 2b � �3,a � 1,

�y � 2�2

1�

�x � 3�2

3� 1

3�y2 � 4y � 4� � �x2 � 6x � 9� � 0 � 12 � 9 � 3

3y2 � x2 � 6x � 12y � 0

58. Vertices:

Asymptotes:

Vertical transverse axis

Slopes of asymptotes:

Thus, . Therefore,

y2

9�

x2

1� 1.

b � 1

±ab

� ±3

a � 3

y � ±3x

�0, ±3�57. Vertices:

Asymptotes:

Horizontal transverse axis

Center:

,

Therefore,x2

1�

y2

9� 1.

±ba

� ±b1

� ±3 ⇒ b � 3a � 1

�0, 0�

y � ±3x

�±1, 0�

368 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

59. Vertices:

Point on graph:

Vertical transverse axis

Center:

Therefore, the equation is of the form

Substituting the coordinates of the point , we have

or

Therefore, the equation is y2

9�

�x � 2�2

9�4� 1.

b2 �94

.259

�4b2 � 1

�0, 5�

y2

9�

�x � 2�2

b2 � 1.

a � 3

�2, 0�

�0, 5��2, ±3�

61. Center:

Vertex:

Focus:

Vertical transverse axis

Therefore,y2

4�

x2

12� 1.

b2 � c2 � a2 � 12c � 4,a � 2,

�0, 4��0, 2��0, 0�

63. Vertices:

Asymptotes:

Horizontal transverse axis

Center:

Slopes of asymptotes:

Thus, Therefore,

�x � 3�2

9�

�y � 2�2

4� 1.

b � 2.

±ba

� ±23

a � 3

�3, 2�

y � 4 �23

xy �23

x,

�0, 2�, �6, 2�

65. (a)

At

At

or

At

or 2x � 3�3y � 3 � 0

y � �3 ��2�3

9�x � 6��6, ��3 �:

2x � 3�3y � 3 � 0

y � �3 �2�3

9�x � 6��6, �3 �:

y� �±6

9�3�

±2�39

y � ±�3,x � 6:

x9y

� y�2x9

� 2yy� � 0,x2

9� y2 � 1, (b) From part (a) we know that the slopes of the normal lines

must be

At

or

At

or 9x � 2�3y � 60 � 0

y � �3 �9

2�3�x � 6��6, ��3 �:

9x � 2�3y � 60 � 0

y � �3 � �9

2�3�x � 6��6, �3 �:

�9��2�3 �.

60. Vertices:

Foci:

Vertical transverse axis

Center:

Therefore,y2

9�

�x � 2�2

16� 1.

b2 � c2 � a2 � 16c � 5,a � 3,

�2, 0�

�2, ±5��2, ±3�

62. Center:

Vertex:

Focus:

Horizontal transverse axis

Therefore,x2

9�

y2

16� 1.

b2 � c2 � a2 � 16c � 5,a � 3,

�5, 0��3, 0��0, 0�

64. Focus:

Asymptotes:

Horizontal transverse axis

Center: since asymptotes intersect at the origin.

Slopes of asymptotes: and

Solving these equations, we have and Therefore, the equation is

x2

64�

y2

36� 1.

b2 � 36.a2 � 64

c2 � a2 � b2 � 100

b �34

a±ba

� ±34

c � 10

�0, 0�

y � ±34

x

�10, 0�

Section 10.1 Conics and Calculus 369

66. (a)

At

At or

At or

(b) From part (a) we know that the slopes of the normallines must be

At : or

At or 3x � 4y � 36 � 0y � 6 �34

�x � 4��4, �6�:

3x � 4y � 36 � 0y � 6 � �34

�x � 4��4, 6�

�3�4.

4x � 3y � 2 � 0y � 6 � �43

�x � 4��4, �6�:

4x � 3y � 34 � 0y � 6 � �43

�x � 4��4, 6�:

y� �±2�4�

6� ±

43

y � ±6,x � 4:

y� �4x2y

�2xy

2yy� � 4x � 0,y2 � 2x2 � 4,y2

4�

x2

2� 1,

68.

Hyperbola

4�x �12�2

� y2 � 4

4�x2 � x �14� � y2 � 3 � 1

4x2 � y2 � 4x � 3 � 0

70.

Parabola

25�x �15�2

� 200� y � 1�

25�x2 �25x �

125� � 200y � 119 � 1

25x2 � 10x � 200y � 119 � 0

72.

Parabola

� y � 2�2 � x � 9

y2 � 4y � 4 � x � 5 � 4

y2 � 4y � x � 5

74.

Ellipse

2x2 � � y �32�

2

�94

2x2 � y2 � 3y � 0

2x2 � 2xy � 3y � y2 � 2xy

2x�x � y� � y�3 � y � 2x�

76.

Ellipse

�x � 3�2

4�

�y � 2�2

9� 1

9�x � 3�2 � 4�y � 2�2 � 36

9�x � 3�2 � 36 � 4�y � 2�2

67.

Ellipse

�x � 3�2 � 4�y � 2�2 � 4

�x2 � 6x � 9� � 4� y2 � 4y � 4� � �21 � 9 � 16

x2 � 4y2 � 6x � 16y � 21 � 0

69.

Parabola

� y � 2�2 � 4�x � 1�

y2 � 4y � 4 � 4x � 4

y2 � 4y � 4x � 0

71.

Circle (Ellipse)

4x2 � 4� y � 2�2 � 1

4x2 � 4� y2 � 4y � 4� � �15 � 16

4x2 � 4y2 � 16y � 15 � 0

73.

Circle (Ellipse)

9�x � 2�2 � 9� y �13�2

� 3

9�x2 � 4x � 4� � 9� y2 �23y �

19� � �34 � 36 � 1

9x2 � 9y2 � 36x � 6y � 34 � 0

75.

Hyperbola

�x � 1�2

2�

�y � 1�2

3� 1

3�x � 1�2 � 2�y � 1)2 � 6

3�x � 1�2 � 6 � 2� y � 1�2

77. (a) A parabola is the set of all points that areequidistant from a fixed line (directrix) and a fixedpoint (focus) not on the line.

(b) or

(c) See Theorem 10.2.

�y � k�2 � 4p�x � h��x � h�2 � 4p�y � k�

�x, y�

370 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

78. (a) An ellipse is the set of all points , the sum ofwhose distance from two distinct fixed points (foci)is constant.

(b) or �x � h�2

b2 ��y � k�2

a2 � 1�x � h�2

a2 ��y � k�2

b2 � 1

�x, y�

80.

For the ellipse is nearly circular.

For the ellipse is elongated.e 1,

e 0,

0 < e < 1e �ca

, c � �a2 � b2,

82. Assume that the vertex is at the origin.

(a)

(b) The deflection is 1 cm when

−8 −4 4 8x

1001

1005

1004

1002

1003( )−8, 100

3( )0, 1003( )8,

y

y �2

100 ⇒ x � ±�128

3 ±6.53 meters.

x2 � 4�16003 �y �

64003

y

16003

� p

82 � 4p� 3100�

x2 � 4py

79. (a) A hyperbola is the set of all points for which theabsolute value of the difference between the distancesfrom two distance fixed points (foci) is constant.

(b) or

(c) or y � k ±ab

�x � h�y � k ±ba

�x � h�

�y � k�2

a2 ��x � h�2

b2 � 1�x � h�2

a2 ��y � k�2

b2 � 1

�x, y�

81. Assume that the vertex is at the origin.

The pipe is located

meters from the vertex.94

94

� p

�3�2 � 4p�1�

x

( 3, 1)− (3, 1)1

1

2

2

3

3−1−2−3

Focus

y x2 � 4py

83.

The equation of the tangent line is

or

Let Then:

Therefore, is the x-intercept.

x

( , )x ax0 02

x0

2, 0( )

y ax= 2

y

�x0

2, 0�

x �x0

2

ax02 � 2ax0x

�ax02 � 2ax0x � 2ax0

2

y � 0.

y � 2ax0x � ax02.

y � ax02 � 2ax0�x � x0�

y� � 2ax

y � ax2

Section 10.1 Conics and Calculus 371

(b)

At the slope is At the slope is2: Solving for x,

Point of intersection: �3, �3�

y � �3.

x � 3

�3x � �9

�x � 2x � 9

y � 2x � 9.�6, 3�,y � �x.�1:�0, 0�,

dydx

�12

x � 1

2x � 4 � 4 dydx

� 0

x2 � 4x � 4y � 084. (a) Without loss of generality, place the coordinate systemso that the equation of the parabola is and,hence,

Therefore, for distinct tangent lines, the slopes areunequal and the lines intersect.

y� � � 12p�x.

x2 � 4py

85. (a) Consider the parabola Let be the slope ofthe one tangent line at and therefore, isthe slope of the second at Differentiating,

and we have:

Substituting these values of x into the equation we have the coordinates of the points of tangency

and and the equationsof the tangent lines are

and

The point of intersection of these lines is

and is on the directrix, .

x

x py2 = 4

(2 , )pm pm0 02

y p= −

p m2( 1)0 −m0

, −p( )

m0 m0( )2p p− , 2

y

y � �p�p�m02 � 1�m0

, �p�

�y �p

m02� �

�1m0

�x �2pm0

�.

�y � pm02� � m0�x � 2pm0�

��2p�m0, p�m02��2pm0, pm0

2�4py,

x2 �

�1m0

�1

2px2 or x2 �

�2pm0

.

m0 �1

2px1 or x1 � 2pm0

2x � 4py� or y� �x

2p,

�x2, y2�.�1�m0�x1, y1�

m0x2 � 4py. (b)

Vertex:

At At

Tangent line at

Tangent line at

Since the lines are perpendicular.

Point of intersection:

Directrix: and the point of intersection lies

on this line.

�12

, 0�y � 0

y � 0

x �12

�52

x � �54

�2x � 1 �12

x �14

m1m2 � ��2��12� � �1,

y �54

�12

�x � 3� ⇒ 2x � 4y � 1 � 0.

�3, 54�:

y � 5 � �2�x � 2� ⇒ 2x � y � 1 � 0.

��2, 5�:

dydx

�12

.�3, 54�,

dydx

� �2.��2, 5�,

dydx

�12

x � 1

2x � 4 � 4 dydx

� 0

�2, 1�

�x � 2�2 � 4�y � 1�

x2 � 4x � 4y � 8 � 0

372 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

86. The focus of is The distance from a point on the parabola, and the focus, , is

Since d is minimized when is minimized, it is sufficient to minimize the function

implies that

This is a minimum by the First Derivative Test. Hence, the closest point to the focus is the vertex, �0, 0�.

x3

16� x � x� x2

16� 1� � 0 ⇒ x � 0.f��x� � 0

f��x� � 2x � 2�x2

8� 2��x

4� �x3

16� x.

f �x� � x2 � �x2

8� 2�

2

.

x

(0, 2) ( )x, x2

8

1 2 3

1

3

4

−1

−2

−2−3

x y2 = 8

yd 2

d ���x � 0�2 � �x2

8� 2�

2

.

�0, 2��x, x2�8�,�0, 2�.x2 � 8y � 4�2�y

87.

At the point of tangency on the mountain, Also,

Choosing the positive value for we have

Thus,

The closest the receiver can be to the hill is �2�3�3� � 1 0.155.

2�3

3� x0.

3 � 2�33

� 1 � x0

�1

3 � 2�3� x0 � 1

�1

x0 � 1� 3 � 2�3

m �0 � 1x0 � 1

� �1

x0 � 1

x

( 1, 1)− ( , )x y1 1( , 0)x0

1

2

1

−1

−1

−2

−2

ym � 1 � 2��1 � �3� � 3 � 2�3

x1 � �1 � �3.x1,

x1 ��2 ± �22 � 4�1���2�

2�1� ��2 ± 2�3

2� �1 ± �3

x12 � 2x1 � 2 � 0

�x12 � x1 � 1 � �2x1

2 � x1 � 1

�x1 � x12� � 1 � �1 � 2x1��x1 � 1�

y1 � 1x1 � 1

� 1 � 2x1

m �y1 � 1x1 � 1

.m � 1 � 2x1.�x1, y1�

dydx

� 1 � 2x

y � x � x2

Section 10.1 Conics and Calculus 373

88. (a)

(b) 350

2005 13

A � 0.75t2 � 9.2t � 301 (c)

Women are spending more time watching TV each year.

y

x2 4 6 8 10 12 14−2

2

4

6

8

10

12

14

6 ≤ t ≤ 12dAdt

� 1.5t � 9.2,

90.

� 2�5 � ln�2 � �5 � 5.916

�14

�4�20 � 4 ln4 � �20 � 4 ln 2�

�14� y�4 � y2 � 4 lny � �4 � y2�4

0

s � 4

0�1 � �y2

4 � dy �12 4

0

�4 � y2 dy

1 � �x��2 � 1 �y2

4

x� �12

y

x �14

y2

89. Parabola

Vertex:

Since the value is positive when we have

�16�4 � 3�3 � 2��

3 15.536 square feet

� 2�16 �6412

� 16�3 � 2�48 � 32 arcsin 12�

� 2�4x �x3

12� 4�3x �

12�x�64 � x2 � 64 arcsin

x8��

4

0

A � 2 4

0��4 �

x2

4 � � ��4�3 � �64 � x2 �� dx

y � �4�3 � �64 � x2.x � 0,y-

y � 4 �x2

4

x2 � �4�y � 4�

p � �1

42 � 4p�0 � 4�

x−2−4−6

−2

−4

−6

−8

2 4 6

8

yx2 � 4p�y � 4�

�0, 4�Circle

Center:

Radius: 8

(Center is on the negative y-axis.)

y � �4�3 ± �64 � x2

x2 � �y � 4�3 �2� 64

k � �4�3

k2 � 48

42 � �0 � k�2 � 64

x2 � �y � k�2 � 64

�0, k�

374 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

91. (a) Assume that

(b)

(Formula 26)

� 10�2�13 � 9 ln�2 � �133 �� 128.4 m

� 20�13 � 90 ln�60 � 30�1390 �

�1

90�1800�13 � 902 ln�60 � 30�13 � � 902 ln 90�

�1

90�60�11,700 � 902 ln�60 � �11,700 � � 902 ln 90�

�2

90 12�x�902 � x2 � 902 lnx � �902 � x2�60

0

S � 2 60

0�1 � � 1

90x�

2

dx �2

90 60

0

�902 � x2 dx

f��x� �1

90xf �x� �

1180

x2,

20 � a�60�2 ⇒ a �2

360�

1180

⇒ y �1

180x2

y � ax2.

15−15−30−45−60

15

5

45 6030

20

10

x

( 60, 20)− (60, 20)

y

92.

��

15��100 � r2�3�2 � 1000� � � �

10�

23

�100 � x2�3�2�r

0

� 2� r

0 x�100 � x2

10 dxS � 2� r

0x�1 � � x

10�2

dx

y� �x

10

y �x2

20

x2 � 20y

94.

�83�ph3�2

� �4�p�23�y3�2�

h

0

� 4�p h

0y1�2 dy

A � 2 h

0

�4py dy 95. (a) At the vertices we notice that the string is horizontal and has a length of 2a.

(b) The thumbtacks are located at the foci and the length ofstring is the constant sum ofthe distances from the foci.

FocusFocus

VertexVertex

93.

As p increases, the graph becomes wider.

x168−8−16

24

y p = 1

p = 2

p =12

p =32

p =14

x2 � 4py, p �14

, 12

, 1, 32

, 2

96.

The tacks should be placed 1.5 feet from the center. Thestring should be feet long.2a � 5

c ���52�

2

� �2�2 �32

b � 2,a �52

, 97.

x

16 17

141312

1

4 3 2

8 9

5 6 7

6 5

10 11

4

12 3

15

9 8

11 10

131412

1517 16

7

y

Section 10.1 Conics and Calculus 375

98.

Least distance:

Greatest distance: a � c � 152,096,286.6 km

a � c � 147,099,713.4 km

c 2,498,286.6

0.0167 �c

149,598,000

e �ca

99.

x( , 0)a

A P

c

y

e �ca

��A � P��2�A � P��2

�A � PA � P

c � a � P �A � P

2� P �

A � P2

a �A � P

2

A � P � 2a

e �ca

100.

�122,881131,119

0.9372

��123,000 � 4000� � �119 � 4000��123,000 � 4000� � �119 � 4000�

e �A � PA � P 101. e �

A � PA � P

�35.29 � 0.5935.29 � 0.59

� 0.9671

102.

As and we have

or the circle x2 � y2 � a2.x2

a2 �y2

a2 � 1

1 � e2 → 1e → 0,

x2

a2 �y2

a2�1 � e2� � 1

x2

a2 �y2

a2�a2 � c2��a2 � 1

x2

a2 �y2

a2�b2�a2� � 1

x2

a2 �y2

b2 � 1 103.

At

The equation of the tangent line is It

will cross the -axis when and y �23�8� � 3 �

253 .x � 0y

y � 3 �23�x � 8�.

y� �8

12�

23

��8, 3�:

y� ��52x102y

��x4y

2x102 �

2yy�

52 � 0

x2

102 �y2

52 � 1

104.

(Recall: Area of ellipse is .)

� 90� �1445 �0.5�6 � �2.5�2 arcsin

15� 354.3 ft3 � 90� �

1445

� �y��2.5�2 � y2 � �2.5�2 arcsin y

2.5�0.5

0

�abV � ���4.5��2.5�2 ��16� � 16 0.5

0 2

95��2.5�2 � y2 dy

V � �Area of bottom��Length� � �Area of top��Length�

x � ±95��2.5�2 � y2

x2 � �4.5�2�1 �y2

�2.5�2�x

3 ft

5 ft

9 ft

yx2

�4.5�2 �y2

�2.5�2 � 1

376 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

105.

when is undefined when

At or

Endpoints of major axis:

At or

Endpoints of minor axis:

Note: Equation of ellipse is �x � 3�2

9�

�y � 2�2

16� 1

�0, �2�, ��6, �2�

�6.y � �2, x � 0

��3, 2�, ��3, �6�

�6.x � �3, y � 2

y � �2.y�x � �3.y� � 0

y� ���32x � 96�

18y � 36

y��18y � 36� � ��32x � 96�

32x � 18yy� � 96 � 36y� � 0

16x2 � 9y2 � 96x � 36y � 36 � 0 106.

when undefined when At or 6.

Endpoints of major axis:

At or

Endpoints of minor axis:

Note: Equation of ellipse is �x � 2�2

4�

�y � 3�2

9� 1

�0, 3�, ��4, 3�

�4.y � 3, x � 0

��2, 0�, ��2, 6�

x � �2, y � 0y � 3.y�x � �2.y� � 0

y� ���18x � 36�

8y � 24

�8y � 24�y� � ��18x � 36�

18x � 8yy� � 36 � 24y� � 0

9x2 � 4y2 � 36x � 24y � 36 � 0

107. (a)

(b) Disk:

(c) Shell:

�4�

3 6 � �3 ln�2 � �3 � 34.69

�8�

2�3��3y�1 � 3y2 � ln�3y � �1 � 3y2�1

0

� 8� 1

0

�1 � 3y2 dy S � 2�2�� 1

02�1 � y2

�1 � 3y2

�1 � y2 dy

�1 � �x��2 ��1 �4y2

1 � y2 ��1 � 3y2

�1 � y2

x� ��2y

�1 � y2

x � 2�1 � y2

� �2�

3 ��4 � x2�3�2�2

0�

16�

3� �� 2

0�2x�4 � x2�1�2 dx V � 2� 2

0 x�4 � x2 dx

�2�

9�9 � 4�3�� 21.48

��

2�3��3x�16 � 3x2 � 16 arcsin��3x4 ��2

0

� � 2

0

�16 � 3x2 dx S � 2�2�� 2

0y��16 � 3x2

4y � dx

��16 � 3x2

4y�1 � �y��2 ��1 �

x2

16 � 4x2

y� ��x

2�4 � x2

y �12�4 � x2

�12

��4x �13

x3�2

0�

8�

3 V � 2� 2

0 14

�4 � x2� dx

�or, A � �ab � ��2��1� � 2��� �x�4 � x2 � 4 arcsin�x2��

2

0� 2�A � 4 2

0 12�4 � x2 dx

Section 10.1 Conics and Calculus 377

108. (a)

(b) Disk:

(c) Shell:

�8�

9�7 3�7�12� � 81 ln�3�7 � 12 � � 81 ln 9 168.53

�169 � �

2�7���7y�81 � 7y2 � 81 ln�7y � �81 � 7y2�3

0

� 4� 3

0 49�81 � 7y2 dy

S � 2�2�� 3

0 43�9 � y2 �9�9 � y2� � 16y2

9�9 � y2� dy

�1 � �x��2 ��1 �16y2

9�9 � y2�

x� ��4y

3�9 � y2

x �43�9 � y2

� 3����12��

23��16 � x2�3�2�

4

0� 64�V � 4� 4

0 x�3

4�16 � x2� dx

�3�

8�7�48�7 � 256 arcsin �74 � 138.93 �

3�

8�7��7x�256 � 7x2 � 256 arcsin �7x16 �

4

0

�3�

4 4

0

�256 � 7x2 dx� 4� 4

0 34�16 � x2

�256 � 7x2

4�16 � x2 dxS � 2�2�� 4

0 34�16 � x2�16�16 � x2� � 9x2

16�16 � x2� dx

�1 � �y��2 ��1 �9x2

16�16 � x2�

y� ��3x

4�16 � x2

y �34�16 � x2

�9�

8 ��16x �13

x3�4

0� 48� V � 2� 4

0

916

�16 � x2� dx

�32�x�16 � x2 � 16 arcsin

x4�

4

0� 12�A � 4 4

0 34�16 � x2 dx

109. From Example 5,

For we have

28�1.3558� 37.9614

C � 4�7� ��2

0�1 �

2449

sin2 � d�

a � 7, b � 5, c � �49 � 25 � 2�6, e �ca

�2�6

7.

x2

25�

y2

49� 1,

C � 4a ��2

0

�1 � e2 sin2 � d�

378 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

110. (a) (b) Slope of line through and

Slope of line through and

At

(c)

Since the tangent line to an ellipse at a point makes equal angles with the lines through and the foci.PP � ,

� arctan� b2

y0c�

�a2y0

2 � b2x02 � b2x0c

a2x0 y0 � a2cy0 � b2x0y0�

a2b2 � b2x0cx0 y0�a2 � b2� � a2cy0

�b2�a2 � x0c�y0c�x0c � a2� �

b2

y0c

tan �m1 � m

1 � m1m�

y0

x0 � c� ��

b2x0

a2y0�

1 � � y0

x0 � c���b2x0

a2y0�

�a2y0

2 � b2x0�x0 � c�a2y0�x0 � c� � b2x0 y0

� arctan��b2

y0c� � �arctan� b2

y0c�

�a2y0

2 � b2x02 � b2x0c

x0y0�a2 � b2� � a2y0c�

a2b2 � b2x0cx0y0c

2 � a2y0c�

b2�a2 � x0c�y0c�x0c � a2� � �

b2

y0c

tan �m2 � m

1 � m2m�

y0

x0 � c� ��

b2x0

a2y0�

1 � � y0

x0 � c���b2x0

a2y0�

�a2y0

2 � b2x0�x0 � c�a2y0�x0 � c� � b2x0y0

y� � �b2

a2 �x0

y0� m.P,

y� � �xb2

ya2

m2 �y0

x0 � c�x0, y0�:�c, 0�2x

a2 �2yy�

b2 � 0

m1 �y0

x0 � c�x0, y0�:��c, 0�

x2

a2 �y2

b2 � 1

111. Area circle

Area ellipse

Hence, the length of the major axis is 2a � 40.

2�100�� � 10�a ⇒ a � 20

� �ab � �a�10�

� �r2 � 100�

113. The transverse axis is horizontal since and are the foci(see definition of hyperbola).

Center:

Therefore, the equation is

�x � 6�2

9�

�y � 2�2

7� 1.

b2 � c2 � a2 � 72a � 6,c � 4,

�6, 2�

�10, 2��2, 2�

112. (a) Hence,

(b)

(c) As approaches 0, the ellipse approaches a circle.e

−3 9

−1

7

�x � 2�2

4�

�y � 3�2

4�1 � e2� � 1

�x � h�2

a2 ��y � k�2

a2�1 � e2� � 1.

�x � h�2

a2 ��y � k�2

b2 � 1

e �ca

��a2 � b2

a ⇒ �ea�2 � a2 � b2.

Section 10.1 Conics and Calculus 379

115.

c � 6 ⇒ b � �11 16 17141312

8 9

56 7

1011

41 2 3

15

14 3 2

6 5

9 8

11 10

131412

1517 16

7

2a � 10 ⇒ a � 5114. The transverse axis is vertical since and are the foci.

Center:

Therefore, the equation is

�y � �3�2��2

1�

�x � 3�2

5�4� 1.

b2 � c2 � a2 �54

2a � 2,c �32

,

��3, 32�

��3, 3���3, 0�

116. Center:

Horizontal transverse axis

Foci:

Vertices:

The difference of the distances from any point on the hyperbola is constant. At a vertex, this constant difference is

Now, for any point on the hyperbola, the difference of the distances between and the two foci must also be

Since we have �x2�a2� � �y2�b2� � 1.a2 � b2 � c2,

x2

a2 �y2

c2 � a2 � 1

x2�c2 � a2� � a2y2 � a2�c2 � a2�

x2c2 � 2a2cx � a4 � a2�x2 � 2cx � c2 � y2�

x( , 0)−c ( , 0)−a

( , )x y

( , 0)c

( , 0)a

y ��xc � a2� � a��x � c�2 � y2

�4xc � 4a2 � 4a��x � c�2 � y2

�x � c�2 � y2 � 4a2 � 4a��x � c�2 � y2 � �x � c�2 � y2

��x � c�2 � y2 � 2a � ��x � c�2 � y2

��x � c�2 � �y � 0�2 � ��x � c�2 � �y � 0�2 � 2a

2a.�x, y��x, y�

�a � c� � �c � a� � 2a.

�±a, 0�

�±c, 0�

�0, 0�

117. Time for sound of bullet hitting target to reach

Time for sound of rifle to reach

Since the times are the same, we have:

x2

c2vs2�vm

2 �y2

c2�vm2 � vs

2��vm2 � 1

�1 �vm

2

vs2�x2 � y2 � � vs

2

vm2 � 1�c2

��x � c�2 � y2 �vm

2x � vs2c

vsvm

4c2

vm2 �

4cvmvs

��x � c�2 � y2 ��x � c�2 � y2

vs2 �

�x � c�2 � y2

vs2

2cvm

���x � c�2 � y2

vs

���x � c�2 � y2

vs

��x � c�2 � y2

vs

�x, y�:x

( , 0)−c

( , )x y

( , 0)ctargetrifle

y�x, y�: 2cvm

���x � c�2 � y2

vs

380 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

118.

When we have

x 110.3 miles.

x2 � 932�1 �752

13,851�y � 75,

x2

932 �y2

13,851� 1

b � �1502 � 932 � �13,851

a � 93,2a � 0.001�186,000�,c � 150,

120.

x0xa2 �

y0yb2 � 1

a2b2 � b2x0x � a2y0y

b2x02 � a2y0

2 � b2x0x � a2y0y

a2y0y � a2y02 � b2x0x � b2x0

2

y � y0 �b2x0

a2y0�x � x0�

2xa2 �

2yy�

b2 � 0 or y� �b2xa2y

x2

a2 �y2

b2 � 1119. The point lies on the line between and Thus, The point also lies on the hyperbola

Using substitution, we have:

Choosing the positive value for we have:

and

y �160 � 96�2

7 3.462

x ��90 � 96�2

7 6.538

x

��180 ± 192�2

14�

�90 ± 96�27

x ��180 ± �1802 � 4�7���1476�

2�7�

7x2 � 180x � 1476 � 0

16x2 � 9�10 � x�2 � 576

x2

36�

�10 � x�2

64� 1

�x2�36� � �y2�64� � 1.y � 10 � x.

�10, 0�.�0, 10��x, y�

121. Let

There are four points of intersection:

—CONTINUED—

x2

a2 � b2 �2y2

b2 � 1 ⇒ 2xc2 �

4yy�

b2 � 0 ⇒ y�h �b2x2c2y

x2

a2 �2y2

b2 � 1 ⇒ 2xa2 �

4yy�

b2 � 0 ⇒ y�e � �b2x

2a2y

���2ac

�2a2 � b2, ±

b2

�2�2a2 � b2�� �2ac

�2a2 � b2, ±

b2

�2�2a2 � b2�,

y2 �b4

2�2a2 � b2� ⇒ y � ±b2

�2�2a2 � b2

2y2

b2 � 1 �1a2� 2a2c2

2a2 � b2� ⇒ 2y2

b2 �b2

2a2 � b2

x2 �2a2�a2 � b2�

2a2 � b2 ⇒ x � ±�2a�a2 � b2

�2a2 � b2� ±

�2ac

�2a2 � b2

1 �x2

a2 �x2

a2 � b2 � 1 ⇒ 2 � x2� 1a2 �

1a2 � b2�

x2

a2 � b2 �2y2

b2 � 1 ⇒ 2y2

b2 �x2

a2 � b2 � 1

c2 � a2 � b2. x2

a2 �2y2

b2 � 1 ⇒ 2y 2

b2 � 1 �x2

a2.

Section 10.1 Conics and Calculus 381

121. —CONTINUED—

At the slopes of the tangent lines are:

and

Since the slopes are negative reciprocals, the tangent lines are perpendicular. Similarly, the curves areperpendicular at the other three points of intersection.

y�h �

b2� �2ac

�2a2 � b2�2c2� b2

�2�2a2 � b2��

ac.y�e �

�b2� �2ac

�2a2 � b2�2a2� b2

�2�2a2 � b2�� �

ca

� �2ac

�2a2 � b2, b2

�2�2a2 � b2�,

122. (Assume and see (b) below)

(a) If we have

which is the standard equation of a circle.

�x �D2A�

2

� �y �E

2C�2

�RA

A � C,

�x � � D

2A��2

C�

�y � � E2C��

2

A�

RAC

A�x2 �DA

x �D2

4A2� � C�y2 �EC

y �E2

4C2� � �F �D2

4A�

E2

4C� R

A�x2 �DA

x� � C�y2 �EC

y� � �F

C � 0;A � 0 Ax2 � Cy2 � Dx � Ey � F � 0

(b) If we have

If we have

These are the equations of parabolas.

C�y �E

2C�2

� �F � Dx �E2

4C.

A � 0,

A�x �D2A�

2

� �F � Ey �D 2

4A.

C � 0,

(c) If we have

which is the equation of an ellipse.

�x � � D2A��

2

RA ��y � � E

2C��2

RC � 1

AC < 0,

(d) If we have

which is the equation of a hyperbola.

�x � � D2A��

2

RA ��y � � E

2C��2

RC � ±1

AC < 0,

123. False. See the definition of aparabola.

124. True 125. True

126. False. yields two intersecting lines:

y � 1 � ± �x � 1�

y2 � x2 � 2x � 2y � 0 127. True 128. True

129. Let be the equation of the ellipse with Let be the foci,

Let be a point on the tangent line at as indicated in the figure.

Slope at

—CONTINUED—

P�x, y� y� � �b2xa2y

2xb2 � 2yy�a2 � 0

y

x

x2

a2

y2

b2+ = 1

F1 F2c

b

−c

a

−a

−b

d

(u, v)

P(x, y)

x2b2 � y2a2 � a2b2

P�x, y�,�u, v�c2 � a2 � b2.

�±c, 0�a > b > 0.x2

a2 �y2

b2 � 1

382 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

129. —CONTINUED—

Now,

Since there is a right angle at

We have two equations:

Multiplying the first by and the second by and adding,

Similarly,

From the figure, and Thus, and

Let and

Finally,

� a2b2, a constant!

�a4b4

a2b2�a2 � b2 � x2 � y2� � �a2 � b2 � x2 � y2�

�a4b4

b2�a2b2 � a2y2� � a2�a2b2 � b2x2� � �a2 � b2 � x2 � y2�

�a4b4

b2�b2x2� � a2�a2y2� � �a2 � b2 � x2 � y2�

d2r1r2 �a4b4

x2b4 � y2a4 � �a2 � b2 � x2 � y2�

� a2 � b2 � x2 � y2

� 2a2 � x2 � y2 � c2

�12

�4a2 � �x � c�2 � y2 � �x � c�2 � y2�

r1r2 �12

��r1 � r2�2 � r12 � r2

2�

r1 � r2 � 2a.r2 � PF2,r1 � PF1

d2 �a4b4

x2b4 � y2a4

x2b4d2 � y2a4d2 � a4b4

cos2 � � sin2 � �x2d2

a4 �y2d2

b4 � 1

sin � �ydb2.cos � �

xda2v � d sin �.u � d cos �

u �xd2

a2 .

v �yd2

b2 .

yd2 � b2v

y�u2 � v2� � b2v

a2v2y � a2u2y � a2b2v

u,v

a2uy � b2vx � 0.

a2vy � b2ux � a2b2

vb2x � a2uy.

vu

�a2yb2x

�u, v�,

a2b2 � a2vy � b2ux

y2a2 � x2b2 � a2vy � b2ux

y2a2 � a2vy � �b2x2 � b2xu

y � vx � u

� �b2xa2y

Section 10.2 Plane Curves and Parametric Equations 383

Section 10.2 Plane Curves and Parametric Equations

130. Consider circle and hyperbola

Let and be points on the circle and

hyperbola, respectively. We need to minimize the distancebetween these 2 points:

The tangent lines at and are both perpendicular to and hence parallel.

The minimum value is �3 � 1�2 � �3 � 1�2 � 8.

y � x,�3, 3��1, 1�

�Distance�2 � f �u, v� � �u � v�2 � ��2 � u2 �9v�

2

.

�v, 9v��u, �2 � u2�

y

x−2 2 3 4 5

−2

2

3

4

5

x2 + y2 = 2

9v

9x

( )v,

u, 2 − u2( )

y = y �9x.x2 � y2 � 2

1.

(a)

(c)

−1 3

−4

2

y � 1 � tx � �t,

(b)

(d)

y � 1 � x2, x ≥ 0

x2 � t

321

1

−1

−2

−3

−1x

y

t 0 1 2 3 4

x 0 1 2

y 1 0 �3�2�1

�3�2

3.

2x � 3y � 5 � 0

y � 2�x � 13 � � 1

y � 2t � 1

2

4

2

4 2x

y x � 3t � 1

2.

(a)

(b)

x

1

2

3

−2

1 2 3 5

y

0 ≤ x ≤ 4 �2 ≤ y ≤ 2

x � 4 cos2 � y � 2 sin �

(c)

(d)

(e) The graph would be oriented in the opposite direction.

x � 4 � y2, �2 ≤ y ≤ 2

x4

�y2

4� 1

y2

4� sin2 �

x4

� cos2 �

−1 5

−3

3

0

x 0 2 4 2 0

y 0 2�2��2�2

2�

4�

4�

2�

4.

2y � 3x � 13 � 0

y � 2 � 3�3 � x2 �

y � 2 � 3t

x

2

4

6

2 4 6 8

y x � 3 � 2t

384 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

9.

3

2

4

−1

−2

4 5 632

1

x

y

y � x2 � 2, x ≥ 0

y � t � 2

x � �t, t ≥ 0 11.

4

2

22x

y

y �x � 1

x

y �t

t � 1

x � t � 1

13.

y � �x2 � 2� � �x � 4�2

y � �t � 2�4

8

1284 4x

yx � 2t

8.

Subtracting the second equation from the first, we have

or

Since the discriminant is

the graph is a rotated parabola.

B2 � 4AC � ��2�2 � 4�1��1� � 0,

y ��x � y�2

4�

x � y2

x

−1

2

3

4

−1 2 3 4

yt �x � y

2.x � y � 2t

y � t2 � tx � t2 � t,

t 2 0 1 2

x 2 0 0 2 6

y 6 2 0 0 2

�1�

5.

4

422x

y

y � �x � 1�2

y � t2

x � t � 1 6.

For the orientation is rightto left.

For the orientation is left toright.

−1−1

1

2

3

4

5

6

1 2 3 4 5 6

y

x

t > 0,

t < 0,

y � �x2�

2

� 1 �x2

4� 1, x ≥ 0

y � t4 � 1

x � 2t2 7.

implies

1

321123x

y

y �12x2�3

t � x1�3x � t3

y �12t2

x � t3

10.

−3 −2 −1−1

1

2

3

−2

−3

1 2 3

y

x

y � 3 � x4, x ≥ 0

y � 3 � t

x � 4�t, t ≥ 0

12.

implies

y �1

x � 1� 1

t �1

x � 1x � 1 �

1t

y � t � 1x

1

2

−3

−2

yx � 1 �1t

Section 10.2 Plane Curves and Parametric Equations 385

21.

−6

−4

6

4

x2

16�

y2

4� 1

y2

4� cos2 2�

x2

16� sin2 2�

y � 2 cos 2�

x � 4 sin 2�

14.

x

2

3

4

5

1

2 4 531

y

x � ��y � 2� � 1� � �y � 3�y � t � 2

x � �t � 1� 15.

3

2

4

5

−13 421−2 −1

1

x

y

y � x3 � 1, x > 0

y � e3t � 1

x � et, x > 0 16.

−3 −2 −1−1

1

2

3

−2

−3

1 3

y

x

y � x�2 � 1 �1x2 � 1, x > 0

y � e2t � 1

x � e�t, x > 0

17.

2

3

31 2

2

3

1

x

y

�x� ≥ 1, �y� ≤ 1

y �1x

xy � 1

0 ≤ � < �

2,

2 < � ≤ �

y � cos �

x � sec � 18.

x

2

3

4

1

2 431

y

x ≥ 0

y � x � 1

sec2 � � tan2 � � 1

y � sec2 �

x � tan2 � 19.

Squaring both equations andadding, we have

4

42

2

2

4 2

4

x

y

x2 � y2 � 9.

y � 3 sin �x � 3 cos �,

20.

−6 −4−2

2

4

4 6

y

x

x2

4�

y2

36� 1 ellipse

�x2�

2

� �y6�

2

� cos2 � sin2 � � 1

y � 6 sin �

x � 2 cos � 22.

−2 2

−3

3

y � ±4x�1 � x2

1 � x2 � sin2 �

y � 4 sin � cos �

y � 2 sin 2�

x � cos �

386 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

23.

−1

−4

8

2

�x � 4�2

4�

�y � 1�2

1� 1

�y � 1�2

1� sin2 �

�x � 4�2

4� cos2 �

y � �1 � sin �

x � 4 � 2 cos �

25.

−2

−5

10

3

�x � 4�2

4�

�y � 1�2

16� 1

�y � 1�2

16� sin2 �

�x � 4�2

4� cos2 �

y � �1 � 4 sin �

x � 4 � 2 cos �

24.

−1 7

−5

3

�x � 4�2 � �y � 1�2 � 4

� y � 1�2 � 4 sin2 �

�x � 4�2 � 4 cos2 �

y � �1 � 2 sin �

x � 4 � 2 cos �

26.

−3 3

−2

2

x2 � y2 � 1

y2 � tan2 �

x2 � sec2 �

y � tan �

x � sec �

28.

−2 2

−1.5

1.5

x2�3 � y2�3 � 1

y2�3 � sin2 �

x2�3 � cos2 �

y � sin3 �

x � cos3 �

27.

−9

−6

9

6

x2

16�

y2

9� 1

y2

9� tan2 �

x2

16� sec2 �

y � 3 tan �

x � 4 sec �

29.

−1

−2

5

2

y � 3 ln 3�x � ln x

y � 3 ln t

x � t3 30.

−2 3

−1

3

y �e2x

r�

14

e2x

t �ex

2

y � t2

x � ln 2t

Section 10.2 Plane Curves and Parametric Equations 387

31.

y > 0

x > 0

y �1x3

3�y �1x

et � 3�y

et �1x

y � e3t

−1

−1

5

3x � e�t 32.

y � �x, x > 0

y > 0

y2 � x

y � et

−1 3

−1

4x � e2t

33. By eliminating the parameters in (a) – (d), we get They differ from each other in orientation and in restricteddomains. These curves are all smooth except for (b).

(a)

(c)

x1 32

1

2

3

4

−1

y

x > 0 y > 1

x � e�t y � 2e�t � 1

x1 2

1

2

3

−2 −1

−1

y

y � 2t � 1x � t,

y � 2x � 1.

(b)

when

(d)

x1 32

1

2

3

4

−1

y

x > 0 y > 1

x � et y � 2et � 1

x1 2

1

2

3

−2 −1

−1

y

� � 0, ±�, ±2�, . . . .dxd�

�dyd�

� 0

�1 ≤ x ≤ 1 �1 ≤ y ≤ 3

x � cos � y � 2 cos � � 1

388 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

34. By eliminating the parameters in (a) – (d), we get They differ from each other in orientation and in restricteddomains. These curves are all smooth.

(a)

x1 3

1

3

−1−3

−3

y

y � 2 sin �x � 2 cos �,

x2 � y2 � 4.

(b)

x1 3

1

2

−1

−2

−1

y

x ≥ 0, x � 2 y � 0

x ��4t2 � 1

�t� ��4 �1t2 y �

1t

(c)

x31 2

1

2

3

y

x ≥ 0 y ≥ 0

x � �t y � �4 � t (d)

x

1

3

−3 −2 −1

y

�2 < x ≤ 0 y > 0

x � ��4 � e2t y � et

37. (a)

−6

−4

6

4

−6

−4

6

4 (b) The orientation of the second curve is reversed.

(c) The orientation will be reversed.

(d) Many answers possible. For example,and x � 1 � 2t.x � 1 � t,y � 1 � 2t,

x � 1 � t,

35. The curves are identical on They are both smooth. Represent y � 2�1 � x2�0 < � < �.

36. The orientations are reversed. The graphs are the same. They are both smooth.

38. The set of points corresponding to the rectangular equation of a set of parametric equations does not show the orientation of the curve nor any restriction on the domain of the original parametric equations.

�x, y�

40.

�x � h�2 � �y � k�2 � r 2

cos2 � � sin2 � ��x � h�2

r2 ��y � k�2

r 2 � 1

sin � �y � k

r

cos � �x � h

r

y � k � r sin �

x � h � r cos �39.

y � y1 � m�x � x1�

y � y1 �y2 � y1

x2 � x1�x � x1�

y � y1 � � x � x1

x2 � x1��y2 � y1�

x � x1

x2 � x1� t

y � y1 � t�y2 � y1�

x � x1 � t�x2 � x1�

Section 10.2 Plane Curves and Parametric Equations 389

42.

�x � h�2

a2 ��y � k�2

b2 � 1

y � k

b� tan �

x � h

a� sec �

y � k � b tan �

x � h � a sec �41.

�x � h�2

a2 ��y � k�2

b2 � 1

y � k

b� sin �

x � h

a� cos �

y � k � b sin �

x � h � a cos �

43. From Exercise 39 we have

Solution not unique

y � �2t.

x � 5t

45. From Exercise 40 we have

Solution not unique

y � 1 � 4 sin �.

x � 2 � 4 cos �

47. From Exercise 41 we have

Center:Solution not unique

�0, 0�

y � 3 sin �.

x � 5 cos �

c � 4 ⇒ b � 3a � 5,

49. From Exercise 42 we have

Center:Solution not unique

�0, 0�

y � 3 tan �.

x � 4 sec �

c � 5 ⇒ b � 3a � 4,

51.

Example

x � t � 3, y � 3t � 11

x � t, y � 3t � 2

y � 3x � 2

53.

Example

x � tan t, y � tan3 t

x � 3�t, y � t

x � t, y � t3

y � x3

44. From Exercise 39 we have

Solution not unique

y � 4 � 6t.

x � 1 � 4t

46. From Exercise 40 we have

Solution not unique

y � 1 � 3 sin �.

x � �3 � 3 cos �

48. From Exercise 41 we have

Center:Solution not unique

�4, 2�

y � 2 � 4 sin �.

x � 4 � 5 cos

c � 3 ⇒ b � 4a � 5,

50. From Exercise 42 we have

Center:Solution not uniqueThe transverse axis is vertical,therefore, and are interchanged.yx

�0, 0�

y � sec �.

x � �3 tan �

c � 2 ⇒ b � �3a � 1,

52.

Example

x � �t, y �2

�t � 1

x � t, y �2

t � 1

y �2

x � 154.

Example

x � t3, y � t6

x � t, y � t 2

y � x2

55.

Not smooth at � � 2n�

−2 16

−1

5

y � 2�1 � cos ��

x � 2�� � sin �� 57.

−2 7

−1

5

y � 1 �32 cos �

x � � �32 sin �56.

Not smooth at x � �2n � 1��

−6 6

−2

6

y � 1 � cos �

x � � � sin �

390 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

59.

Not smooth at andor � �

12n�.�0, ±3�,

�x, y� � �±3, 0�

−6 6

−4

4

y � 3 sin3 �

x � 3 cos3 �58.

−9 9

−3

9

y � 2 � 4 cos �

x � 2� � 4 sin � 60.

Smooth everywhere

−0

5

4

� �

y � 2 � cos �

x � 2� � sin �

61.

Smooth everywhere

y � 2 sin2 �−6 6

−4

4x � 2 cot �

63. See definition on page 709.

62.

Smooth everywhere

y �3t2

1 � t3−3 3

−2

2x �3t

1 � t3

64. Each point in the plane is determined by the plane curve For each plot As increases, the curve is traced out in a specific direction called the orientation of the curve.t

�x, y�.t,x � f �t�, y � g�t�.�x, y�

65. A plane curve C, represented by , is smooth if and arecontinuous and not simultaneously 0. See page 714.

g�f�x � f �t�, y � g�t�

67. When the circle has rolled radians, we know that the center is at

or

or

Therefore, and y � a � b cos �.x � a� � b sin �

�AP� � �b cos �cos � � �cos�180� � �� � �AP��b

�BD� � b sin �sin � � sin�180� � �� � �C�b

� �BD�b

b

a

P

A C

B D

θ

x

y�a�, a�.�

68. Let the circle of radius 1 be centered at C. A is the point of tangency on the line OC. is the point on the curve being traced out as the

angle changes and Form the right triangle The angle and

Hence, y � 3 sin � � sin 3�.x � 3 cos � � cos 3�,

y � EC � CD � 3 sin � � cos�3� ��

2� � 3 sin � � sin 3�

x � OE � Ex � 3 sin��

2� �� � sin�3� �

2� � 3 cos � � cos 3�

�DCP � � ��

2� �� � � � � ��

2� � 3� � ��

2�.

OCE � ���2� � ��CDP.

�AP � ⇒ � 2�.�AB � 2��AB � �AP,�P � �x, y�OC � 3.AC � 1,OA � 2,

x

3

2

1

1

θ

α

AD

B E x

C

P x y= ( , )

y

66. (a) Matches (iv) because is on the graph.

(c) Matches (ii) because and

(e) Matches (vi) because undefined at � � 0.

1 ≤ y ≤ 3.�1 ≤ x ≤ 0

�0, 2� (b) Matches (v) because is on the graph.

(d) Matches (iii) because is on the graph.

(f) Matches (i) because for all y.x � � y � 2�2 � 1

�4, 0�

�1, 0�

Section 10.2 Plane Curves and Parametric Equations 391

70. False. Let and Then and is not afunction of x.

yx � y2y � t.x � t269. False

The graph of the parametric equations is only a portion ofthe line y � x.

y � t2 ⇒ y ≥ 0

x � t2 ⇒ x ≥ 0

72. (a)

(b)

tan and

Hence,

y � 5 � �80 sin�45���t � 16t2.

x � �80 cos�45���t

v02 �

320.005

� 6400 ⇒ v0 � 80.

0.005 �16 sec2���4�

v02 �

16v0

2�2�

� � 1 ⇒ � ��

4,h � 5,

y � 5 � x � 0.005x2 � h � �tan ��x �16 sec2 �

v02 x2

y � h � �tan ��x �16 sec2 �

v02 x2

t �x

v0 cos � ⇒ y � h � �v0 sin �� x

v0 cos �� 16� x

v0 cos ��2

y � h � �v0 sin ��t � 16t2

x � �v0 cos ��t

(c)

(d) Maximum height:

Range: 204.88

y � 55 �at x � 100�

−5

0 250

80

71. (a)

(b)

It is not a home run—when

(c)

Yes, it’s a home run when x � 400, y > 10.

00

400

60

x � 400, y < 10.

00

400

30

� 3 � �4403

sin ��t � 16t2

y � h � �v0 sin ��t � 16t2

x � �v0 cos ��t � �4403

cos ��t

100 mi�hr ��100��5280�

3600�

4403

ft�sec (d) We need to find the angle (and time ) such that

From the first equation Substitutinginto the second equation,

We now solve the quadratic for tan :

tan � � 0.35185 ⇒ � � 19.4�

16�12044 �

2

tan2 � � 400 tan � � 7 � 16�12044 �

2

� 0

� 400 tan � � 16�12044 �

2

�tan2 � � 1�.

7 � 400 tan � � 16�12044 �

2

sec2 �

10 � 3 � �4403

sin ��� 1200440 cos �� � 16� 1200

440 cos ��2

t � 1200�440 cos �.

y � 3 � �4403

sin ��t � 16t2 � 10.

x � �4403

cos ��t � 400

t�

Section 10.3 Parametric Equations and Calculus

392 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

1.dy

dx�

dy�dt

dx�dt�

�4

2t�

�2

t

3.

�Note: x � y � 1 ⇒ y � 1 � x and dyd�

� �1�

dydx

�dy�d�

dx�d��

�2 cos � sin �2 sin � cos �

� �1

5.

Lined 2ydx2 � 0

dydx

�dy�dtdx�dt

�32

y � 3t � 1x � 2t,

7.

when

concave upwardsd 2ydx2 � 2

t � �1.dydx

�2t � 3

1� 1

y � t2 � 3tx � t � 1,

9.

when

when

concave downward

� ��

4.

d 2ydx2 �

csc2 ��2 sin �

��csc3 �

2� ��2

� ��

4.� �cot � � �1

dydx

�2 cos �

�2 sin �

y � 2 sin �x � 2 cos �,

11.

when

when

concave downward

� ��

6. � �2 cot3 � � �6�3

d 2ydx2 �

d�dydx�

dxd�

��2 csc � cot �

sec � tan �

� ��

6. �

2 sec �tan �

� 2 csc � � 4

dydx

�2 sec2 �

sec � tan �

y � 1 � 2 tan �x � 2 � sec �,

2.dy

dx�

dy�dt

dx�dt�

�1

�1�3�t�2�3� �3t2�3

4. � �14

e�3��2 ��1

4e3��2 dydx

�dy�d�

dx�d��

��1�2�e���2

2e�

6.

when

concave upwardsd 2ydx2 �

3��t

1��2�t � � 6

t � 1.dydx

�3

1��2�t� � 6�t � 6

y � 3t � 1x � �t,

8.

concave downward

d 2ydx2 �

�2�2���2t � 3�2

2t � 3�

�4�2t � 3�3 �

�427

when t � 0.

dydx

�2

2t � 3�

23

when t � 0.

x � t2 � 3t � 2, y � 2t

10.

is undefined when

is undefined when � � 0.d 2ydx2 �

3 csc2 ��sin �

��3

sin3 ��

d 2ydx2

� � 0.dydx

�3 cos ��sin �

� �3 cot � �dydx

y � 3 sin �x � cos �,

12.

when

when

concave downward

t � 2. ��1

�t � 1�3�2 � �1

d 2ydx2 �

��t � 1��2�t � � �t�1�2�t � 1 ����t � 1�1��2�t �

t � 2. ��t

�t � 1� �2

dydx

�1��2�t � 1 �

1��2�t �

y � �t � 1x � �t,

Section 10.3 Parametric Equations and Calculus 393

14.

when

when

concave downward

� � �. ��1

�1 � cos ��2 � �14

d 2ydx2 �

��1 � cos �� cos � � sin2 ���1 � cos ��2

�1 � cos ��

� � �.dydx

�sin �

1 � cos �� 0

y � 1 � cos �x � � � sin �,13.

when

when

concave upward

� ��

4. �

sec4 � csc �3

�4�2

3

d 2ydx2 �

�sec2 ��3 cos2 � sin �

�1

3 cos4 � sin �

� ��

4. � �tan � � �1

dydx

�3 sin2 � cos �

�3 cos2 � sin �

y � sin3 �x � cos3 �,

15.

At , and

Tangent line:

At and

Tangent line:

At and

Tangent line:

�3x � 8y � 10 � 0

y �12

� ��38

�x � 2�3 �

dydx

� ��38

.� ��

6,2�3,

12,

y � 2 � 0

dydx

� 0.� ��

2,�0, 2�,

3�3x � 8y � 18 � 0

y �32

�3�3

8 x �2

�3

dydx

�3�3

8.� �

2�

3�2

�3,

32,

dydx

�4 sin � cos ��2 csc2 �

� �2 sin3 � cos �

y � 2 sin2 �x � 2 cot �,

17.

(a)

(b) At and

(c) At

(d)

−5

−4

5

(4, 3)

10

y � 2x � 5.

y � 3 � 2�x � 4��4, 3�,dydx

� 2.

dydx

� 2.dydt

� 4,dxdt

� 2,

�x, y� � �4, 3�,t � 2,

−4

6−6

10

t � 2y � t2 � 1,x � 2t,

16.

At and is undefined.

Tangent line:

At and

Tangent line:

At and

Tangent line:

2�3x � 3y � 4�3 � 3 � 0

y � 2 �2�3

3 x �4 � 3�3

2

dydx

�2�3

3.� �

7�

6,4 � 3�3

2, 2,

y � 5

dydx

� 0.� ��

2,�2, 5�,

x � �1

dydx

� � 0,��1, 3�,

dydx

�2 cos �3 sin �

�23

cot �

y � 3 � 2 sin �x � 2 � 3 cos �,

18.

(a)

(b) At and

(c) At

(d)

−4

5−3

4

y � �x � 2.

y � 2 � �1�x � 0��0, 2�,dydx

� �1.

dydx

� �1.dydt

� �1,dxdt

� 1,

�x, y� � �0, 2�,t � 1,

−4

5−3

4

t � 1y �1t

� 1,x � t � 1,

394 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

19.

(a)

(b) At and

(c) At

(d)

−1

−3

8

(4, 2)

5

y � 2

y � 2 � 0�x � 4��4, 2�,dydx

� 0.

dydx

� 0.dydt

� 0,dxdt

� �3,

�x, y� � �4, 2�,t � �1,

−1 8

−3

5

t � �1y � t3 � 3t,x � t2 � t � 2,

21. crosses itself at the origin,

At this point, or

At and Tangent Line

At and Tangent Liney ��34

x.t � �, dydx

� �34

y �34

x.t � 0: dydx

�34

dydx

�3 cos t

4 cos 2t

t � �.t � 0

�x, y� � �0, 0�.x � 2 sin 2t, y � 3 sin t

20.

(a)

(b) At and

(c) At

(d)

−6 6

−4

4

y �34

x � 3�2.

y �3

�2�

34x �

4

�2�4

�2,

3

�2,dydx

�34

.

dydx

�34

.dydt

� �3�2

2,

dxdt

� �2�2,

�x, y� � �4

�2,

3

�2,� �3�

4,

−6 6

−4

4

� �3�

4y � 3 sin �,x � 4 cos �,

23. crosses itself at the point

At this point, or

At and Tangent Line

At and or Tangent Liney � 3x � 5.y � 1 � 3�x � 2�t � 2, dydt

�93

� 3

y � 1.t � �1, dydx

� 0

dydx

�3t2 � 32t � 1

t � 2.t � �1

�x, y� � �2, 1�.x � t2 � t, y � t3 � 3t � 1

22. crosses itself at a pointon the -axis: The corresponding -values are

At

Tangent line:

At

Tangent line:

y � �2�

x �4�

y � 0 � �2�

�x � 2�

t � ��

2: dy

dx� �

2�

.

y �2�

x �4�

y � 0 �2�

�x � 2�

t ��

2: dy

dx�

2�

.

dydx

�2 � � cos t

� sin tdxdt

� � sin t,dydt

� 2 � � cos t,

t � ±��2.t�2, 0�.x

y � 2t � � sin tx � 2 � � cos t,

Section 10.3 Parametric Equations and Calculus 395

24. crosses itself at The corresponding -values are

At

Tangent line:

At

Tangent line: y � ��66

x � 6

dydx

� �2�612

� ��66

.t � ��6,

y ��66

x � 6

y � 6 ��66

�x � 0�

dydx

�2�612

��66

.t � �6,

dydx

�2t

3t2 � 6

t � ±�6.t�0, 6�.y � t2x � t3 � 6t,

25.

Horizontal tangents: when

Points: where n is an integer. Points shown:

Vertical tangents: when

Note: corresponds to the cusp at

Points: Points shown: 5�

2, 1�

3�

2, �1,�

2, 1,��1�n�1�2n � 1��

2, ��1�n�1

dydx

�� sin �

� cos �� tan � � 0 at � � 0

�x, y� � �1, 0�.� � 0

±5�

2, . . .±

3�

2,� � ±

2,

dxd�

� � cos � � 0

�1, �2����1, ��,�1, 0�,�1, 2n����1, �2n � 1���,

±3�, . . . ±2�,±�,� �dyd�

� � sin � � 0

y � sin � � � cos �x � cos � � � sin �,

27.

Horizontal tangents: when

Point:

Vertical tangents: none

−2

−1

4(1, 0)

3

dxdt

� �1 � 0;

�1, 0�

t � 0dydt

� 2t � 0

y � t2x � 1 � t,26.

Horizontal tangents: when

Points: where is an integer

Points shown:

Vertical tangents: nonedxd�

� 2 � 0;

�4�, 0��2�, 4�,�0, 0�,

n�2�2n � 1��, 4��4n�, 0�,

±2�, . . .

±�,� � 0,dyd�

� 2 sin � � 0

y � 2�1 � cos ��x � 2�,

28.

Horizontal tangents: when

Point:

Vertical tangents: nonedxdt

� 1 � 0;

�12

, �94

t � �32

dydt

� 2t � 3 � 0

y � t2 � 3tx � t � 1,

396 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

29.

Horizontal tangents: when

Points:

Vertical tangents: none

−4

−3

5

(0, 2)−

(2, 2)3

dxdt

� �1 � 0;

�2, 2��0, �2�,

t � ±1.dydt

� 3t2 � 3 � 0

y � t3 � 3tx � 1 � t,

33.

Horizontal tangents: when

Points:

Vertical tangents: when

Points:

0

−3

8

(4, 2)−

(2, 1)− (6, 1)−

(4, 0)

1

�2, �1��6, �1�,

� � 0, �.dxd�

� �2 sin � � 0

�4, �2��4, 0�,

� ��

2,

3�

2.

dyd�

� cos � � 0

y � �1 � sin �x � 4 � 2 cos �,

32.

Horizontal tangents: when

Points:

Vertical tangents: when

Points: ��1, 0��1, 0�,

� � 0, �.dxd�

� �sin � � 0

�22

, �2��22

, 2,�22

, 2, ��22

, �2,

� ��

4,

3�

4,

5�

4,

7�

4.

dyd�

� 4 cos 2� � 0

y � 2 sin 2�x � cos �,

30.

Horizontal tangents: when

Points:

Vertical tangents: when

Point: 74

, �118

t �12

.dxdt

� 2t � 1 � 0

�4, 2��2, �2�,

t � ±1.dydt

� 3t2 � 3 � 0

y � t3 � 3tx � t2 � t � 2,

31.

Horizontal tangents: when

Points:

Vertical tangents: when

Points:

−6

−4

6

(0, 3)−

(0, 3)

(3, 0)( 3, 0)−

4

��3, 0��3, 0�,

� � 0, �.dxd�

� �3 sin � � 0

�0, �3��0, 3�,

� ��

2,

3�

2.

dyd�

� 3 cos � � 0

y � 3 sin �x � 3 cos �,

34.

Horizontal tangents: when

Since at and exclude them.

Vertical tangents: when

Point: �4, 0�

�.� � 0,

dxd�

� �8 cos � sin � � 0

3��2,��2� 0dx�d�

� ��

2,

3�

2.

dyd�

� 2 cos � � 0

y � 2 sin �x � 4 cos2 �,

35.

Horizontal tangents: none

Vertical tangents: when

Points: ��1, 0��1, 0�,

x � 0, �.dxd�

� sec � tan � � 0

dyd�

� sec2 � � 0; −6

−4

6( 1, 0)− (1, 0)

4y � tan �x � sec �,

Section 10.3 Parametric Equations and Calculus 397

36.

Horizontal tangents: when

Since at these values, exclude them.

Vertical tangents: when

Point: �0, 0�

�Exclude 0, �.�

3�

2.� �

2,

dxd�

� �2 cos � sin � � 0

� 0dx�d�

x � 0, �.dyd�

� �sin � � 0

y � cos �x � cos2 �, 37.

Concave upward for

Concave downward for t < 0

t > 0

�6t2 � 2

8t3�

1 � 3t2

4t3

d2ydx2 � �2t�6t� � �3t2 � 1�2

4t2

2t�

dydx

�3t2 � 1

2t

y � t3 � tx � t2,

38.

Concave upward for

Concave downward for t < 0

t > 0

d2ydx2 �

3�22t

�34t

dydx

�2t � 3t2

2t� 1 �

32

t

y � t2 � t3x � 2 � t2, 39.

Because

Concave upward for t > 0

d2ydx2 > 0t > 0,

�4t

�2t � 1�3

�4

�2t � 1�2 �t

2t � 1

d2ydx2 � ��2t � 1�2 � �2t � 1�2

�2t � 1�2 ��2 �1t

dydx

�2 � �1�t�2 � �1�t� �

2t � 12t � 1

t > 0y � 2t � ln t,x � 2t � ln t,

40.

Because

Concave downward for t > 0

d2ydx2 < 0t > 0,

d2ydx2 � �

1�t3

2t� �

12t4

dydx

�1�t2t

�1

2t2

t > 0y � ln t,x � t2, 41.

Concave upward on

Concave downward on 0 < t < ��2

��2 < t < �

d2ydx2 � �

sec2tcos t

� �1

cos3t

dydx

� �sin tcos t

� �tan t

0 < t < �y � cos t,x � sin t,

42.

Concave upward on

Concave downward on 0 < t < �

� < t < 2�

d2ydx2 �

�1�2� csc2 t�2 sin t

��1

4 sin3 t

dydx

��cos t2 sin t

� �12

cot t

0 < t < 2�y � sin t,x � 2 cos t, 43.

� �2

1

�4t2 � t � 4 dt

� �2

1

�4 � 8t � 4t2 � 9t dt

� �2

1

��2 � 2t�2 � �3t1�2�2 dt

s � �b

a�dx

dt2

� dydt

2 dt

dxdt

� 2 � 2t, dydt

� 3t1�2

1 ≤ t ≤ 2y � 2t3�2,x � 2t � t2,

398 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

44.

� �6

1�1

t2� 1 dt

s � �b

a�dx

dt2

� dydt

2 dt

dxdt

�1t,

dydt

� 1

1 ≤ t ≤ 6y � t � 1,x � ln t, 45.

� �2

�2

�e2t � 4 dt

s � �b

a�dx

dt2

� dydt

2 dt

dxdt

� et, dydt

� 2

�2 ≤ t ≤ 2y � 2t � 1,x � et � 2,

46.

� ��

0

�3 � 2 cos t � 2 sin t dt

� ��

0

��1 � cos t�2 � �1 � sin t�2 dt

s � �b

a�dx

dt2

� dydt

2 dt

dxdt

� 1 � cos t, dydt

� 1 � sin t

0 ≤ t ≤ �y � t � cos t,x � t � sin t, 47.

� 2�5 � ln�2 � �5 � � 5.916

� �t�t2 � 1 � ln t � �t2 � 1 �2

0

s � 2�2

0

�t2 � 1 dt

dxdt

2

� dydt

2

� 4t2 � 4 � 4�t2 � 1�dydt

� 2,dxdt

� 2t,

0 ≤ t ≤ 2y � 2t,x � t2,

49.

� ���2e�t���2

0� �2�1 � e���2� � 1.12

� ���2

0

�2e�2t dt � ��2���2

0e�t��1� dt

s � ���2

0�dx

dt2

� dydt

2

dt

dydt

� e�t�cos t � sin t�dxdt

� �e�t�sin t � cos t�,

0 ≤ t ≤ �

2y � e�t sin t,x � e�t cos t,

51.

du �3

�t dtu � 6�t,

�1

12�ln��37 � 6� � 6�37� � 3.249

�1

12�ln��1 � u2 � u� � u�1 � u2�6

0

�16�

6

0

�1 � u2 du

s � �1

0� 1

4t� 9 dt �

12�

1

0 �1 � 36t

�t dt

dydt

� 3dxdt

�1

2�t,y � 3t � 1,x � �t,

48.

� ���1 � 36t2�3�2

54 �0

�1�

�154

�1 � 373�2� � 4.149

s � �0

�1

�4t2 � 144t4 dt � �0

�1�2t�1 � 36t2 dt

dxdt

2

� dydt

2

� 4t2 � 144t4dxdt

� 2t, dydt

� 12t2,

x � t2 � 1, y � 4t3 � 3, �1 ≤ t ≤ 0

50.

� �12

ln13 �

12

ln�3� � 0.549

� ��12

ln t � 1t � 1 �1�2

0

� �1�2

0� 1

�1 � t2�2 dt � �1�2

0

11 � t2

dt

s � �1�2

0�dx

dt2

� dydt

2

dt

dydt

�12

�2t1 � t2 �

t1 � t2

dxdt

�1

�1 � t2,

0 ≤ t ≤12

y � ln�1 � t 2,x � arcsin t,

Section 10.3 Parametric Equations and Calculus 399

52.

� � t 5

10�

16t3�

2

1�

779240

� �2

1t4

2�

12t4 dt

� �2

1 �t4

2�

12t4

2

dt

S � �2

1�1 � t 4

2�

12t4

2

dt �

dydt

�t4

2�

12t4

dxdt

� 1,y �t5

10�

16t3,x � t,

54.

� 4a���2

0 d� � �4a��

��2

0� 2�a

S � 4���2

0

�a2 sin2 � � a2 cos2 � d�

dyd�

� a cos �dxd�

� �a sin �,y � a sin �,x � a cos �,

53.

� 6a���2

0 sin 2� d� � ��3a cos 2��

��2

0� 6a

� 12a���2

0 sin � cos ��cos2 � � sin2 � d�

s � 4���2

0

�9a2 cos4 � sin2 � � 9a2 sin4 � cos2 � d�

dyd�

� 3a sin2 � cos �

dxd�

� �3a cos2 � sin �,y � a sin3 �,x � a cos3 �,

55.

� ��4�2a�1 � cos ���

0� 8a

� 2�2a��

0

sin �

�1 � cos � d�

� 2�2a��

0

�1 � cos � d�

s � 2��

0

�a2 �1 � cos � �2 � a2 sin2 � d�

dyd�

� a sin �dxd�

� a�1 � cos ��,

y � a�1 � cos ��,x � a�� � sin ��,

57.

(a)

(b) Range: 219.2 ft,

(c)

for

� 230.8 ft

s � �45�16

0

��90 cos 30��2 � �90 sin 30� � 32t�2 dt

t �4516

.y � 0

dydt

� 90 sin 30� � 32tdxdt

� 90 cos 30�,

t �4516

00

240

35

y � �90 sin 30��t � 16t2x � �90 cos 30��t,56.

� �2�

0 � d� � ��2

2 �2�

0� 2� 2

S � �2�

0

��2 cos2 � � �2 sin2 � d�

dyd�

� � sin �

dxd�

� � cos �y � sin � � � cos �,x � cos � � � sin �,

400 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

58.

By the First Derivative Test, maximizes the range

To maximize the arc length, we have

Using a graphing utility, we see that is a maximum of approximately 303.67 feet at� � 0.9855�56.5��.

s

�2025

8 sin � �

202516

cos2 � ln�1 � sin �1 � sin ��

s � ��90�16�sin �

0

��90 cos ��2 � �90 sin � � 32t�2 dt

dxdt

� 90 cos �, dydt

� 90 sin � � 32t.

�x � 253.125 feet�.� ��

4 �45��

x��� �902

322 cos 2� � 0 ⇒ � �

4

�902

16 sin � cos � �

902

32 sin 2�

x � �90 cos ��t � �90 cos �� 9016

sin �

y � 0 ⇒ �90 sin ��t � 16t2 ⇒ t � 0, 9016

sin �

59.

(a)

(c)

� 8�1

0

�t 8 � 4t 6 � 4t 5 � 4t 3 � 4t 2 � 1�1 � t 3�2 dt � 6.557

� 2�1

0� 16

�1 � t3�4 �t 8 � 4t 6 � 4t 5 � 4t 3 � 4t2 � 1� dts � 2�1

0��4�1 � 2t3�

�1 � t3�2 �2

� �4t�2 � t3��1 � t3�2 �

2

dt

6−6

−4

4

x3 � y3 � 4xy

y �4t2

1 � t3x �4t

1 � t3,

(b)

when or

Points: 4 3�23

, 4 3�4

3 � �1.6799, 2.1165��0, 0�,

t � 3�2.t � 0� 0 �4t�2 � t3��1 � t3�2

dydt

��1 � t3��8t� � 4t2�3t2�

�1 � t3�2

60.

(a)

(c) Arc length over �

4≤ t ≤

2: 4.5183

6

6

−2

−6

��

2≤ � ≤

2y � 4 sin2 �,x � 4 cot � �

4tan �

,

(b)

for

Horizontal tangent at

(Function is not defined at � � 0�

�x, y� � �0, 4�� � ±�

2

� � 0, ±�

2dyd�

� 0

dxd�

� �4 csc2 �

dyd�

� 8 sin � � cos �

Section 10.3 Parametric Equations and Calculus 401

61. (a)

�− �3

−1

3

−1

3

�− �3

0 ≤ t ≤ �0 ≤ t ≤ 2�

y � 1 � cos�2t�y � 1 � cos t

x � 2t � sin�2t�x � t � sin t (b) The average speed of the particle on thesecond path is twice the average speed ofa particle on the first path.

(c)

The time required for the particle totraverse the same path is t � 4�.

y � 1 � cos�12t�

x �12t � sin�1

2t�

62. (a) First particle:

Second particle:

4

4−

6 6−

0 ≤ t ≤ 2�y � 3 cos t,x � 4 sin t,

4

4−

6− 6

0 ≤ t ≤ 2�y � 4 sin t,x � 3 cos t, (b) There are 4 points of intersection.

(c) Suppose at time that

and

and

Yes, the particles are at the same place at the same timefor 3.7851. The intersection pointsare and

(d) The curves intersect twice, but not at the same time.

��2.4, �2.4��2.4, 2.4�t � 0.6435,tan t �

34.

tan t �34tan t �

34

4 sin t � 3 cos t3 cos t � 4 sin t

t

63.

� 103.6249

� 2��2

0

�17 �t � 1� dt � 8��17

S � 2��2

0 �t � 1��42 � 12 dt

dydt

� 1y � t � 1,

dxdt

� 4x � 4t, 64.

� 159.6264

� ��4

0 �t � 2��t2 � 4 dt

S � ��4

0�t � 2��t2

4� 1 dt

dydt

� 1y � t � 2,

dxdt

�12

tx �14

t2,

65.

� 5.3304

S � 2����2

0 cos ��4 cos2 � sin2 � � sin2 � d�

dyd�

� �sin �y � cos �,

dxd�

� �2 cos � sin �x � cos2 �, 66.

� 23.2433

� 2����2

0 �� � cos ���3 � 2 cos � � 2 sin � d�

S � 2����2

0 �� � cos ����1 � cos ��2 � �1 � sin ��2 d�

dyd�

� 1 � sin �y � � � cos �,

dxd�

� 1 � cos �x � � � sin �,

67.

(a)

� �2�5�t 2�4

0� 32��5

S � 2��4

0 2t�1 � 4 dt � 4�5��4

0 t dt

dydt

� 2dxdt

� 1,y � 2t,x � t,

(b)

� ��5� t2�4

0� 16��5

� 2�5��4

0 t dtS � 2��4

0 t�1 � 4 dt

402 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

70.

��

9�173�2 � 23�2� � 23.48

S � 2��2

1 13

t3�t4 � 1 dt ��

9��x4 � 1�3�2�2

1

dxdt

� t2, dydt

� 1

x �13

t3, y � t � 1, 1 ≤ t ≤ 2, y-axis69.

� 32����2

0 cos � d� � �32� sin ��

��2

0� 32�

S � 2����2

0 4 cos ����4 sin ��2 � �4 cos ��2 d�

dyd�

� 4 cos �dxd�

� �4 sin �,y � 4 sin �,x � 4 cos �,

71.

�12�a2

5 �sin5 ����2

0�

125

�a2� 12a2����2

0 sin4 � cos � d�S � 4����2

0 a sin3 ��9a2 cos4 � sin2 � � 9a2 sin4 � cos2 � d�

dyd�

� 3a sin2 � cos �dxd�

� �3a cos2 � sin �,y � a sin3 �,x � a cos3 �,

72.

(a)

(b)

� 2�a2 �2�ab2

�a2 � b2 ln a � �a2 � b2

b � 2�a2 � �b2

e ln 1 � e1 � e

�2a�

c�c �b2 � c2 � b2 ln c � �b2 � c2 � b2 ln b�

�2a�

c �c sin ��b2 � c2 sin2 � � b2 ln c sin � � �b2 � c2 sin2 � ���2

0

� 4����2

0 a cos ��b2 � c2 sin2 � d� �

4a�

c ���2

0 c cos ��b2 � c2 sin2 � d�

S � 4����2

0 a cos ��a2 sin2 � � b2 cos2 � d�

e ��a2 � b2

a�

ca

: eccentricity � 2�b2 � 2�a2b

�a2 � b2 arcsin�a2 � b2

a � 2�b2 � 2�abe arcsin�e�

��2ab�

e �e cos ��1 � e2 cos2 � � arcsin�e cos �����2

0�

2ab�

e�e�1 � e2 � arcsin�e��

� 4����2

0 ab sin ��1 � a2 � b2

a2 cos2 � d� ��4ab�

e ���2

0��e sin ���1 � e2 cos2 � d�

S � 4����2

0 b sin ��a2 sin2 � � b2 cos2 � d�

dyd�

� b cos �dxd�

� �a sin �,y � b sin �,x � a cos �,

73.

See Theorem 10.7.

dydx

�dy�dtdx�dt

74. (a) 0

(b) 4

68.

(a)

� �2�5��4t � t 2��2

0� 8��5

S � 2��2

0�4 � 2t��1 � 4 dt

dydt

� �2dxdt

� 1,y � 4 � 2t,x � t,

(b) S � 2��2

0 t�1 � 4 dt � ��5�t2�

2

0� 4��5

Section 10.3 Parametric Equations and Calculus 403

78. (a)

(b) S � 2��b

a

f �t��dxdt

2

� dydt

2

dt

S � 2��b

a

g�t��dxdt

2

� dydt

2

dt

80.

� 2�r2�1 � cos ��

� ��2�r2 cos ��

0

� 2�r2��

0 sin d

x

θ

yS � 2���

0 r sin �r2 sin2 � r2 cos2 d

y � r sin x � r cos ,

81.

� 8��sin3 � cos �4

�38

sin � cos � �38

����2

0�

3�

2

� 8���2

0 sin4 � d� A � ���2

0 2 sin2 � tan ��4 sin � cos �� d�

dxd�

� 4 sin � cos �

y � 2 sin2 � tan �

x

2

−2

−1

−1

1

1−2

0 ≤ θ < π2

y x � 2 sin2 �

79. Let y be a continuous function of on Suppose that and Then using integration by substitution, and

�b

a

y dx � �t2

t1

g�t�f�t� dt.

dx � f�t� dtf �t2� � b.f �t1� � a,y � g�t�,x � f �t�,

a ≤ x ≤ b.x

82.

A � 2�0

��2�2 sin2 ����2 csc2 �� d� � �8�0

��2 d� � ��8��

0

��2� 4�

dxd�

� �2 csc2 �y � 2 sin2 �,x � 2 cot �,

83. is area of ellipse (d).�ab 84. is area of asteroid (b).38�a2 85. is area of cardioid (f).6�a2

86. is area of deltoid (c).2�a2 87. is area of hourglass (a).83ab 88. is area of teardrop (e).2�ab

75. One possible answer is the graphgiven by

x

4

4

2

2

−4

−4

−2

−2

y

y � �t.x � t,

76. One possible answer is the graphgiven by

x4321

4

3

2

1

−4 −3

−3

−2

−2

−4

y

y � �t.x � �t,

77.

See Theorem 10.8.

s � �b

a�dx

dt2

� dydt

2

dt

404 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

89.

�x, y� � 34

, 85

y �1A�

2

0 y2

2 dx �

332�

4

0�4 � t�2

1

2�t dt �

364�

4

0�16t�1�2�8t1�2 � t3�2� dt �

364�32�t �

163

t�t �25

t2�t�4

0�

85

x �1A�

2

0 yx dx �

316�

4

0�4 � t��t 1

2�t dt �3

32�4

0�4 � t� dt � � 3

324t �t2

2�4

0�

34

A � �2

0 y dx � �4

0�4 � t� 1

2�t dt �

12�

4

0 �4t�1�2 � t1�2� dt � �1

28�t �23

t�t�4

0�

163

0 < t < 4y � 4 � t,x � �t,

90.

Let then and

�x, y� � 83�

, 8

3�

y �1

2��0

4��t �2�

1

2�4 � t dt � �1

4��0

4

t

�4 � t dt � �

14��

�2�8 � t�3

�4 � t�0

4�

83�

x �1��

0

4

�4 � t�t�1

2�4 � t dt � �1

2��0

4

�t dt � ��1

2� 23

t3�2�0

4�

83�

�t � �4 � u2.du � �1��2�4 � t � dtu � �4 � t,

A � �0

4

�t�1

2�4 � t dt � �2

0

�4 � u2 du �12�u�4 � u2 � 4 arcsin

u2�

2

0� �

0 ≤ t ≤ 4dxdt

� �1

2�4 � t,y � �t,x � �4 � t,

91.

(Sphere) � �54���cos � �cos3 �

3 �0

��2� 36�

� �54��0

��2�1 � cos2 �� sin � d�

� �54��0

��2sin3 � d�

V � 2��0

��2�3 sin ��2��3 sin �� d�

dxd�

� �3 sin �y � 3 sin �,x � 3 cos �, 92.

� �18���cos � �cos3 �

3 �0

��2� 12�

� �18��0

��2sin3 � d�

V � 2��0

��2�3 sin ��2��sin �� d�

dxd�

� �sin �y � 3 sin �,x � cos �,

93.

(a)

(b) At

Tangent line:

—CONTINUED—

y � a1 ��32 � �2 � �3�x � a�

6�

12

dydx

�1�2

1 � �3�2� 2 � �3.y � a1 �

�32 ,x � a�

6�

12,� �

6,

�cos � � 1

a�1 � cos ��3 ��1

a�cos � � 1�2

d2ydx2 � ��1 � cos �� cos � � sin ��sin ��

�1 � cos ��2 ���a�1 � cos ���

dydx

�a sin �

a�1 � cos �� �sin �

1 � cos �

dxd�

� a�1 � cos ��dyd�

� a sin �,

y � a�1 � cos ��x � a�� � sin ��,

Section 10.3 Parametric Equations and Calculus 405

93. —CONTINUED—

(c)

Points of horizontal tangency:

(d) Concave downward on all open -intervals:. . ., ��2�, 0�, �0, 2��, �2�, 4��, . . .

�x, y� � �a�2n � 1��, 2a�

⇒ sin � � 0, 1 � cos � � 0dydx

�sin �

1 � cos �� 0 (e)

� �4a cos�

2�2�

0� 8a

� 2a �2�

0 sin

2 d�

� a�2�

0

�4 sin2��2 d�

� a �2�

0

�2 � 2 cos � d�

s � �2�

0

�a2 sin2 � � a2�1 � cos ��2 d�

94.

(a)

(b)

(c) at

y ��33

x �53

y �83

��33

�x � �3�

dydx

�2

2�3�

�33

t � 1.�x, y� � ��3, 83�

��2�3t2 � 6�3

�12t2��2�3t� � �t2 � 312t3

d2ydx2 � �2�3�t���2t� � �3 � t2�2�3

12t2 ���2�3t�

dydx

�3 � t2

2�3tdydt

� 3 � t2,dxdt

� 2�3t,

6

18

−6

0

y � 3t �13t3x � t2�3,

(d)

(e) S � 2��3

0 �3t �

13t3��t3 � 3� dt � 81�

� �3

�3 �t3 � 3� dt � 36

� �3

�3

��t2 � 3�2 dt

� �3

�3

�t4 � 6t2 � 9 � 12t2 dt

s � �3

�3

�12t2 � �3 � t�2 dt

95.

� r�sin � � � cos ��

y � v � w � r sin � � r� cos �

� r�cos � � � sin ��

rr

t

v

w

u

θ

( , )x y

θ

θx

y x � t � u � r cos � � r� sin �

406 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

96. Let’s focus on the region above the -axis. From Exercise 95,the equation of the involute from to is

At the string is fully extended and has length So, the area

of region A is

We now need to find the area of region B.

Hence, the far right point on the involute is

The area of the region is given by

where and

Thus, we can calculate

Since the area of is we have

Total area covered �56

�3. � 2�14

�3 ��

6��2 � 3� �

2���2,C � D

�0

�sin � � � cos ��� cos � d� ��

6��2 � 3�.

dx � � cos � d�.y � sin � � � cos �

�����2

��� y dx � �����2

��0 y dx � ���0

��� y dx

B � C � D

���2, 1�.

�� � 0 is cusp.�dxd�

� �sin � � sin � � � cos � � � cos � � 0 ⇒ � ��

2.

14���2� �

14�3.

�.��1, ��,

0 ≤ � ≤ �.

y � sin � � � cos �

x � cos � � � sin �

��1, ���1, 0�

x

π( (−1,

(−1 − , 0)π(−1, 0) (1, 0)

π( (,12A

D CB

yx

97. (a)

(b)

The graph is the circle except the point

(c) As increases from to 0, the speed increases, and as increases from 0 to 20, the speed decreases.t�20t

Verify: x2 � y2 � 1 � t2

1 � t22

� 2t1 � t2

2

�1 � 2t2 � t4 � 4t2

�1 � t2�2 ��1 � t2�2

�1 � t2�2 � 1

��1, 0�.x2 � y2 � 1,�for �� < t < ��

�20 ≤ t ≤ 20y �2t

1 � t2,x �1 � t2

1 � t2,

−3

−2

3

2

(b)

Same as the graph in (a), but has the advantage ofshowing the position of the object and any given time t.

1200

60

0 ≤ ty � t � 12 tanh t

12,x � 12 sech

t12

,98. (a)

—CONTINUED—

1200

60

0 < x ≤ 12

y � �12 ln12 � �144 � x2

x � �144 � x2

Section 10.4 Polar Coordinates and Polar Graphs 407

Section 10.4 Polar Coordinates and Polar Graphs

98. —CONTINUED—

(c)

Tangent line:

-intercept:

Distance between and

d � 12 for any t ≥ 0.

d ���12 sech t012�

2

� ��12 tanh t012�

2

� 12�x, y�:�0, t0�

�0, t0�y

y � t0 � �sinh t012�x

y � �t0 � 12 tanh t012� � �sinh

t012�x � 12 sech

t012�

x

8

4

16

10

24

122 4 6

12

( , )x y

(0, )y0

ydydx

�1 � sech2�t�12�

�sech�t�12� tan�t�12� � �sinh t

12

99. False. d 2ydx2 �

ddt�

g��t�f��t� �

f��t� �f��t�g��t� � g��t� f��t�

f��t�3 100. False. Both and are zero when Byeliminating the parameter, we have which doesnot have a horizontal tangent at the origin.

y � x2�3

t � 0.dy�dtdx�dt

2.

22,( )−

1 20

π2

�x, y� � ���2, �2�

y � �2 sin�7�

4 � � �2

x � �2 cos�7�

4 � � ��2

��2, 7�

4 �

4.

01

(0, 0)

π2

�x, y� � �0, 0�

y � 0 sin��7�

6 � � 0

x � 0 cos��7�

6 � � 0

�0, �7�

6 �

1.

01 2 3

3π6

4,( )

π2

�x, y� � �0, 4�

y � 4 sin��

2� � 4

x � 4 cos��

2� � 0

�4, �

2� 3.

01

π3

− −4,( )

π2

�x, y� � ��2, 2�3 �

y � �4 sin���

3� � 2�3

x � �4 cos���

3� � �2

��4, ��

3�

5.

0

2, 2.36( )

1

π2

�x, y� � ��1.004, 0.996�

y � �2 sin�2.36� � 0.996

x � �2 cos�2.36� � �1.004

��2, 2.36� 6.

1 2

( 0.0024, 3)−

0

π2

�x, y� � ��0.0024, 3�

y � �3 sin��1.57� � 3

x � �3 cos��1.57� � �0.0024

��3, �1.57�

408 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

7.

x

1

2

3

4

1−1

−1

−2−3−4

y

(−3.54, 3.54)

�x, y� � ��3.5355, 3.5355�

�r, �� � �5, 3�

4 � 9.

x21 3

1

−1

−1

−2

−3

(2.804, 2.095)−

y

�x, y� � �2.804, �2.095�

�r, �� � ��3.5, 2.5�

11.

x1 2

2

1(1, 1)

y

���2, 5�

4 �� ��

4,

5�

4, ��2,

4�,

tan � � 1

r � ±�2

�x, y� � �1, 1�

15.

� ��2, 5��6�

�r, �� � �2, 11��6�

tan � � ��3�3

r � �3 � 1 � 2

−1 1 2

−1

−2

1

x

y

( 3 , −1)

�x, y� � ��3, �1�

14.

(4, −2)

−1

1 2 3 4 5

−2

−3

−4

−5

y

x

�2�5, �0.464�, ��2�5, 2.678�� � �0.464

tan � � �24

� �12

r � ±�16 � 4 � ±2�5

�x, y� � �4, �2�

8.

x21

2

1

−2 −1

−1

−2

( 1.7321, 1)−

y

�x, y� � ��1.7321, 1�

�r, �� � ��2, 11�

6 �

10.

(2.2069, 7.9494)

1 2−1−2−2

−4

−6

−8

2

4

6

8

x

y

�x, y� � �2.2069, 7.9494�

�r, �� � �8.25, 1.3� 12.

undefined

1 2 3x

−2

−1

−3

−5

−4

−2 −1

(0, 5)−

y

��5, �

2�� ��

2,

3�

2, �5,

3�

2 �,

tan �

r � ±5

�x, y� � �0, �5�

16

� ��2�3, 5��6�

�r, �� � �2�3, 11��6�

tan � � ��3�3

r � �9 � 3 � 2�3

y

x−1 1 2 3 4

−1

−2

−3

1

2

3, − )) 3

�x, y� � �3, �3�

13.

x−4 −3 −2 −1 1

5

4

3

2

1

( 3, 4)−

y

� � 2.214, 5.356, �5, 2.214�, ��5, 5.356�

tan � � �43

r � ±�9 � 16 � ±5

�x, y� � ��3, 4�

Section 10.4 Polar Coordinates and Polar Graphs 409

17.

�r, �� � �3.606, �0.588�

�x, y� � �3, �2�

19.

�r, �� � �2.833, 0.490�

�x, y� � �52, 43�

18.

�r, �� � �6, 0.785�

�x, y� � �3�2, 3�2�

20.

�r, �� � �5, �1.571�

�x, y� � �0, �5�

21. (a)

x1 2 3 4

1

2

3

4 (4, 3.5)

y�x, y� � �4, 3.5� (b)

0

(4, 3.5)

1

π2

�r, �� � �4, 3.5�

22. (a) Moving horizontally, the -coordinate changes. Moving vertically, the -coordinate changes.

(b) Both r and values change.

(c) In polar mode, horizontal (or vertical) changes result in changes in both and �.r

yx

23. circle

Matches (c)

r � 2 sin � 24.

Rose curve

Matches (b)

r � 4 cos 2� 25.

Cardioid

Matches (a)

r � 3�1 � cos �� 26.

Line

Matches (d)

r � 2 sec �

27.

r � a

0a

π2

x2 � y2 � a2

29.

r � 4 csc �

r sin � � 4

042

π2

y � 4

28.

r � 2a cos �

r�r � 2a cos �� � 0

r2 � 2ar cos � � 0

a 2a0

π2

x2 � y2 � 2ax � 0

30.

r � 10 sec �

r cos � � 10

2 4 6 8 120

π2

x � 10

31.

r ��2

3 cos � � sin �

r�3 cos � � sin �� � �2

3r cos � � r sin � � 2 � 0

021

π2

3x � y � 2 � 0

410 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

35.

2

1

21

112

2

x

y

x2 � y2 � 9

r2 � 9

r � 3

37.

x

12

122

1

y

x2 � y2 � y � 0

x2 � �y �12�

2

�14

x2 � y2 � y

r2 � r sin �

r � sin �

39.

�x2 � y2 � arctan yx

tan�x2 � y2 �yx

tan r � tan �

x

π

π

2

ππ 2

π2

π−

y r � �

33.

021 3 5 74 6

π2

r � 9 csc2 � cos �

r �9 cos �sin2 �

r2 sin2 � � 9r cos �

y2 � 9x32.

2 40

π2

� 8 csc 2�

r2 � 4 sec � csc �

�r cos ���r sin �� � 4

xy � 4

34.

1 20

π2

r2 � 9 cos 2�

r2r2 � 9�cos 2�� � 0

�r2�2 � 9�r2 cos2 � � r2 sin2 �� � 0

�x2 � y2�2 � 9�x2 � y2� � 0

36.

x

1

−1

1−1

y

x2 � y2 � 4

r2 � 4

r � �2 38.

−2 −1 1 2 3 4 6

−2

1

2

3

4

−3

−4

y

x

�x �52�

2

� y2 � �52�

2

x2 � 5x �254

� y2 �254

x2 � y2 � 5x

r2 � 5r cos �

r � 5 cos �

Section 10.4 Polar Coordinates and Polar Graphs 411

40.

x

2

−2

21

1

−2 −1

−1

y

y � ��33

x

yx

� ��33

tan � � tan 5�

6

� �5�

642.

x21−1

1

3

y

y � 2 � 0

y � 2

r sin � � 2

r � 2 csc �

44.

−10

−18

10

3

0 ≤ � < 2�

r � 5�1 � 2 sin ��

46.

10−4

−6

6

0 ≤ � < 2�

r � 4 � 3 cos �

41.

2

1

3

21x

y

x � 3 � 0

x � 3

r cos � � 3

r � 3 sec �

43.

−12 6

−6

6

0 ≤ � < 2�

r � 3 � 4 cos � 45.

−4 5

−2

4

0 ≤ � < 2�

r � 2 � sin �

47.

Traced out once on

−10

−5

5

5

�� < � < �

r �2

1 � cos �48.

Traced out once on

3−3

−1

3

0 ≤ � ≤ 2�

r �2

4 � 3 sin �

50.

6−6

−4

4

0 ≤ � < 4�

r � 3 sin�5�

2 �49.

−3

−2

3

2

0 ≤ � < 4�

r � 2 cos�3�

2 � 51.

−3 3

−2

2

0 ≤ � < �

2

r2 � �2�sin 2�

r1 � 2�sin 2�

r2 � 4 sin 2�

412 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

52.

Graph as

It is traced out once on

2−2

−1.5

1.5

0, ��.

r1 �1

��, r2 � �

1

��.

r2 �1�

.

53.

�x � h�2 � � y � k�2 � h2 � k2

�x2 � 2hx � h2� � � y2 � 2ky � k2� � 0 � h2 � k2

x2 � y2 � 2hx � 2ky � 0

x2 � y2 � 2�hx � ky�

r2 � 2h�r cos �� � k�r sin ��

r2 � 2r�h cos � � k sin ��

r � 2�h cos � � k sin ��

Radius:

Center: �h, k�

�h2 � k2

54. (a) The rectangular coordinates of are The rectangular coordinates of are

(b) If the points lie on the same line passing through the origin. In this case,

(c) If then and the Pythagorean Theorem!

(d) Many answers are possible. For example, consider the two points and

Using and ,

You always obtain the same distance.

� �5.d ����1�2 � �2�2 � 2��1��2� cos�� �5�

2 ��r2, �2� � 2, �5��2��r1, �1� � ��1, ��

d ��1 � 22 � 2�1��2� cos�0 ��

2� � �5

�r2, �2� � �2, ��2�.�r1, �1� � �1, 0�

d � �r12 � r2

2,cos��1 � �2� � 0�1 � �2 � 90,

� ��r1 � r2�2 � �r1 � r2�.d � �r1

2 � r22 � 2r1r2 cos�0�

�1 � �2,

d � �r12 � r2

2 � 2r1r2 cos��1 � �2�

� r12 � r2

2 � 2r1r2 cos��1 � �2�

� r22 �cos2 �2 � sin2 �2� � r1

2 �cos2 �1 � sin2 �1� � 2 r1r2�cos �1 cos �2 � sin �1 sin �2�

� r22 cos2 � 2 � 2r1r2 cos �1 cos � 2 � r1

2 cos2 �1 � r22 sin2 � 2

2 � 2r1r2 sin �1 sin � 2 � r12 sin2 �1

� �r2 cos �2 � r1 cos �1�2 � �r2 sin �2 � r1 sin �1�2

d 2 � �x2 � x1�2 � � y2 � y1�2

�r2 cos �2, r2 sin �2�.�r2, �2��r1 cos �1, r1 sin �1�.�r1, �1�

Section 10.4 Polar Coordinates and Polar Graphs 413

55.

��20 � 16 cos �

2� 2�5 � 4.5

d ��42 � 22 � 2�4��2� cos�2�

3�

6�

�4, 2�

3 �, �2, �

6�

57.

� �53 � 28 cos��0.7� � 5.6

d � �22 � 72 � 2�2��7� cos�0.5 � 1.2�

�7, 1.2��2, 0.5�,

59.

At

At

At dydx

� 0.��1, 3�

2 �,

dydx

� �23

.�2, ��,

dydx

� 0.�5, �

2�,

�2 cos ��3 sin � � 1�3 cos 2� � 2 sin �

�2 cos ��3 sin � � 1�

6 cos2 � � 2 sin � � 3

dydx

�3 cos � sin � � cos ��2 � 3 sin ��3 cos � cos � � sin ��2 � 3 sin ��

r � 2 � 3 sin �

61. (a), (b)

Tangent line:

(c) At � ��

2,

dydx

� �1.0.

y � �x � 3

y � 3 � �1�x � 0�

�r, �� � �3, �

2� ⇒ �x, y� � �0, 3�

−8

−4

4

4

r � 3�1 � cos ��

56.

��109 � 60 cos �

6� �109 � 30�3 � 7.6

d ��102 � 32 � 2�10��3� cos�7�

6� ��

�10, 7�

6 �, �3, ��

58.

� �160 � 96 cos 1.5 � 12.3

d � �42 � 122 � 2�4��12� cos�2.5 � 1�

�12, 1��4, 2.5�,

60.

At

At is undefined.

At dydx

� 0.�4, 3�

2 �,

dydx�3,

7�

6 �,

dydx

� �1.�2, 0�,

dydx

��2 cos � sin � � 2 cos ��1 � sin ���2 cos � cos � � 2 sin ��1 � sin ��

r � 2�1 � sin ��

62. (a), (b)

Tangent line:

(c) At does not exist (vertical tangent).dydx

� � 0,

x � 1

�r, �� � �1, 0� ⇒ �x, y� � �1, 0�

4−8

−4

4

r � 3 � 2 cos �

63. (a), (b)

Tangent line:

y � ��3x �92

y �94

� ��3�x �3�3

4 ��r, �� � �3�3

2,

3� ⇒ �x, y� � �3�34

, 94�

−4

−1

5

5r � 3 sin � (c) At � ��

3,

dydx

� ��3 � �1.732.

414 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

64. (a), (b)

Tangent line:

(c) At � ��

4,

dydx

� �1.

y � �x � 4�2

y � 2�2 � �1�x � 2�2 �

�r, �� � �4, �

4� ⇒ �x, y� � �2�2, 2�2 �

8−8

−6

6

r � 4 65.

Horizontal tangents:

Vertical tangents: �32

, 7�

6 �, �32

, 11�

6 �

sin � � 1 or sin � � �12 ⇒ � �

2,

7�

6,

11�

6

� �2 sin � � 1��sin � � 1� � 0

� 2 sin2 � � sin � � 1

� �sin � � sin2 � � sin2 � � 1

dxd�

� ��1 � sin �� sin � � cos � cos �

�2, 3�

2 �, �12

, �

6�, �12

, 5�

6 �

cos � � 0 or sin � �12 ⇒ � �

2,

3�

2,

6,

5�

6

� cos ��1 � 2 sin �� � 0

dyd�

� �1 � sin �� cos � � cos � sin �

r � 1 � sin �

67.

Horizontal: �1, 3�

2 ��5, �

2�,

� ��

2,

3�

2

� 3 cos � � 0

dyd�

� �2 csc � � 3� cos � � ��2 csc � cot �� sin �

r � 2 csc � � 366.

Horizontal:

Vertical: �a�22

, �

4�, �a�22

, 3�

4 ��0, 0�, �a,

2�

7�

45�

4,

3�

4,� �

4,sin � � ±

1

�2,

dxd�

� �a sin2 � � a cos2 � � a�1 � 2 sin2 �� � 0

3�

2�,

2,� � 0,

� 2a sin � cos � � 0

dyd�

� a sin � cos � � a cos � sin �

r � a sin �

68.

Horizontal: ��2a4

, �

4�, ��2a4

, 3�

4 �, �0, 0�

� ��

4,

3�

4tan2 � � 1,� � 0,

� 2a sin � cos ��cos2 � � sin2 �� � 0

� 2asin � cos3 � � sin3 � cos �

dyd�

� a sin � cos3 � � �2a sin2 � cos � � a cos3 � sin �

r � a sin � cos2 � 69.

Horizontal tangents:

�r, �� � �0, 0�, �1.4142, 0.7854�, �1.4142, 2.3562�

−3 3

−2

2

r � 4 sin � cos2 �

Section 10.4 Polar Coordinates and Polar Graphs 415

70.

Horizontal tangents: �r, �� � �2.061, ±0.452�

4−2

−2

2

r � 3 cos 2� sec � 71.

Horizontal tangents: �r, �� � �7, �

2�, �3, 3�

2 �

−12 12

−6

10

r � 2 csc � � 5

73.

Circle

Center:

Tangent at the pole: � � 0

�0, 32�

r �32

x 2 � �y �32�

2

�94

x2 � y 2 � 3y

r2 � 3r sin �

01 2 3

π2

r � 3 sin �

75.

Cardioid

Symmetric to -axis, � ��

2y

0321

π2

r � 2�1 � sin ��

72.

Horizontal tangents:

��0.423, 0.072��r, �� � �1.894, 0.776�, �1.755, 2.594�, �1.998, �1.442�,

3−3

−2

2

r � 2 cos�3� � 2�

74.

Circle:

Center:

Tangent at pole: � ��

2

�32

, 0�

r �32

�x �32�

2

� y2 �94

x2 � y2 � 3x

r2 � 3r cos �

1 2 40

π2 r � 3 cos �

76.

Cardioid

Symmetric to polar axis since is a function of cos �.r

10

π2

r � 3�1 � cos ��

0

r 0 3 692

32

�2�

3�

2�

3�

416 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

77.

Rose curve with three petals

Symmetric to the polar axis

Relative extrema:

Tangents at the pole: � ��

6,

2,

5�

6

�2, 0�, ��2, �

3�, �2, 2�

3 �0

2

π2

r � 2 cos�3��

0

r 2 0 0 2 0 �2�2��2

�5�

62�

3�

2�

3�

4�

6�

78.

Rose curve with five petals

Symmetric to

Relative extrema occur when

at

Tangents at the pole:

01

π2

5,

2�

5,

3�

5,

4�

5� � 0,

� ��

10,

3�

10,

5�

10,

7�

10,

9�

10.

drd�

� �5 cos�5�� � 0

� ��

2

r � �sin�5��

80.

Rose curve with four petals

Symmetric to the polar axis, and pole

Relative extrema:

Tangents at the pole:

and given the same tangents.7�

4� �

5�

4

� ��

4,

3�

4

�3, 0�, ��3, �

2�, �3, ��, ��3, 3�

2 �

� ��

2,

20

π2

r � 3 cos 2�

79.

Rose curve with four petals

Symmetric to the polar axis, and pole

Relative extrema:

Tangents at the pole:

give the same tangents.

03

π2

�3��2�� � �,

� � 0, �

2

�±3, �

4�, �±3, 5�

4 �

� ��

2,

r � 3 sin 2�

81.

Circle radius: 5

x2 � y2 � 250

642

π2

r � 5 82.

Circle radius: 2

x2 � y2 � 4

10

π2

r � 2

Section 10.4 Polar Coordinates and Polar Graphs 417

83.

Cardioid

0642 10

π2

r � 4�1 � cos �� 84.

Cardioid

1 20

π2r � 1 � sin �

87.

Horizontal line

021

π2

y � 3

r sin � � 3

r � 3 csc �

85.

Limaçon

Symmetric to polar axis0

2

π2

r � 3 � 2 cos �

0

r 1 2 3 4 5

�2�

3�

2�

3�

86.

Limaçon

Symmetric to � ��

2

02 4

π2r � 5 � 4 sin �

0

r 9 7 5 3 1

2�

6�

6�

2�

88.

Line

01

π2

2y � 3x � 6

2r sin � � 3r cos � � 6

r �6

2 sin � � 3 cos �

89.

Spiral of Archimedes

Symmetric to

Tangent at the pole:

021 3

π2

� � 0

� ��

2

r � 2�

0

r 0 3�5�

22�

3�

2�

2

3�

25�

4�

3�

4�

2�

4�

90.

Hyperbolic spiral

10

π2

r �1�

r2

3�

45�

1�

43�

2�

4�

3�

25�

4�

3�

4�

2�

4�

418 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

91.

Lemniscate

Symmetric to the polar axis, and pole

Relative extrema:

Tangents at the pole: � ��

4,

3�

4

�±2, 0�

� ��

2,

0 ≤ � ≤ 2�r � 2�cos 2�,

01

π2

r2 � 4 cos�2��

0

r 0±�2±2

4�

6�

93. Since

the graph has polar axis symmetry and the tangentsat the pole are

Furthermore,

as

as

Also,

Thus, as x ⇒ �1.r ⇒ ±�

r �2x

1 � x.

rx � 2x � r

r � 2 �1

cos �� 2 �

rr cos �

� 2 �rx

� ⇒ ��

2�.r ⇒ �

� ⇒ �

2�r ⇒ ��

−6

−4

6

x = 1−4

� ��

3,

��

3.

r � 2 � sec � � 2 �1

cos �,

92.

Lemniscate

Symmetric to the polar axis, and pole

Relative extrema:

Tangent at the pole: � � 0

�±2, �

2�� �

2,

20

π2

r2 � 4 sin �

0

r 0 0±�2±2±�2

�5�

6�

2�

6�

94. Since

the graphs has symmetry with respect to Furthermore,

as

as

Also,

Thus, as

4−4

−2

y = 1

4

y ⇒ 1.r ⇒ ±�

r �2y

y � 1.

ry � 2y � r

r � 2 �1

sin �� 2 �

rsin �

� 2 �ry

� ⇒ ��.r ⇒ �

� ⇒ 0�r ⇒ �

� � ��2.

r � 2 � csc � � 2 �1

sin �,

Section 10.4 Polar Coordinates and Polar Graphs 419

95.

Hyperbolic spiral

as

−3 3

−1

y = 2

3

lim�→0

2 sin �

�� lim

�→0 2 cos �

1� 2

y �2 sin �

r �2�

⇒ � �2r

�2 sin �r sin �

�2 sin �

y

� ⇒ 0r ⇒ �

r �2�

96.

Strophoid

as

as

3−3

−2

x = 2− 2

lim�→±��2

�4 cos2 � � 2� � �2

x � 4 cos2 � � 2

r cos � � 4 cos2 � � 2

r � 2 cos 2� sec � � 2�2 cos2 � � 1� sec �

� ⇒ ���

2r ⇒ �

� ⇒ ��

2r ⇒ ��

r � 2 cos 2� sec �

97. The rectangular coordinate system consists of all points of the form where is the directeddistance from the -axis to the point, and is the directed distance from the -axis to the point.Every point has a unique representation.

The polar coordinate system uses to designate the location of a point.

is the directed distance to the origin and is the angle the point makes with the positive -axis,measured clockwise.

Points do not have a unique polar representation.

x�r

�r, ��

xyyx�x, y�

98.

x2 � y2 � r2, tan � �yx

x � r cos �, y � r sin �

100. Slope of tangent line to graph of at is

If and then is tangent at the pole.� � �f���� 0,f ��� � 0

dydx

�f ���cos � � f����sin �

�f ���sin � � f����cos �.

�r, ��r � f ���

99. circle

line� � b,

r � a,

101.

(a)

01 2

π2

0 ≤ � ≤ �

2

r � 4 sin �

(b)

01 2

π2

2 ≤ � ≤ � (c)

01 2

π2

��

2 ≤ � ≤

2

420 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

102.

(a)

(b)

The graph of is rotated through theangle ��4.

r � 61 � cos �

15−9

−6

12

� ��

4, r � 6�1 � cos�� �

4��

15−9

−9

9

� � 0, r � 61 � cos �

r � 61 � cos�� � ��

(c)

The graph of is rotated through theangle ��2.

r � 61 � cos �

12−12

−3

15

� 61 � sin �

� 6�1 � cos � cos �

2� sin � sin

2�

r � 6�1 � cos�� ��

2��

� ��

2

104. (a)

(c)

r � f �sin�� �3�

2 �� � f �cos ��

� cos �

sin�� �3�

2 � � sin � cos�3�

2 � � cos � sin�3�

2 � � f ��cos ��

r � f �sin�� ��

2�� � �cos �

sin�� ��

2� � sin � cos��

2� � cos � sin��

2� (b)

� f ��sin ��

r � f sin�� � ��

� �sin �

sin�� � �� � sin � cos � � cos � sin �

103. Let the curve be rotated by to form the curve If is a point on then is on That is,

Letting or we see that

g��� � g��1 � �� � f ��1� � f �� � ��.

�1 � � � �,� � �1 � �,

g��1 � �� � r1 � f ��1�.

r � g���.�r1, �1 � ��r � f ���,�r1, �1�

φ

φ

θ

( , + )r

( , )r

0

θ

θ

π2

r � g���.�r � f ���

105.

(a)

—CONTINUED—

−6 6

−4

4

r � 2 � sin�� ��

4� � 2 ��22

�sin � � cos ��

r � 2 � sin �

(b)

−6 6

−4

4

r � 2 � sin�� ��

2� � 2 � ��cos �� � 2 � cos �

Section 10.4 Polar Coordinates and Polar Graphs 421

105. —CONTINUED—

(c)

−6 6

−4

4

r � 2 � sin�� � �� � 2 � ��sin �� � 2 � sin � (d)

−6 6

−4

4

r � 2 � sin�� �3�

2 � � 2 � cos �

106.

(a)

(c)

3−3

−2

2

r � 4 sin�� �2�

3 � cos�� �2�

3 �

3−3

−2

2

r � 4 sin�� ��

6� cos�� ��

6�r � 2 sin 2� � 4 sin � cos �

(b)

(d)

3−3

−2

2

r � 4 sin�� � �� cos�� � �� � 4 sin � cos �

3−3

−2

2

r � 4 sin�� ��

2� cos�� ��

2� � �4 sin � cos �

107. (a)

021

π2

r � 1 � sin � (b)

Rotate the graph of

through the angle ��4.

r � 1 � sin �0

21

π2

r � 1 � sin�� ��

4�

108. By Theorem 9.11, the slope of the tangent line through and is

This is equal to

Equating the expressions and cross-multiplying, you obtain

tan �ff�

�r

dr�d�.

f �cos2 � � sin2 �� � f� tan �cos2 � � sin2 ��

f cos2 � � f cos � sin � tan � f� sin � cos � � f� sin2 � tan � �f sin2 � � f sin � cos � tan � f� sin � cos � � f� cos2 � tan

� f cos � � f� sin ���cos � � sin � tan � � �sin � � cos � tan ���f sin � � f� cos ��

�sin � � cos � tan cos � � sin � tan

.tan�� � � �tan � � tan

1 � tan � tan

f cos � � f� sin ��f sin � � f� cos �

.

θ

P r= ( , )θ

A0

ψ

Tangentline

Polar curver f= ( )θ

Radial lineπ2

PA

422 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

109.

At is undefined

−6

−3

3

3

⇒ ��

2.tan � � �,

tan �r

dr�d��

2�1 � cos ��2 sin �

110.

At

2−8

−5

5

� arctan�2 � �2

�2 � � 1.178�� 67.5�

tan �1 � ��2�2�

�2�

2 � �2

�2.� �

3�

4,

tan �r

dr�d��

3�1 � cos ��3 sin �

112.

At

6−6

−4

4

� arctan��32 � � 0.7137�� 40.89�

tan �sin���3�

2 cos���3� ��32

.� ��

6,

tan �r

dr�d��

4 sin 2�

8 cos 2�111.

At

−3

−2

3

2

θ

ψ

� arctan�13� � 18.4

� ��

4, tan � �

13

cot�3�

4 � �13

.

tan �r

dr�d��

2 cos 3�

� 6 sin 3�� �

13

cot 3�

r � 2 cos 3�

113.

At

��

3, �60�

tan �

1 � ��12�

��32

� ��3.� �2�

3,

tan �rdrd�

6�1 � cos ��

�6 sin ��1 � cos ��2

�1 � cos �

�sin �

−20 22

−12

θ

ψ

16

r �6

1 � cos �� 6�1 � cos ���1 ⇒

drd�

�6 sin �

�1 � cos ��2

114. undefined

9−9

−6

6

⇒ ��

2tan �

rdr�d�

�50

115. True 116. True 117. True 118. True

Section 10.5 Area and Arc Length in Polar Coordinates

Section 10.5 Area and Arc Length in Polar Coordinates 423

1.

� 2 ��

��2 sin2 � d�

�12

��

��2 �2 sin ��2 d�

A �1

2 ��

� f ����2 d� 2.

�12

�5��4

3��4 �cos 2��2 d�

A �12

��

� f ����2 d�

3.

�12

�3��2

��2 �1 � sin ��2 d�

A �12

��

� f ����2 d� 4.

�12

��

0 �1 � cos 2��2 d�

A �12

��

� f ����2 d�

5. (a)

A � ��4�2 � 16�

02 4

π2

r � 8 sin � (b)

� 16� � 32�� �sin 2�

2 ���2

0

� 32 ���2

0 �1 � cos 2�� d�

� 64 ���2

0 sin2 � d�

A � 212 ���2

0 �8 sin ��2 d�

7. � 2�� �16

sin 6����6

0�

3 A � 2�1

2 ���6

0 �2 cos 3��2 d��

6. (a)

A � �32

2

�9�

4

02 4

π2

r � 3 cos � (b)

�9�

4 �

92�� �

sin 2�

2 ���2

0

�92

���2

0 �1 � cos 2�� d�

� 9 ���2

0 cos2 � d�

A � 212�

��2

0 �3 cos ��2 d�

8.

� 18��

4� �9�

2

� 18�� �sin 4�

4 ���4

0

� 36���4

0

1 � cos 4�

2 d�

A � 2�1

2���4

0

�6 sin 2��2 d�� � 36���4

0

sin2 2� d�

424 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

9.

�12�� �

14

sin 4����4

0�

8

A � 2�12

���4

0 �cos 2��2 d�� 10.

�12�� �

110

sin�10�����10

0�

20

A � 2�12

���10

0 �cos 5��2 d��

11.

� �32

� � 2 cos � �14

sin 2����2

���2�

3�

2

A � 2�12

���2

���2 �1 � sin ��2 d��

13.

−1 4

−2

2

� �3� � 4 sin � � sin 2���

2��3�

2� � 3�32

A � 212

���

2��3 �1 � 2 cos ��2 d��

15. The area inside the outer loop is

From the result of Exercise 13, the area between the loops is

A � 4� � 3�32 � 2� � 3�3

2 � � � 3�3.

2�12

�2��3

0 �1 � 2 cos ��2 d�� � �3� � 4 sin � � sin 2��

2��3

0�

4� � 3�32

.−1 4

−2

2

12.

� �32

� � 2 cos � �14

sin 2����2

0�

3� � 84

A � 2�12

���2

0 �1 � sin ��2 d��

14.

−8

−12

8

2

� �34� � 48 cos � � 9 sin 2����2

arcsin�2�3�� 1.7635

� ���2

arcsin�2�3��16 � 48 sin � � 361 � cos 2�

2 � d�

� ���2

arcsin�2�3��16 � 48 sin � � 36 sin2 �� d�

A � 2�12�

��2

arcsin�2�3��4 � 6 sin ��2 d��

16. Four times the area in Exercise 15, More specifically, we see that the area inside the outer loop is

The area inside the inner loop is

Thus, the area between the loops is �8� � 6�3 � � �4� � 6�3 � � 4� � 12�3.

2 12

��3��2

7��6 �2�1 � 2 sin ���2 d�� � 4� � 6�3. 4−4

−1

62�12

���2

���6 �2�1 � 2 sin ���2 d�� � ���2

���6 �4 � 16 sin � � 16 sin2 �� d� � 8� � 6�3.

A � 4�� � 3�3 �.

Section 10.5 Area and Arc Length in Polar Coordinates 425

17.

Solving simultaneously,

Replacing by and by in the first equationand solving,Both curves pass through the pole, and respectively.

Points of intersection: 1, �

2, 1, 3�

2 , �0, 0�

�0, 0�,�0, ��,� � 0.cos � � 1,�1 � cos � � 1 � cos �,

� � ���rr

� ��

2,

3�

2.

2 cos � � 0

1 � cos � � 1 � cos �

r � 1 � cos �

r � 1 � cos �

19.

Solving simultaneously,

Replacing r by and by in the first equation and solving,which has no solution. Both curves pass through the pole,

and respectively.

Points of intersection: 2 ��22

, 3�

4 , 2 ��22

, 7�

4 , �0, 0�

�0, ��2�,�0, ��,

sin � � cos � � 2,�1 � cos � � 1 � sin �,� � ���r

� �3�

4,

7�

4.

tan � � �1

cos � � �sin �

1 � cos � � 1 � sin �

r � 1 � sin �

r � 1 � cos �

18.

Solving simultaneously,

Replacing r by and by in the first equationand solving,

Both curves pass through the pole,and respectively.

Points of intersection: �3, 0�, �3, ��, �0, 0�

�0, ��2�,�0, 3��2�,� � ��2.

sin � � 1,�3�1 � sin �� � 3�1 � sin ��,� � ���r

� � 0, �.

2 sin � � 0

3�1 � sin �� � 3�1 � sin ��

r � 3�1 � sin ��

r � 3�1 � sin ��

20.

Solving simultaneously,

Both curves pass through the pole, (0, ), andrespectively.

Points of intersection: 12

, �

3, 12

, 5�

3 , �0, 0�

�0, ��2�,arccos 2�3

� ��

3,

5�

3.

cos � �12

2 � 3 cos � � cos �

r � cos �

r � 2 � 3 cos �

21.

Solving simultaneously,

Both curves pass through the pole, and respectively.

Points of intersection: 32

, �

6, 32

, 5�

6 , �0, 0�

�0, 0�,�0, arcsin 4�5�,

� ��

6,

5�

6.

sin � �12

4 � 5 sin � � 3 sin �

r � 3 sin �

r � 4 � 5 sin � 22.

Solving simultaneously,

Both curves pass through the pole, and respectively.

Points of intersection: 32

, �

3, 32

, 5�

3 , �0, 0�

�0, ��2�,�0, ��,

� ��

3,

5�

3.

cos � �12

1 � cos � � 3 cos �

r � 3 cos �

r � 1 � cos �

426 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

23.

Solving simultaneously, wehave

Points of intersection:

�2, 4�, ��2, �4�

��2 � 2, � � 4.

r � 2

01

π2

r ��

224.

Line of slope 1 passingthrough the pole and a circleof radius 2 centered at thepole.

Points of intersection:

2, �

4, �2, �

4

r � 2

01 3

π2

� ��

4

25.

is the equation of a rose curve with four petals and is symmetric to the polar axis,and the pole. Also, is the equation of a circle of radius 2 centered at the pole.

Solving simultaneously,

Therefore, the points of intersection for one petal are and By symmetry, the other points of intersection are and �2, 23��12�.�2, 19��12�,�2, 17��12�,�2, 11��12�, �2, 13��12�,�2, 7��12�,

�2, 5��12�.�2, ��12�

� ��

12,

5�

12.

2� ��

6,

5�

6

4 sin 2� � 2

r � 2� � ��2,r � 4 sin 2�

r � 2

01 3

π2

r � 4 sin 2�

26.

The graph of is a limaçon symmetric toand the graph of is the horizontal

line Therefore, there are two points of intersection. Solving simultaneously,

� � arcsin �17 � 32 � 0.596.

sin � ��3 ± �17

2

sin2 � � 3 sin � � 2 � 0

3 � sin � � 2 csc �

y � 2.r � 2 csc �� � ��2,

r � 3 � sin �

r � 2 csc �

01 2

π2r � 3 � sin � Points of intersection:

�3.56, 0.596�, �3.56, 2.545�

�17 � 32

, � � arcsin�17 � 32 ,

�17 � 32

, arcsin�17 � 32 ,

Section 10.5 Area and Arc Length in Polar Coordinates 427

27.

The graph of is a limaçon with an inner loop and is symmetricto the polar axis. The graph of is the vertical line Therefore, thereare four points of intersection. Solving simultaneously,

Points of intersection: ��0.581, ±2.607�, �2.581, ±1.376�

� � arccos�2 � �106 � 2.6068.

� � arccos�2 � �106 � 1.376

cos � ��2 ± �10

6

6 cos2 � � 4 cos � � 1 � 0

2 � 3 cos � �sec �

2

x � 1�2.r � �sec ���2�b > a�r � 2 � 3 cos �

r �sec �

2−4 8

−4

r = sec θ

r = 2 + 3 cos θ

24

r � 2 � 3 cos �

29.

Points of intersection:

The graphs reach the pole at different times ( values).

−4 5

−5

r = cos θ

θr = 2 3 sin−

1

�0, 0�, �0.935, 0.363�, �0.535, �1.006�

r � 2 � 3 sin �

r � cos �

28.

The graph of is a cardioid with polar axis symmetry. The graph of

is a parabola with focus at the pole, vertex and polar axis symmetry. Therefore,there are two points of intersection. Solving simultaneously,

Points of intersection: �3�2, 2� � arccos�1 � �2 �� � �4.243, 4.285��3�2, arccos �1 � �2 �� � �4.243, 1.998�,

� � arccos �1 � �2 �. cos � � 1 ± �2

�1 � cos ��2 � 2

3�1 � cos �� �6

1 � cos �

�3, ��,

r � 6��1 � cos ��

r � 3�1 � cos ��

r �6

1 � cos �5

5

−10

−5

r = 61 cos−

r = 3(1 cos )−

θ

θ

r � 3�1 � cos ��

30.

Points of intersection:

The graphs reach the pole at different times ( values).

6−6

−2

r = 4 sin

r = 2 (1 + sin )

θ

θ

6

�0, 0�, 4, �

2r � 2�1 � sin ��

r � 4 sin �

428 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

31. From Exercise 25, the points of intersection for one petal are and The area within one petal is

Total area � 44�

3� �3 �

16�

3� 4�3 �

43

�4� � 3�3�

� 8�� �14

sin 4����12

0� �2��5��12

��12�

4�

3� �3.

� 16 ���12

0 sin2�2�� d� � 2 �5��12

��12 d� (by symmetry of the petal) 6−6

−4

4r = 2 θr = 4 sin 2 � �12

���12

0 �4 sin 2��2 d� �

12

�5��12

��12 �2�2 d� �

12

���2

5��12 �4 sin 2��2 d�

�2, 5��12�.�2, ��12�

33.

9−9

−6

6 r = −3 + 2 sin θ

r = 3 − 2 sin θ

� 2�11� � 12 cos � � sin�2�����2

0� 11� � 24

A � 4�12

���2

0 �3 � 2 sin ��2 d��

35.

−6 6

−3

5

r = 2

r = 4 sin θ

�8�

3� 2�3 �

23

�4� � 3�3 �

� 16�12

� �14

sin�2�����6

0� �4����2

��6

A � 2�12

���6

0 �4 sin ��2 d� �

12

���2

��6 �2�2 d��

37.

�3a2�

2�

a2�

4�

5a2�

4

� a2 �32

� � 2 sin � �sin 2�

4 ��

0�

a2�

4

A � 2�12

��

0 �a�1 � cos ���2 d�� �

a2�

4

32.

(from Exercise 14)

7−7

−7

7

� 18 ���2

0 �1 � sin ��2 d� �

92

�3� � 8�

A � 4�12

���2

0 9�1 � sin ��2 d��

34. and intersect at and

−12

−8

12

8

�59�

2� 30�2 � 50.251

� 592 5�

4 � 30�22

�94 � 59

2 �

4 � 30�22

�94

� �592

� � 30 cos � �94

sin 2��5��4

��4

A � 2�12�

5��4

��4�5 � 3 sin ��2 d��� � 5��4.� � ��4

r � 5 � 3 cos �r � 5 � 3 sin �

36.

4−4

−4

4

� ��2 sin�2�� � 4 cos ����2

��6� 3�3

� ���2

��6 ��4 cos 2� � 4 sin �� d�

A � 2�12

���2

��6 �3 sin ��2 d� �

12

���2

��6 �2 � sin ��2 d��

Section 10.5 Area and Arc Length in Polar Coordinates 429

38.twice area and the lines

2aa0

=

= −

3

π3

πθ

θ

�2�a2

3� 2a2 ��

2�

3�

�34 � �

2�a2 � 3�3a2

6

�2�a2

3� 2a2 �� �

sin 2�

2 ���2

��3

�2�a2

3� 2a2 ���2

��3 �1 � cos 2�� d�

A � �a2 � �

3a2 � 2�12

���2

��3 �2a cos ��2 d��

� ��

3, � �

2.

between r � 2a cos �Area � Area of r � 2a cos � � Area of sector �

40.

a

a

r = a sin θ

r = a cos θ

0

π2

�18

a2� �14

a2

�12

a2��

4�

12�

�12

a2�� �sin 2�

2 �

��4

0

� a2���4

0

1 � cos 2�

2 d�

A � 2�12�

��4

0�a sin ��2 d��

tan � � 1, � � ��4

r � a cos �, r � a sin �

39.

0a 2a

π2

��a2

8�

a2

2 �3�

2�

3�

4� 2� �

a2

2�� � 2�

��a2

8�

a2

2 �32

� � 2 sin � �sin 2�

4 ��

��2

��a2

8�

a2

2 ��

��2 3

2� 2 cos � �

cos 2�

2 d�

A ��a2

8�

12

��

��2 �a�1 � cos ���2 d�

41. (a)

(b)

(c)

� 10�32

� � sin 2� �18

sin 4����2

0�

15�

2

� 10 ���2

0 1 � 2 cos 2� �

1 � cos 4�

2 d�

� 10 ���2

0 �1 � cos 2��2 d�

� 40 ���2

0 cos4 � d�

A � 412 ���2

0 ��6 cos2 ��2 � �4 cos2 ��2� d�

−6 6

−4

a = 4 a = 6

4

�x2 � y2�3�2 � ax2

r3 � ar2 cos2 �

r � a cos2 �

430 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

42. By symmetry, and

[Note: area of circle of radius a]A1 � A6 � A7 � A4 � �a2 �

� a2 ���4

��6 �4 sin2 � � 1� d� � a2 �� � sin 2����4

��6� a2�

12� 1 �

�32

A7 � 212 ���4

��6 ��2a sin ��2 � �a�2� d�

� a2�2� � sin 2����6

0�

�a2

12� a2�

3�

�32 �

�a2

12� a25�

12�

�32

� 2a2 ���6

0 �1 � cos 2�� d� � �a2����4

��6

A6 � 212 ���6

0 �2a sin ��2 d� � 21

2 ���4

��6 a2 d�

�5�a2

12� a2�2� � sin 2��

5��6�

5�a2

12� a2�

3�

�32 � a2�

12�

�32

�5�a2

12� 2a2 ��

5��6 �1 � cos 2�� d�

A5 �12

5�

6 a2 � 212�

5��6 �2a sin ��2 d�

2a

2a

a

a

0

=

=

=

= =

= −

=π3 π

5

4

66

3

r a= 2 sin

r = 2 r a= 2 cos

A5

A2

A3

A4

A1A6

A7

θ

θ

θ

θ θ

θ

θ

θ

π

π

π

θ

θ

2π A3 � A4 �

12

2a2 ��a2

4

�a2

2 �� � sin 2����6

���3� a2�sin 2����4

��6�

a2

2 �

2� �3 � a21 �

�32 � a2 �

4� 1

�a2

2 ���6

���3 �4 cos2 � � 1� d� � 2a2 ���4

��6 cos 2� d�

A1 � A2 �12

���6

���3 ��2a cos ��2 � �a�2� d� �

12

���4

��6 ��2a cos ��2 � �2a sin ��2� d�

A3 � A4.A1 � A2

43.

For

—CONTINUED—

0a

π2

A � �a2

2�

�a2

4

r � a cos �

n � 1:

r � a cos�n��

For

0a

π2

A � 812 ���4

0 �a cos 2��2 d� �

�a2

2

r � a cos 2�

n � 2:

Section 10.5 Area and Arc Length in Polar Coordinates 431

44.

� ���4

0 �sec2 � � 4 � 4 cos2 �� d� � ���4

0 �sec2 � � 4 � 2�1 � cos 2��� d� � �tan � � 2� � sin 2����4

0� 2 �

2

A � 212 ���4

0 �sec � � 2 cos ��2 d�

y 2 �x2�1 � x�

1 � x

y 2�x � 1� � �x2 � x3

�x2 � y 2�x � x2 � y 2 � 2x2

x � 1 � 2 r2 cos2 �

r2 � 1 � 2 x2

x2 � y 2 r cos � � 1 � 2 cos2 �

1

x

−1

y

r � sec � � 2 cos �, ��

2 < � <

2

43. —CONTINUED—

For

In general, the area of the region enclosed by for is if is odd and isif is even.n��a2��2

n��a2��4n � 1, 2, 3, . . .r � a cos�n��

0a

π2

A � 612 ���6

0 �a cos 3��2 d� �

�a2

4

r � a cos 3�

n � 3: For

0a

π2

A � 1612 ���8

0 �a cos 4��2 d� �

�a2

2

r � a cos 4�

n � 4:

45.

(circumference of circle of radius )a

s � �2�

0 �a2 � 02 d� � �a��2�

0� 2�a

r � 0

r � a

47.

� 4�2 ��2 � 0� � 8

� �4�2 �1 � sin ��3��2

��2

� 2�2 �3��2

��2

�cos ��1 � sin �

d�

� 2�2 �3��2

��2 �1 � sin � d�

s � 2 �3��2

��2 ��1 � sin ��2 � �cos ��2 d�

r � cos �

r � 1 � sin �

46.

� ���2

���22a d� � �2a����2

���2� 2�a

s � ���2

���2 ��2a cos ��2 � ��2a sin ��2 d�

r � �2a sin �

r � 2a cos �

432 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

48.

� 64

� �32�2�1 � cos ���

0

� 16�2��

0

sin ��1 � cos �

d�

� 16�2��

0

�1 � cos � �1 � cos ��1 � cos � d�

� 16�2��

0

�1 � cos � d�

� 16��

0

�1 � 2 cos � � cos2 � � sin2 � d�

s � 2��

0

��8�1 � cos ���2 � ��8 sin ��2 d�

r � �8 sin �

r � 8�1 � cos ��, 0 ≤ � ≤ 2�

49.

Length � 4.16

−1 2

−1

4

r � 2�, 0 ≤ � ≤ �

250.

Length exact �3 ��� 1.73

4−3

−3

3

r � sec �, 0 ≤ � ≤ �

351.

Length � 0.71

−0.5

−0.5

0.5

0.5

r �1�

, � ≤ � ≤ 2�

52.

Length � 31.31

5−25

−5

10

r � e�, 0 ≤ � ≤ � 53.

Length � 4.39

−1

−1

2

1

r � sin�3 cos ��, 0 ≤ � ≤ � 54.

Length � 7.78

4−2

−2

2

r � 2 sin�2 cos ��, 0 ≤ � ≤ �

55.

� 36�

� �36� sin2 ����2

0

� 72����2

0sin � cos � d�

S � 2����2

06 cos � sin ��36 cos2 � � 36 sin2� d�

r � �6 sin �

r � 6 cos � 56.

� ��a2� �sin 2�

2 ���2

0�

�2a2

2

� 2� a2 ���2

0 cos2 � d� � �a2 ���2

0 �1 � cos 2�� d�

S � 2� ���2

0 a cos � �cos ���a2 cos � � a2 sin2 � d�

r � �a sin �

r � a cos �

Section 10.5 Area and Arc Length in Polar Coordinates 433

57.

�2��1 � a2

4a2 � 1 �e�a � 2a�

� 2��1 � a2 � e2a�

4a2 � 1 �2a cos � � sin ���

��2

0

� 2��1 � a2 ���2

0 e2a� cos � d�

S � 2� ���2

0 ea� cos ���ea��2 � �aea��2 d�

r � aea�

r � ea�

58.

� �2�2�a2 ��

0 �1 � cos ��3�2��sin �� d� � �

4�2�a2

5 ��1 � cos ��5�2�

0�

32�a2

5

S � 2� ��

0 a�1 � cos �� sin ��a2�1 � cos ��2 � a2 sin2 � d� � 2�a2 ��

0 sin ��1 � cos ���2 � 2 cos � d�

r � �a sin�

r � a�1 � cos ��

60.

S � 2� ��

0 � sin ���2 � 1 d� � 42.32

r � 1

r � �

62. The curves might intersect for different values of

See page 741.

�:

59.

� 32� ���4

0 cos 2� sin ��cos2 2� � 4 sin2 2� d� � 21.87

S � 2����4

4 cos 2� sin ��16 cos2 2� � 64 sin2 � 2� d�

r � �8 sin 2�

r � 4 cos 2�

61.

Arc length � ��

�f ���2 � f���2 d� � ��

��r2 � dr

d�2 d�

Area �12�

� f ����2 d� �12�

r2 d�

63. (a) is correct: s � 33.124. 64. (a)

(b) S � 2���

f ��� cos ��f ���2 � f���2 d�

S � 2���

f ��� sin ��f ���2 � f���2 d�

65. Revolve about the line

� 40�2

� 4��5� � 2 sin ��2�

0

� 4� �2�

0 �5 � 2 cos �� d�

S � 2� �2�

0 �5 � 2 cos ���22 � 02 d�

f ��� � 2, f��� � 0

0

π2

5

r = 2

r = 5sec�

�5 − 2cos

r � 5 sec �.r � 2

434 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

66. Revolve about the line where

� 2�a�2�b� � 4� 2ab

� 2�a�b� � a sin ��2�

0

S � 2� �2�

0 �b � a cos ���a2 � 02 d�

f��� � 0

f ��� � a

0

a

a

2a

b

π2

b > a > 0.r � b sec �r � a

67.

(a)

(Area circle )

(b)

� �r2 � �42 � 16�

� 16�� �sin 2�

2 ��

0� 16 � A �

12

��

0r2 d� �

12

��

0 64 cos2 � d� � 32 ��

0 1 � cos 2�

2 d�

r � 8 cos �, 0 ≤ � ≤ �

0.2 0.4 0.6 0.8 1.0 1.2 1.4

A 6.32 12.14 17.06 20.80 23.27 24.60 25.08

�(c), (d) For of area

For of area

For of area

(e) No, it does not depend on the radius.

�12� � 37.70�: 2.7334

���2��8� � 25.13�: 1.5712

�4� � 12.57�: 0.4214

68.

(a)

Note: radius of circle is

(b)

(c), (d) For of area

For of area

For of area �12 94 � � 3.5343�: � � ��2 � 1.571

2

�14 94 � � 1.7671�: � � 1.151

4

�18 94 � � 0.8836�: � � 0.881

8

32

⇒ A � �32

2��

�94�� �

12

sin 2���

0�

94

� A �12

��

0r2 d� �

92

��

0 sin2 � d� �

94

��

0 �1 � cos 2�� d�

r � 3 sin �, 0 ≤ � ≤ �

0.2 0.4 0.6 0.8 1.0 1.2 1.4

A 0.0119 0.0930 0.3015 0.6755 1.2270 1.9401 2.7731

69.

x2 � y2 � bx � ay � 0 represents a circle.

x2 � y2 � ay � bx

r2 � ar sin � � br cos �

r � a sin � � b cos �

Section 10.5 Area and Arc Length in Polar Coordinates 435

70. Circle

�12

�� � sin2 ���

0 ��

2

�12

��

0 �1 � 2 sin � cos �� d�

�12

��

0 �sin � � cos ��2 d�

A �12

��

r2 d�

r � sin � � cos �, Converting to rectangular form:

Circle of radius and center

Area � �� 1�2�

2

��

2

�12

, 12�

1�2

�x �12�

2

� �y �12�

2

�12

�x2 � x �14� � � y2 � y �

14� �

12

x2 � y2 � y � x

r2 � r sin � � r cos �

71. (a)

As increases, the spiral opens more rapidly. If the spiral is reflected aboutthe -axis.

(b) crosses the polar axis for and integer. To see this

for The points are

(c)

�12

ln��4�2 � 1 � 2�� � ��4�2 � 1 21.2563

s � �2�

0 ��2 � 1 d� �

12

�ln��x2 � 1 � x� � x�x2 � 1�2�

0

f ��� � �, f���� � 1

n � 1, 2, 3, . . ..�r, �� � �an�, n��,� � n�.

r � a� ⇒ r sin � � y � a� sin � � 0

n� � n�,� ≥ 0,r � a�,

y� < 0,a

12

14

−12

−10

r � �, � ≥ 0

72.

9.3655

�32

e2�3 �32

�32

�32

e�2�3 �32

�e2�3 � e�2�3 � 2�

� �32

e�3�2�

0� �3

2e�3� 0

�2�

�12

�2�

0 e�3 d� �

12

�0

�2�

e�3 d�

A �12

�2�

0 �e�6�2 d� �

12

�0

�2�

�e�6�2 d�

1 2 30

π2

= −2

= 2

π

π = 0

−3

−2

2�

��

r � e�6

(d)

��3

6 �2�

0�

43

�3

�12

�2�

0 �2 d�

A �12

��

r2 dr

73. The smaller circle has equation

The area of the shaded lune is:

�a2

2 ��

4�

12� �

4

� �a2

2 �� �sin 2�

2 � � ���4

0

� ��4

0 �a2

2�1 � cos 2�� � 1� d�

A � 2�12��

�4

0 ��a cos ��2 � 1� d�

y

x

r = 1 π4

r = a cos θ

22

r � a cos �. This equals the area of the square,

Smaller circle: r � �2 cos �

a � �2

a2 �4 � 2�

2 � �� 2

�a2 � 2a2 � 2� � 4 � 0

a2

2 ��

4�

12� �

4�

12

��22 �

2

�12

.

436 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

Section 10.6 Polar Equations of Conics and Kepler’s Laws

76. False. and have only one point ofintersection.

g��� � sin 2�f ��� � 075. False. and have the same graphs.g��� � �1f ��� � 1

77. In parametric form,

Using instead of we have and Thus,

and

It follows that

Therefore, s � ��

�� f ����2 � � f�����2 d�.

�dxd��

2

� �dyd��

2

� � f ����2 � � f�����2.

dyd�

� f���� sin � � f ��� cos �.dxd�

� f���� cos � � f ��� sin �

y � r sin � � f ��� sin �.x � r cos � � f ��� cos �t,�

s � �b

a

��dxdt�

2

� �dydt�

2

dt.

74.

(a) (b) (c)

Hence,

r �3 cos � sin �

cos3 � � sin3 �

�r cos ��3 � �r sin ��3 � 3�r cos ���r sin ��

x3 � y3 � 3xy.

3xy �27t3

�1 � t3�2

A �12�

�2

0 r2 d� �

32

2

3

−2

−3

x3 � y3 �27�t3 � t6��1 � t3�3 �

27t3

�1 � t3�2

x �3t

1 � t3, y �

3t2

1 � t3

1.

(a) parabola

(b) ellipse

(c) hyperbolae � 1.5, r �3

1 � 1.5 cos ��

62 � 3 cos �

,

e � 0.5, r �1

1 � 0.5 cos ��

22 � cos �

,

e � 1, r �2

1 � cos �, −4 8

−4

e = 1.5e = 0.5

e = 1.0

4r �2e

1 � e cos �

2.

(a) parabola

(b) ellipse

(c) hyperbolae � 1.5, r �3

1 � 1.5 cos ��

62 � 3 cos �

,

e � 0.5, r �1

1 � 0.5 cos ��

22 � cos �

,

e � 1, r �2

1 � cos �, 3−9

−4

e = 1.5

e = 0.5

e = 1

4r �2e

1 � e cos �

Section 10.6 Polar Equations of Conics and Kepler’s Laws 437

3.

(a) parabola

(b) ellipse

(c) hyperbolae � 1.5, r �3

1 � 1.5 sin ��

62 � 3 sin �

,

e � 0.5, r �1

1 � 0.5 sin ��

22 � sin �

,

e � 1, r �2

1 � sin �, −9 9

−8

e = 1.5

e = 1.0 e = 0.54

r �2e

1 � e sin �

4.

(a) parabola

(b) ellipse

(c) hyperbolae � 1.5, r �3

1 � 1.5 sin ��

62 � 3 sin �

,

e � 0.5, r �1

1 � 0.5 sin ��

22 � sin �

,

e � 1, r �2

1 � sin �,

9−9

−3

e = 1.5

e = 0.5

e = 1

9r �

2e1 � e sin �

6.

(a) Because the conic is an ellipse with vertical directrix to the left of the pole.

(b)

The ellipse is shifted to the left. The vertical directrix is tothe right of the pole

The ellipse has a horizontal directrix below the pole.

r �4

1 � 0.4 sin �.

r �4

1 � 0.4 cos �

e � 0.4 < 1,

r �4

1 � 0.4 cos �

(c)

8−8

−5

9

10−10

−7

7

10. Parabola; Matches (e) 11. Ellipse; Matches (b) 12. Hyperbola; Matches (d)

7. Parabola; Matches (c) 8. Ellipse; Matches (f) 9. Hyperbola; Matches (a)

5.

(a) The conic is an ellipse. As the ellipsebecomes more elliptical,and as it becomesmore circular.

e → 0�,

e → 1�,−30

−40

30

e = 0.1

e = 0.75

e = 0.9

e = 0.5

e = 0.25

5

r �4

1 � e sin �

(c) The conic is a hyper-bola. As thehyperbolas opensmore slowly, and as

they openmore rapidly.e →,

e → 1�,

−90

−40

90

e = 1.1

e = 1.5

e = 2

80

(b) The conic is a parabola.−30

−40

30

5

438 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

13.

Parabola because

Distance from pole to directrix:

Directrix:

Vertex:

021

π2�r, �� � ��

12

, 3�

2 �y � 1

d � 1

d � �1e � 1;

r ��1

1 � sin �14.

Parabola because

Directrix:

Distance from pole to directrix:

Vertex:

04�4�8 8

π2�r, �� � �3, 0�

d � 6

x � 6

d � 6e � 1;

r �6

1 � cos �

15.

Ellipse because

Directrix:

Distance from pole to directrix:

Vertices:

01 3

π2

�r, �� � �2, 0�, �6, ��d � 6

x � 6

d � 6e � 12;

r �6

2 � cos ��

31 � �12� cos �

16.

Ellipse because

Directrix:

Distance from pole to directrix:

Vertices:

02−2

�r, �� � �58, �2�, �52, 3�2�d � 53

y � 53

d � 53e � 35 < 1;

r �5

5 � 3 sin ��

11 � �35� sin �

17.

Ellipse because

Directrix:

Distance from pole to directrix:

Vertices:

01 43

π2

�r, �� � �43, �2�, �4, 3�2�d � 4

y � 4

d � 4e � 12;

r �4

2 � sin ��

21 � �12� sin �

r�2 � sin �� � 4 18.

Ellipse because

Directrix:

Distance from pole to drectrix:

Vertices:

02−2 4

�r, �� � �6, 0�, �65, ��d � 3

x � �3

d � 3e � 23 < 1;

r �6

3 � 2 cos ��

21 � �23� cos �

r�3 � 2 cos �� � 6

Section 10.6 Polar Equations of Conics and Kepler’s Laws 439

19.

Hyperbola because

Directrix:

Distance from pole to directrix:

Vertices:

04 6 8

π2

�r, �� � �5, 0�, ��53, ��d � 52

x � 52

d � �52e � 2 > 1;

r �5

�1 � 2 cos ��

�51 � 2 cos �

20.

Hyperbola because

Directrix:

Distance from pole to directrix:

Vertices:

2−20

�r, �� � ��35, �2�, �32, 3�2�d � 67

y � �67

d � �67e � 73 > 1;

r ��6

3 � 7 sin ��

�21 � �73� sin �

21.

Hyperbola because

Directrix:

Distance from pole to directrix:

Vertices:

01

π2

�r, �� � �38, �2�, ��34, 3�2�d � 12

y � 12

d � 12e � 3 > 0;

r �3

2 � 6 sin ��

321 � 3 sin �

22.

Hyperbola because

Directrix:

Distance from pole to directrix:

Vertices:

0

π2

21 3 5

�r, �� � �43, 0�, ��4, ��d � 2

x � 2

d � 2e � 2 > 1;

r �4

1 � 2 cos �

27.

Rotate the graph of

counterclockwise through the angle �

4.

r ��1

1 � sin �

−2 10

−6

2r ��1

1 � sin�� ��

4�28.

Rotate the graph of

counterclockwise through the angle �

3.

r �6

1 � cos �

10−14

−10

6r �6

1 � cos�� ��

3�

23.

Ellipse

−2 2

−2

1

r � 3��4 � 2 sin �� 24.

Hyperbola

−6 6

−4

4

r � �3�2 � 4 sin �� 25.

Parabola

−2 1

−2

2

r � �1�1 � cos �� 26.

Hyperbola

−6 6

−4

4

r � 2�2 � 3 sin ��

440 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

29.

Rotate the graph of

clockwise through the angle �

6.

r �6

2 � cos �

−8 4

−3

5r �6

2 � cos�� ��

6�30.

Rotate graph of

clockwise through angle of 2�3.

r ��6

3 � 7 sin �

−6

−4

6

4r ��6

3 � 7 sin�� � �2�3��

31. Change to r �5

5 � 3 cos�� ��

4�� �

4:� 32. Change to r �

2

1 � sin�� ��

6�� �

6:�

33. Parabola

r �ed

1 � e cos ��

11 � cos �

e � 1, x � �1, d � 1

34. Parabola

r �ed

1 � e sin ��

11 � sin �

e � 1, y � 1, d � 1

35. Ellipse

�1

2 � sin�

�12

1 � �12� sin�

r �ed

1 � e sin �

e �12

, y � 1, d � 1

36. Ellipse

�6

4 � 3 sin�

�2�34�

1 � �34� sin�

r �ed

1 � e sin �

e �34

, y � �2, d � 2

37. Hyperbola

r �ed

1 � e cos��

21 � 2 cos�

e � 2, x � 1, d � 1

38. Hyperbola

�3

2 � 3 cos�

�32

1 � �32� cos�

r �ed

1 � e cos �

e �32

, x � �1, d � 1

39. Parabola

Vertex:

e � 1, d � 2, r �2

1 � sin�

�1, ��

2�40. Parabola

Vertex:

r �ed

1 � e cos��

101 � cos�

e � 1, d � 10

�5, ��

41. Ellipse

Vertices:

�16

5 � 3 cos�

�165

1 � �35� cos�

r �ed

1 � e cos�

e �35

, d �163

�2, 0�, �8, ��

Section 10.6 Polar Equations of Conics and Kepler’s Laws 441

42. Ellipse

Vertices:

�8

3 � sin�

�83

1 � �13� sin�

r �ed

1 � e sin�

e �13

, d � 8

�2, �

2�, �4, 3�

2 �43. Hyperbola

Vertices:

�9

4 � 5 sin�

�94

1 � �54� sin�

r �ed

1 � e sin �

e �54

, d �95

�1, 3�

2 �, �9, 3�

2 �44. Hyperbola

Vertices:

�10

2 � 3 cos�

�5

1 � �32� cos�

r �ed

1 � e cos�

e �32

, d �103

�2, 0�, �10, 0�

45. Ellipse if parabola if hyperbola if e > 1.e � 1,0 < e < 1,

46. is a parabola with horizontal directrix above the pole.

(a) Parabola with vertical directrix to left of pole. (b) Parabola with horizontal directrix below pole.

(c) Parabola with vertical directrix to right of pole. (d) Parabola (b) rotated counterclockwise �4.

r �4

1 � sin �

47. (a) Hyperbola

(b) Ellipse

(c) Parabola

(d) Rotated hyperbola �e � 3�

�e � 1�

�e �1

10< 1�

�e � 2 > 1� 48. If the foci are fixed and then To see this,compare the ellipses

d � 54.e � 14,r �516

1 � �14� cos �,

d � 1e � 12,r �12

1 � �12� cos �,

d →.e → 0,

49.

�b2

1 � �ca�2 cos2 ��

b2

1 � e2 cos2�

r2 �a2b2

a2 � �b2 � a2� cos2 ��

a2b2

a2 � c2 cos2 �

r2�a2 � cos2 ��b2 � a2�� � a2b2

r2�b2 cos2 � � a2�1 � cos2 ��� � a2b2

b2r2 cos2 � � a2r2 sin2 � � a2b2

x2b2 � y2a2 � a2b2

x2

a2 �y2

b2 � 1 50.

��b2

1 � e2 cos2 �

r2 �a2b2

�a2 � c2 cos2 ��

b2

�1 � �c2a2� cos2 �

r2��a2 � cos2��a2 � b2�� � a2b2

r2�b2 cos2� � a2�1 � cos2 �� � a2b2

b2r2 cos2 � � a2r2 sin2 � � a2b2

x2b2 � y2a2 � a2b2

x2

a2 �y2

b2 � 1

51.

r2 �9

1 � �1625� cos2 �

a � 5, c � 4, e �45

, b � 3 52.

r2 ��9

1 � �2516� cos2 �

a � 4, c � 5, b � 3, e �54

53.

r2 ��16

1 � �259� cos2 �

a � 3, b � 4, c � 5, e �53

54.

r2 �1

1 � �34� cos2 �

a � 2, b � 1, c � �3, e ��32 55.

� 9 ��

0

1�2 � cos��2 d� � 10.88

A � 2�12

��

0 � 3

2 � cos��2

d�

442 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

56. A � 2�12

��2

��2 � 2

3 � 2 sin��2

d� � 4 ��2

��2

1�3 � 2 sin��2 d� � 3.37

57. Vertices:

When

Distance between earth and the satellite is miles.r � 4000 � 11,015

r � 15,015.� � 60� ��

3,

r �7979.21

1 � 0.93717 cos ��

1,046,226,000131,119 � 122,881 cos �

d �131,119�1 � e2�

2e� 8514.1397

131,119 � d� e1 � e

�e

1 � e� � d� 2e1 � e2�

2a � 2�65,559.5� �ed

1 � e�

ed1 � e

r �ed

1 � e� � �:r �

ed1 � e

,� � 0:

r �ed

1 � e cos �

e �ca

�122,881131,119

� 0.93717

c � 65,559.5 � 4119 � 61,440.5

a �127,000 � 4119

2� 65,559.5

�119 � 4000, �� � �4119, ��

(123,000 � 4000, 0� � �127,000, 0� 58. (a)

When

Therefore,

Thus,

(b) The perihelion distance is

When

The aphelion distance is

When r ��1 � e2�a

1 � e� a�1 � e�.� � 0,

a � c � a � ea � a�1 � e�.

r ��1 � e2�a

1 � e� a�1 � e�.� � �,

a � c � a � ea � a�1 � e�.

r ��1 � e2�a

1 � e cos�.

a�1 � e2� � ed.

a�1 � e��1 � e� � ed

a�1 � e� �ed

1 � e

� � 0, r � c � a � ea � a � a�1 � e�.

r �ed

1 � e cos�

59.

Perihelion distance:

Aphelion distance: a�1 � e� � 152,098,320 km

a�1 � e� � 147,101,680 km

r ��1 � e2�a

1 � e cos ��

149,558,278.11 � 0.0167 cos �

e � 0.0167a � 1.496 � 108, 60.

Perihelion distance:

Aphelion distance: a�1 � e� � 1,504,343,400 km

a�1 � e� � 1,349,656,600 km

r ��1 � e2�a

1 � e cos ��

1,422,807,9881 � 0.0542 cos �

e � 0.0542a � 1.427 � 109,

61.

Perihelion distance:

Aphelion distance: a�1 � e� � 7,375,412,800 km

a�1 � e� � 4,436,587,200 km

r ��1 � e2�a

1 � e cos ��

5,540,410,0951 � 0.2488 cos �

e � 0.2488a � 5.906 � 109, 62.

Perihelion distance

Aphelion distance � a�1 � e� � 69,816,296 km

� a�1 � e� � 46,003,704 km

r ��1 � e2�a

1 � e cos ��

55,462,065.541 � 0.2056 cos �

e � 0.2056a � 5.791 � 107,

Section 10.6 Polar Equations of Conics and Kepler’s Laws 443

63.

(a)

(b)

By trial and error,

because the rays in part (a) are longer than those in part (b).0.9018 > �9 � 0.3491

� � � � 0.9018

12�

r2 d� � 9.368 � 108

248�12�

�9

0 r2 d�

12�

2�

0 r2 d� � 21.89 yrs

A �12�

�9

0 r2 d� � 9.368 � 1018 km2

r �5.540 � 109

1 � 0.2488 cos �

(c) For part (a)

Average per year

For part (b)

Average per year �4.133 � 109

21.89� 1.89 � 108 kmyr

s � ���0.9018

�r2 � �drd��2 d� � 4.133 � 109

�2.563 � 109

21.89� 1.17 � 108 kmyr

s � ��9

0

�r2 � �drd��2 d� � 2.563 � 109 km

64.

(a)

length of minor axis: 2b � 24.97 au

b2 � a2 � c2 ⇒ b � 12.4844

e �ca

⇒ c � 124.375

e � 0.995a �12

�250� � 125 au,

(b)

(c) Perihelion distance:

Aphelion distance: a�1 � e� � 249.375 au

a�1 � e� � 0.625 au

r ��1 � e2�a

1 � e cos ��

1.2468751 � 0.995 cos �

65.

1 � e1 � e

1 �ca

1 �ca

�a � ca � c

�r1

r0

e �ca

�r1 � r0

r1 � r0

r1 � r0 � 2ar1 � r0 � 2c,r0 � a � c,r1 � a � c, 66. For a hyperbola,

and

Thus and

e � 1e � 1

�ca � 1ca � 1

�c � ac � a

�r1

r0

e �ca

�r1 � r0

r1 � r0

r1 � r0 � 2a.r1 � r0 � 2c

r1 � c � a.r0 � c � a

67. and

Points of intersection:

At At

At At

Therefore, at we have and at we have The curves intersect at right angles.

m1m2 � 1��1� � �1.�ed, ��m1m2 � ��1��1� � �1,�ed, 0�

�ed, ��, dydx

� �1.�ed, 0�, dydx

� 1.

dydx

�� ed

1 � sin���cos�� � � ed cos�

�1 � sin��2��sin��

� �ed1 � sin���sin�� � � ed cos�

�1 � sin��2��cos��r2:

�ed, ��, dydx

� 1.�ed, 0�, dydx

� �1.

dydx

�� ed

1 � sin���cos�� � � �ed cos�

�1 � sin��2��sin��

� �ed1 � sin���sin�� � � �ed cos�

�1 � sin��2��cos��r1:

�ed, 0�, �ed, ��

r2 �ed

1 � sin�r1 �

ed1 � sin�

Review Exercises for Chapter 10

1.

Ellipse

Vertex:

Matches (e)

�1, 0�.

4x2 � y2 � 4 2.

Hyperbola

Vertex:

Matches (c)

�1, 0�

4x2 � y2 � 4 3.

Parabola opening to left.

Matches (b)

y2 � �4x

4.

Hyperbola

Vertex:

Matches (d)

�0, 2�

y2 � 4x2 � 4 5.

Ellipse

Vertex:

Matches (a)

�0, 1�

x2 � 4y2 � 4 6.

Parabola opening upward.

Matches (f )

x2 � 4y

7.

Circle

Center:

Radius: 1

�12

, 34�

21

1

1

2

x

31,,

42

y

�x �12�

2

� �y �34�

2

� 1

�x2 � x �14� � �y2 �

32

y �9

16� �3

16�

14

�9

16

16x2 � 16y2 � 16x � 24y � 3 � 0 8.

Parabola

Vertex: ��2, 6�

x

12

16

8 12−4

y

�y � 6�2 � 4�2��x � 2� y2 � 12y � 36 � 8x � 20 � 36

y2 � 12y � 8x � 20 � 0

444 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

68. (Parabolas)

To find the intersection points:

For the first parabola,

Similarly for the second parabola,

Since the product of the slopes is they intersect at right angles.�1,

dydx

�sin �

1 � cos �.

dydx

�r1 cos � � r�1 sin �

�r1 sin � � r�1 cos ��

c � cos ��1 � cos �� � c � sin2 ��c � sin ��1 � cos �� � c � sin � cos �

�1 � cos �

�sin �.

dr2

d��

�d � sin ��1 � cos ��2

dr1

d��

c � sin ��1 � cos ��2,

r1 �c

1 � �c � dc � d�

�c�c � d�

2c�

c � d2

� r2

cos � �c � dc � d

c � c � cos � � d � d � cos �

c

1 � cos ��

d1 � cos �

r2 �d

1 � cos �r1 �

c1 � cos �

,

Review Exercises for Chapter 10 445

9.

Hyperbola

Center:

Vertices:

Asymptotes:

y � 3 ± �32

�x � 4�

��4 ± �2, 3���4, 3� 6

4

2

246x

y

�x � 4�2

2�

�y � 3�2

3� 1

3�x2 � 8x � 16� � 2�y2 � 6y � 9� � �24 � 48 � 18

3x2 � 2y2 � 24x � 12y � 24 � 0 10.

Ellipse

Center:

Vertices: �2, ±1��2, 0�

2

−2

31

1

−1

x

(2, 0)

y �x � 2�2

1�4�

y2

1� 1

4�x2 � 4x � 4� � y2 � �15 � 16

4x2 � y2 � 16x � 15 � 0

11.

Ellipse

Center:

Vertices: �2, �3 ±�22 �

�2, �3�

321

1

1

2

3

4

x

(2, −3)

y �x � 2�2

1�3�

�y � 3�2

1�2� 1

3�x2 � 4x � 4� � 2�y2 � 6y � 9� � �29 � 12 � 18

3x2 � 2y2 � 12x � 12y � 29 � 0

12.

Hyperbola

Center:

Vertices:

Asymptotes: y � 1 ± �x �12�

�12

± �2, 1�

�12

, 1�

−2 1 3 4−1

−2

1

3

4

x

y

�x � �1�2��2

2�

�y � 1�2

2� 1

4�x2 � x �14� � 4�y2 � 2y � 1� � 11 � 1 � 4

4x2 � 4y2 � 4x � 8y � 11 � 0 13. Vertex:

Directrix:

Parabola opens to the right.

y2 � 4y � 12x � 4 � 0

�y � 2�2 � 4�3��x � 0�p � 3

x � �3

�0, 2�

14. Vertex:

Focus:

Parabola opens downward.

x2 � 8x � 8y � 0

�x � 4�2 � 4��2��y � 2�p � �2

�4, 0�

�4, 2� 15. Vertices:

Foci:

Horizontal major axis

Center:

�x � 2�2

25�

y2

21� 1

a � 5, c � 2, b � �21

�2, 0�

�0, 0�, �4, 0�

��3, 0�, �7, 0�

446 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

16. Center:

Solution points:

Substituting the values of the coordinates of the givenpoints into

we obtain the system

Solving the system, we have

and �x2

4 � � �3y2

16 � � 1.b2 � 4,a2 �163

4b2 � 1.� 1

b2� � � 4a2� � 1,

�x2

b2� � � y2

a2� � 1,

�1, 2�, �2, 0�

�0, 0� 17. Vertices:

Foci:

Center:

Horizontal transverse axis

x2

16�

y2

20� 1

a � 4, c � 6, b � �36 � 16 � 2�5

�0, 0��±6, 0�

�±4, 0�

18. Foci:

Asymptotes:

Center:

Vertical transverse axis

asymptote

y2

1024�17�

x2

64�17� 1

⇒ b2 �6417

⇒ a2 �102417

b2 � c2 � a2 � 64 � �4b�2 ⇒ 17b2 � 64

→ a � 4by �ab

x � 4x

c � 8

�0, 0�

y � ±4x

�0, ±8� 19.

By Example 5 of Section 10.1,

C � 12 ��2

0 �1 � �5

9� sin2 � d� 15.87.

x2

9�

y2

4� 1, a � 3, b � 2, c � �5, e �

�53

20.

By Example 5 of Section 10.1,

C � 20 ��2

0 �1 �

2125

sin2 � d� 23.01.

x2

4�

y2

25� 1, a � 5, b � 2, c � �21, e �

�215

21. has a slope of 1. The perpendicular slope is

when and

Perpendicular line:

4x � 4y � 7 � 0

y �54

� �1�x �12�

y �54

.x �12

dydx

� 2x � 2 � �1

y � x2 � 2x � 2

�1.y � x � 2

22. has slope The perpendicular slope is

Parabola

Perpendicular line:

y �12

x �28748

y �9516

�12�x �

112�

y � �9516

x �1

12,

6x �12

y� � �6x � 1 �12

y � �3x2 � x � 6

12

.�2.2x � y � 5 23.

(a)

Focus:

(b)

S � 2�100

0x�1 �

x2

10,000 dx 38,294.49

�1 � �y��2 ��1 �x2

10,000

y� �1

100x

y �1

200x2

�0, 50�x2 � 4�50�yx2 � 200y

y �1

200x2

Review Exercises for Chapter 10 447

24. (a)

(b)

(c) You want of the total area of covered. Find h so that

By Newton’s Method, Therefore, the total height of the water is

(d)

Total area � 24� � 353.656 429.054

353.656

� 16 ��2

0 ��1 � � 7

16� sin2�� d� �16� �from Example 5 of Section 9.1�

Area of sides � �Perimeter��Length�

Area of ends � 2�12�� � 24�

1.212 � 3 � 4.212 ft.h 1.212.

h�9 � h2 � 9 arcsin �h3� �

9�

4.

12 �y�9 � y2 � 9 arcsin �y

3��h

0�

9�

8

h

0 �9 � y2 dy �

9�

8

2 h

0 43�9 � y2 dy � 3�

x

2

−2

2

4

−4

−2

h

Area of filledtank abovex-axis is 3 .π

Area of filledtank belowx-axis is 6 .π

x y= 9 − 243

y12�34

�83

�62.4��32 �

9�

2 � �32 ��

9�

2 �� �83

�62.4��27�

2 � 7057.274

�83

�62.4��32 �y�9 � y2 � 9 arcsin

y3� �

13

�9 � y2�3�2�3

�3

F � 2�62.4� 3

�3 �3 � y� 4

3�9 � y2 dy �

83

�62.4��33

�3 �9 � y2 dy � 3

�3 y�9 � y2 dy�

V � ��ab��Length� � 12��16� � 192� ft3

25.

Line

2

4

−1

−2

5321−1

1

x

y

4y � 3x � 11 � 0

y � �34

x �114

t �x � 1

4 ⇒ y � 2 � 3�x � 1

4 �x � 1 � 4t, y � 2 � 3t 26.

Parabola

−1 1

1

2

3

4

5

6

7

2 3 4 5 6 7

y

x

t � x � 4 ⇒ y � �x � 4�2

x � t � 4, y � t2 27.

Circle

−2

−4

42−2−4

2

4

x

y

x2 � y2 � 36

�x6�

2

� �y6�

2

� 1

x � 6 cos �, y � 6 sin �

448 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

28.

Ellipse

�x � 3�2

9�

�y � 2�2

25� 1

−2

−2

1234567

−3

−1 1 2 3 4 5 6 7 8

y

x

�x � 33 �

2

� �y � 25 �

2

� 1

x � 3 � 3 cos �, y � 2 � 5 sin � 29.

Hyperbola

4

8

−2

−4

842−4

2

x

y �x � 2�2 � �y � 3�2 � 1

�x � 2�2 � sec2 � � 1 � tan2 � � 1 � �y � 3�2

x � 2 � sec �, y � 3 � tan �

30.

x2�3 � y2�3 � 52�3

�x5�

2�3

� �y5�

2�3

� 1

−6 −4

−4

2

4

6

−6

2 4 6

y

x

x � 5 sin3 �, y � 5 cos3 � 31.

(other answers possible)

y � 2 � �2 � 6�t � 2 � 4t

x � 3 � �3 � ��2��t � 3 � 5t

32.

y � 3 � 2 sin tx � 5 � 2 cos t,

�x � 5�2 � �y � 3�2 � 22 � 4

�x � h�2 � �y � k�2 � r2 33.

Let and

Then and y � 4 � 3 sin�.x � �3 � 4 cos�

�y � 4�2

9� sin2 �.

�x � 3�2

16� cos2 �

�x � 3�2

16�

�y � 4�2

9� 1

34.

Let and

Then and y � 4 sec�.x � 3 tan�

x2

9� tan2 �.

y2

16� sec2 �

y2

16�

x2

9� 1b2 � c2 � a2 � 9,c � 5,a � 4, 35.

y � sin 3� � 5 sin�−7 8

−5

5x � cos 3� � 5 cos�

36. (a)

(b)

� 8x

� 8�2 cot ��

� 16 cos �sin �

� 16 csc2 � � sin � � cos �

�4 � x2�y � �4 � 4 cot2 ��4 sin � cos �

−12

−4

12

4

x � 2 cot �, y � 4 sin � cos �, 0 < � < � 37.

(a)

No horizontal tangents

(b)

(c)

5321

1

2

4

5

x

y

y � 2 �34

�x � 1� ��3x � 11

4

t �x � 1

4

dydx

� �34

y � 2 � 3t

x � 1 � 4t

Review Exercises for Chapter 10 449

38.

(a) when

Point of horizontal tangency:

(b)

(c)

x5

5

2

2

1

1 3

3

4

4

6

6

y

y � �x � 4�2

t � x � 4

�4, 0�

t � 0.dydx

�2t1

� 2t � 0

y � t2

x � t � 4 39.

(a)

No horizontal tangents,

(b)

(c)

6

4

224 4

2

x

y

y �2x

� 3

t �1x

�t � 0�

dydx

�2

�1�t2 � �2t2

y � 2t � 3

x �1t

40.

(a)

No horizontal tangents,

(b)

(c)

x

2

3

4

1

1

2−2 −1

y

y �1x2

t �1x

�t � 0�

dydx

�2t

�1�t2 � �2t3

y � t2

x �1t

41.

(a) when

Point of horizontal tangency:

(b)

,

(c)

x−2 −1 2 3

−2

−1

3

2

y

�x � 0��4x2

�1 � x�2 � 4x�1 � x� �4x2

�5x � 1��x � 1�

y �1

12 �

1 � xx ��1

2 �1 � x

x � � 2�

2t � 1 �1x ⇒ t �

12 �

1x

� 1�

�13

, �1�

t � 1.dydx

��2t � 2��t2 � 2t�2

�2�2t � 1�2

��t � 1��2t � 1�2

t2�t � 2�2 � 0

y �1

t2 � 2t

x �1

2t � 1

450 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

42.

(a)

when

Point of horizontal tangency:

(b)

(c)

x42−2

2

y

y �1

��x � 1��2�2 � 2��x � 1��2� �4

�x � 3��x � 1�

t �x � 1

2

�1, �1�

t � 1.�1 � t

t2�t � 2�2 � 0

dydx

���t2 � 2t��2�2t � 2�

2

y �1

t2 � 2t

x � 2t � 1 43.

(a) when

Points of horizontal tangency:

(b)

(c)8

4

4

4 8x

y

�x � 3�2

4�

�y � 2�2

25� 1

�3, 7�, �3, �3�

� ��

2,

3�

2.

dydx

�5 cos�

�2 sin�� �2.5 cot� � 0

y � 2 � 5 sin�

x � 3 � 2 cos�

44.

(a) when

Points of horizontal tangency:

(b)

(c)

−4

−4

−2

−2

2

2

4

4

x

y

�x6�

2

� �y6�

2

� 1

�0, 6�, �0, �6�

� ��

2,

3�

2.

dydx

�6 cos�

�6 sin�� �cot� � 0

y � 6 sin�

x � 6 cos� 45.

(a)

when

But, at Hence no points of

horizontal tangency.

(b)

(c)

42

4

4 2

4

x

y

x2�3 � �y4�

2�3

� 1

� � 0, �.dydt

�dxdt

� 0

� � 0, �.

��4 sin�

cos�� �4 tan� � 0

dydx

�12 sin2 � cos�

3 cos2 ���sin��

y � 4 sin3 �

x � cos3 �

Review Exercises for Chapter 10 451

46.

(a)

No horizontal tangents

(b)

(c)

x2

2

31

1

3

y

y � e�ln x � eln�1�x� �1x, x > 0

t � ln x

dydx

��e�t

et � �1

e2t � �1x2

y � e�t

x � et 47.

for

Horizontal tangent at

No vertical tangents

�x, y� � �4, 0�t � 0:

t � 0.dydt

� 0

dydt

� 2tdxdt

� �1,

y � t2x � 4 � t,

48.

for

Horizontal tangents:

No vertical tangents

�1.1835, 1.0887�

�x, y� � ���63

� 2, 23�6 �

2�69 �t � �

�63

:

�2.8165, �1.0887�

�x, y� � ��63

� 2, 2�6

9�

23�6�t �

�23

:

t � ±�23

�±�6

3dydt

� 0

dydt

� 3t2 � 2dxdt

� 1,

y � t3 � 2tx � t � 2,

49.

for

Horizontal tangents:

for

Vertical tangents: �x, y� � �4, 1�, �0, 1�

� ��

2,

3�

2, . . .

dxd�

� 0

�x, y� � �2, 2�, �2, 0�

� � 0, �, 2�, . . .dyd�

� 0

dyd�

� �sin �dxd�

� 2 cos �,

y � 1 � cos �x � 2 � 2 sin �, 50.

for

Horizontal tangents:

for

Vertical tangents: �x, y� � �0, 0�, �4, 0�

� � 0, �, 2�, . . .dxd�

� 0

�2 ± �2, �2��x, y� � �2 ± �2, 2�,

� ��

4,

3�

4,

5�

4,

7�

4, . . .

dyd�

� 0

dyd�

� 4 cos 2�dxd�

� 2 sin �,

y � 2 sin 2�x � 2 � 2 cos �,

452 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

51.

(a), (c)

(b) At and dydx

� �14

.� ��

6,

dxd�

� �4, dyd�

� 1,

−3 3

−2

2

y � sin 2� � 2 sin� cos�

x � cot� 52.

(a), (c)

(b) At

and dydx

0.441.dydt

� 0.5,

� ��

6,

dxd�

1.134, �2 ��32 �,

−8 8

−4

8

y � 2 � cos�

x � 2� � sin�

53.

� r �

0 � d� �

r2��2�

0�

12

�2r

s � r �

0 ��2 cos2 � � �2 sin2 � d�

dyd�

� r� sin�

dxd�

� r� cos�

y � r�sin� � � cos��

x � r�cos� � � sin�� 54.

(one-half circumference of circle)

s � �

0 �36 sin2 � � 36 cos2 � d� � �6��

0� 6�

dyd�

� 6 cos�

dxd�

� �6 sin�

y � 6 sin�

x � 6 cos�

55.

(a)

(b) S � 2�2

0 �10 dt � 2��10�t�2

0� 4��10

S � 2�2

0 3t�10 dt � 6�10 ��t2

2�2

0� 12�10 �

��dxdt�

2� �dy

dt�2

� �1 � 9 � �10dydt

� �3,dxdt

� 1,

x � t, y � 3t, 0 ≤ t ≤ 256.

(a)

(b)

Note: The surface is a hemisphere:12

�4� �22�� � 8���

S � 2���2

0 2 cos ��2�d� � 8��sin ����2

0 � 8�

S � 2���2

0 2 sin ��2�d� � 8���cos ��

��2

0� 8�

��dxd��

2� �dy

d��2

� 2dydt

� 2 cos �,dxdt

� �2 sin �,

0 ≤ � ≤�

2y � 2 sin �,x � 2 cos �,

57.

� 3��

2�

2� � 3�

� 3�� �sin 2�

2 ���2

���2

� 6��2

���2 1 � cos 2�

2 d�

A � b

a

y dx � �2

���2 2 cos ��3 cos �� d�

y � 2 cos �x � 3 sin �, 58.

� ��� �sin 2�

2 �0

�� �

� �0

1 � cos 2�

2 d�

A � b

a

y dx � 0

sin ���2 sin �� d�

Review Exercises for Chapter 10 453

59.

� �0, 3�

�x, y� � �3 cos �

2, 3 sin

2�( )

1 2 30

π2

3, π2

�r, �� � �3, �

2� 60.

0

π2

−4, π116

−1−2−3−4−5 1−1

1

2

3

4

5

( )

� ��2�3, 2�

�x, y� � ��4 cos 11�

6, �4 sin

11�

6 �

�r, �� � ��4, 11�

6 �

61.

1 2

π2

0

3( , 1.56)

�0.0187, 1.7319�

�x, y� � ��3 cos�1.56�, �3 sin�1.56���r, �� � ��3, 1.56� 62.

0−1 1 2 3

−1

1

2

3

(−2, −2.45)

2π �1.5405, 1.2755�

�x, y� � ��2 cos��2.45�, � sin��2.45��

�r, �� � ��2, �2.45�

63.

2 51

1

43

−2

−5

−1

−4

−3

(4, 4)−

x

y

��4�2, 3�

4 � �r, �� � �4�2, 7�

4 �,

� �7�

4

r � �42 � ��4�2 � 4�2

�x, y� � �4, �4� 64.

( 1, 3)−

2

1

1 2 3x

−1

−2

−3

−3 −2 −1

y

�r, �� � ��10, 1.89�, ���10, 5.03� � � arctan��3� 1.89�108.43�

r � ���1�2 � 32 � �10

�x, y� � ��1, 3�

65.

x2 � y2 � 3x � 0

x2 � y2 � 3x

r2 � 3r cos�

r � 3 cos� 66.

x2 � y2 � 100

r2 � 100

r � 10

67.

�x2 � y2 � 2x�2 � 4�x2 � y2�

x2 � y2 � �2�±�x2 � y2 � � 2x

r2 � �2r�1 � cos��

r � �2�1 � cos�� 68.

3x2 � 4y2 � 2x � 1 � 0

4�x2 � y2� � �x � 1�2

2�±�x2 � y2 � � x � 1

2r � r cos� � 1

r �1

2 � cos�

454 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

69.

�x2 � y2�2 � x2 � y2

r 4 � r2 cos2 � � r2 sin2 �

� cos2 � � sin2 � r2 � cos 2� 70.

x � �3 y � 8

r �cos� � �3 sin�� � 8

�4

�1�2� cos� � ��3�2� sin�

r � 4 sec�� ��

3� �4

cos�� � ���3��

71.

y2 � x2�4 � x4 � x�

x3 � xy2 � 4x2 � 4y2

x � 8� x2

x2 � y2� � 4

r cos� � 8 cos2 � � 4

� 4�2 cos2� � 1�� 1cos ��

r � 4 cos 2� sec� 72.

y � �x

yx

� �1

tan� � �1

� �3�

473.

r � a cos2 � sin�

r4 � a�r2 cos2 ���r sin ��

�x2 � y2�2 � ax2y

74.

r � 4 cos�

r2 � 4r cos� � 0

x2 � y2 � 4x � 0 75.

r2 � a2�2

x2 � y2 � a2�arctan yx�

2

76.

r2�2 � a2

�x2 � y2��arctan yx�

2

� a2

77.

Circle of radius 4

Centered at the pole

Symmetric to polar axis,

and pole

2 60

π2

� � ��2,

r � 4 78.

Line

01 2

π2

� ��

1279.

Vertical line

01

π2

r cos� � �1, x � �1

r � �sec� ��1

cos�

80.

Horizontal line

01 2 3 4

π2

r � 3 csc �, r sin� � 3, y � 3 81.

Cardioid

Symmetric to polar axis0

1

π2r � �2�1 � cos��

0

r 0�1�2�3�4

�2�

3�

2�

3�

Review Exercises for Chapter 10 455

82.

Limaçon

Symmetric to polar axis0

2

π2

r � 3 � 4 cos� 83.

Limaçon

Symmetric to polar axis0

2 4

π2r � 4 � 3 cos�

0

r 1 3 5 7�1

�2�

3�

2�

3� 0

r 1 4 7112

52

�2�

3�

2�

3�

84.

Spiral

Symmetric to � � ��20

2 4 8

π2r � 2� 85.

Rose curve with four petals

Symmetric to polar axis, and pole

Relative extrema:

Tangents at the pole:

04

π2� �

4,

3�

4

��3, 0�, �3, �

2�, ��3, ��, �3, 3�

2 �

� ��

2,

r � �3 cos 2�

0

r 0 3�5�

22�

3�

2�

2

3�

25�

4�

3�

4�

2�

4�

86.

Rose curve with five petals

Symmetric to polar axis

Relative extrema:

Tangents at the pole: � ��

10,

3�

10,

2,

7�

10,

9�

10

�1, 0�, ��1, �

5�, �1, 2�

5 �, ��1, 3�

5 �, �1, 4�

5 �0

1

π2

r � cos 5�

87.

Rose curve with four petals

Symmetric to the polar axis, and pole

Relative extrema:

Tangents at the pole: � � 0, �

2

�±2, �

4�, �±2, 3�

4 �

� ��

2,

r � ±2 sin�2��

02

π2

r2 � 4 sin2 2�

456 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

88.

Lemniscate

Symmetric to the polar axis

Relative extrema:

Tangents at the pole: � ��

4,

3�

4

�±1, 0�0

12

π2r2 � cos 2�

0

r 0±�22

±1

4�

6�

89.

Graph of rotatedthrough an angle of ��4

r � 3 sec�

−1

−1

8

5r �3

cos � � ���4�90.

Bifolium

Symmetric to � � ��2 1−1

−0.25

0.75r � 2 sin� cos2 �

91.

Strophoid

Symmetric to the polar axis

r ⇒ � as � ⇒ ���

2

r ⇒ � as � ⇒ ��

2

−6 6

−4

4r � 4 cos 2� sec�

93.

(a) The graph has polar symmetry and the tangents at the pole are

(b)

Horizontal tangents:

When

Vertical tangents:

(c)

−5 1

−2.5

2.5

�12

, ±arccos 14� �0.5, ±1.318�

� � 0, �, � � ±arccos�14�, ��1, 0�, �3, ��sin��4 cos� � 1� � 0, sin� � 0, cos� �

14

,

�3 � �334

, �arccos�1 � �338 �� �2.186, �2.206�.

�3 � �334

, arccos�1 � �338 �� �2.186, 2.206�

�3 � �334

, �arccos�1 � �338 �� ��0.686, �0.568�

�3 � �334

, arccos�1 � �338 �� ��0.686, 0.568�

cos� �1 ± �33

8, r � 1 � 2�1 � �33

8 � �3 � �33

4,

�4 cos2 � � cos� � 2 � 0, cos� ��1 ± �1 � 32

�8�

1 ± �338

dydx

�2 sin2 � � �1 � 2 cos�� cos�

2 sin� cos� � �1 � 2 cos�� sin�

� ��

3, �

3.

r � 1 � 2 cos�

92.

Semicubical parabola

Symmetric to the polar axis

r ⇒ � as � ⇒ ���

2

r ⇒ � as � ⇒ ��

2

5−1

−3

3r � 4�sec� � cos��

Review Exercises for Chapter 10 457

94.

(a)

Tangents at the pole:

(c)

3−3

−2

2

� � 0, �

2

drd�

�4 cos�2��

r

2r�drd�� � 8 cos�2��

r2 � 4 sin�2��(b)

Horizontal tangents:

when

Vertical tangents when

tan 2� tan� � 1, � � 0, �

6, �0, 0�, �±�2�3,

6�cos 2� cos� � sin 2� sin� � 0:

tan� � �tan�2��, � � 0, �

3, �0, 0�, �±�2�3,

3�cos�2�� sin� � sin�2�� cos� � 0,

dydx

� 0

�cos�2�� sin� � sin�2�� cos�

cos�2�� cos� � sin�2�� sin�

dydx

�r cos� � ��4 cos 2� sin���r�

�r sin� � ��4 cos 2� cos���r�

95. Circle:

at

Limaçon:

at

Let be the angle between the curves:

Therefore, � arctan�2�33 � 49.1.

tan ��3 � ��3�9�

1 � �1�3� �2�3

3.

� ��

6,

dydx

��39

dydx

��5 cos� sin� � �4 � 5 sin�� cos�

�5 cos� cos� � �4 � 5 sin�� sin�

r � 4 � 5 sin�

� ��

6,

dydx

� �3dydx

�3 cos� sin� � 3 sin� cos�

3 cos� cos� � 3 sin� sin��

sin 2�

cos2 � � sin2 �� tan 2�

r � 3 sin�

97.

The points and are the two points of intersection (other than the pole). The slope of the graph ofis

At and at The slope of the graph of is

At and at In both cases, and we conclude that

the graphs are orthogonal at and �1, 3��2�.�1, ��2�

m1 � �1�m2�1, 3��2�, m2 � 1�1 � 1.�1, ��2�, m2 � 1��1 � �1

m2 �dydx

�sin2 � � cos��1 � cos��

sin� cos� � sin��1 � cos��.

r � 1 � cos��1, 3��2�, m1 � �1�1 � �1.�1, ��2�, m1 � �1��1 � 1

m1 �dydx

�r� sin� � r cos�

r� cos� � r sin��

�sin2 � � cos��1 � cos���sin� cos� � sin��1 � cos�� .

r � 1 � cos��1, 3��2��1, ��2�

r � 1 � cos�, r � 1 � cos�

98.

The points of intersection are and For

At is undefined and at . For

At and at is undefined. Therefore, the graphs are orthogonal at and �0, 0�.�a��2, ��4�m2�0, 0�,�a��2, ��4�, m2 � 0

m2 �dydx

��a sin2 � � a cos2 �

�a sin� cos� � a cos� sin��

cos 2�

�2 sin� cos�.

r � a cos�,m1 � 0�0, 0�,�a��2, ��4�, m1

m1 �dydx

�a cos� sin� � a sin� cos�

a cos2 � � a sin2 ��

2 sin� cos�

cos 2�.

r � a sin�,�0, 0�.�a��2, ��4�r � a sin�, r � a cos�

96. False. There are an infinite number of polar coordinate representations of a point. For example, the point has polar representations etc.�r, �� � �1, 0�, �1, 2��, ��1, ��,

�x, y� � �1, 0�

458 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

99.

−3 6

−3

3

�9�

2 �A � 2�12

0 �2 � cos��2 d�� 14.14,

r � 2 � cos� 100.

8−8

−12

4

�75�

2 �A � 2�12

3��2

��2 �5�1 � sin���2� d� 117.81,

r � 5�1 � sin��

101.

−3 3

−2

2

A � 2�12

��2

0 4 sin 2� d�� � 4

r2 � 4 sin 2� 102.

−3 6

−3

3

A � 2�12

��3

0 4 d� �

12

��2

��3 �4 cos��2 d�� 4.91

r � 4 cos�, r � 2

103.

−0.5 0.5

−0.1

0.5

� �

32� 0.10,

A � 2�12

��2

0 �sin� cos2 ��2 d��

r � sin� cos2 � 104.

−6 6

−4

4

�4�� 12.57

A � 3�12

��3

0 �4 sin 3��2 d��

r � 4 sin 3�

105.

1.2058 � 9.4248 � 1.2058 11.84

A � 2�12

��12

0 18 sin 2� d� �

12

5��12

��12 9 d� �

12

��2

5��12 18 sin 2� d��

� ��

12

sin 2� �12

9 � r2 � 18 sin 2�−6 6

−4

4r � 3, r2 � 18 sin 2�

106.

5−25

−5

10

A �12

0 �e� �2 d� 133.62

r � e�, 0 ≤ � ≤ �

Review Exercises for Chapter 10 459

107.

� �2�2a��1 � cos ��1�2��

0� 4a

� �2a�

0

sin ��1 � cos �

d�

� �2a�

0 �1 � cos � d�

s � �

0 �a2�1 � cos ��2 � a2 sin2 � d�

drd�

� a sin �

0 ≤ � ≤ �r � a�1 � cos ��, 108.

Using a graphing utility, s 4.8442a.

� a��2

���2 �1 � 3 sin2 2� d�

s � ��2

���2 �a2 cos2 2� � 4a2 sin2 2� d�

drd�

� �2a sin 2�

���4 ≤ � ≤ ��2r � a cos 2�,

109.

�34��17

5 88.08

S � 2���2

0 �1 � 4 cos �� sin ��17 � 8 cos � d�

�f ���2 � f����2 � ��1 � 4 cos ��2 � ��4 sin ��2 � �17 � 8 cos �

f���� � �4 sin �

f ��� � 1 � 4 cos �

110.

� 4�

S � 2���2

0 2 sin � cos ��2� d�

�f ���2 � f����2 � �4 sin2 � � 4 cos2 � � 2

f���� � 2 cos �

f ��� � 2 sin �

111.

Parabola

0842 6

π2r �

21 � sin �

, e � 1 112.

Parabola

20

π2r �

21 � cos �

, e � 1

113.

Ellipse

02

π2

r �6

3 � 2 cos ��

21 � �2�3� cos �

, e �23

114.

Ellipse

1 20

π2

r �4

5 � 3 sin ��

4�51 � �3�5� sin �

, e �35

460 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

115.

Hyperbola

02 3 4

π2

r �4

2 � 3 sin ��

21 � �3�2�sin �

, e �32

116.

Hyperbola

1 20

π2

r �8

2 � 5 cos ��

41 � �5�2�cos �

, e �52

117. Circle

Center: in rectangular coordinates

Solution point:

r � 10 sin�

r2 � 10r sin� � 0

x2 � y2 � 10y � 0

x2 � �y � 5�5 � 25

�0, 0�

�5, �

2� � �0, 5�

118. Line

Slope:

Solution point:

tan � � �3, � ��

3

y � �3 x, r sin� � �3 r cos�,

�0, 0�

�3

119. Parabola

Vertex:

Focus:

r �4

1 � cos�

e � 1, d � 4

�0, 0�

�2, ��

120. Parabola

Vertex:

Focus:

r �4

1 � sin�

e � 1, d � 4

�0, 0�

�2, �

2�

121. Ellipse

Vertices:

Focus:

r ��2

3��52�

1 � �23� cos�

�5

3 � 2 cos�

a � 3, c � 2, e �23

, d �52

�0, 0�

�5, 0�, �1, ��

122. Hyperbola

Vertices:

Focus:

r ��4

3��74�

1 � �43� cos�

�7

3 � 4 cos�

a � 3, c � 4, e �43

, d �74

�0, 0�

�1, 0�, �7, 0�

Problem Solving for Chapter 10

Problem Solving for Chapter 10 461

1. (a)

(b)

Tangent line at

Tangent line at

Tangent lines have slopes of 2 and perpendicular.

(c) Intersection:

Point of intersection, is on directrix y � �1.�3�2, �1�,

x �32 ⇒ �3

2, �1�

10x � 15

8x � 16 � �2x � 1

2x � 4 � �12

x �14

�1�2 ⇒

��1, 14� y �

14

� �12

�x � 1� ⇒ y � �12

x �14

�4, 4� y � 4 � 2�x � 4� ⇒ y � 2x � 4

y� �12

x

2x � 4y�

x2 � 4y

6

4

8

10

−2642−4−6 −2

2

x

y

(4, 4)

(−1, )14

2. Assume

Let be the equation of the focal chord.

First find x-coordinates of focal chord endpoints:

(a) The slopes of the tangent lines at the endpoints are perpendicular because

—CONTINUED—

�1

4p2 ��4p2� � �1�1

4p2 �4p2m2 � 4p2�m2 � 1��12p

�2pm � 2p�m2 � 1� 12p

�2pm � 2p�m2 � 1�

x2 � 4py, 2x � 4py� ⇒ y� �x

2p.

x �4pm ± �16p2m2 � 16p2

2� 2pm ± 2p�m2 � 1

x2 � 4pmx � 4p2 � 0

x2 � 4py � 4p�mx � p�

y � mx � p

y = −p

(0, p)

x2 = 4py

y

x

p > 0.

462 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

2. —CONTINUED—

(b) Finally, we show that the tangent lines intersect at a point on the directrix

Let and

Tangent line at

Tangent line at

Intersection of tangent lines:

Finally, the corresponding -value is which shows that the intersection point lies on the directrix.y � p,y

x � 2pm

2x�4p�m2 � 1� � 16p2m�m2 � 1

2x�b � c� � b2 � c2

2bx � b2 � 2cx � c2

bx2p

�b2

4p�

cx2p

�c2

4p

y �c2

4p�

c2p

�x � c� ⇒ y �cx2p

�c2

4px � c:

y �b2

4p�

b2p

�x � b� ⇒ y �bx2p

�b2

4px � b:

c2

4p� 2pm2 � p � 2pm�m2 � 1

b2

4p� 2pm2 � p � 2pm�m2 � 1

c2 � 8p2m2 � 4p2 � 8p2m�m2 � 1

b2 � 8p2m2 � 4p2 � 8p2m�m2 � 1

c � 2pm � 2p�m2 � 1.b � 2pm � 2p�m2 � 1

y � �p.

3. Consider with focus

Let be point on parabola.

Tangent line at

For

Thus,

is isosceles because

Thus, �FQP � �BPA � �FPQ.

� b � p.

� ��b � p�2

� �4pb � b2 � 2bp � p2

FP � ��a � 0�2 � �b � p�2 � �a2 � b2 � 2bp � p2

FQ � p � b

�FQP

Q � �0, �b�.

x � 0, y � b �a

2p��a� � b �

a2

2p� b �

4pb2p

� �b.

Py � b �a

2p�x � a�

2x � 4py� ⇒ y� �x

2p

P�a, b�

y

x

F

Q

P(a, b)

ABF � �0, p�.x2 � 4py

Problem Solving for Chapter 10 463

4.

Tangent line at

At

To show consider

Thus, and the reflective property is verified.� � �

x0

2

a2 �y0

2

b2 � 1.⇔

x02b2 � a2y0

2 � a2b2⇔

a��x0 � c�2 � y02 � x0c � a2⇔

az � x0c � a2 � 0⇔

�x0 � c�2 � y02 � 2az � �x0

2 � �y0 �b2

y0�

2

� � �c2 �b4

y02�⇔

z2 � 2az � f 2 � d2⇔

��z � 2a�2 � f 2 � d 2��z� � �z2 � f 2 � d 2��z � 2a�⇔

��MF2�2 � f 2 � d2��2 f �MF1�� � ��MF1�2 � f 2 � d 2��2 f �MF2 ��

� � �,

MQ ��x02 � �y0 �

b2

y0�

2

� f

QF2 � QF1 ��c2 �b4

y02 � d

x � 0, y � �b2

y0 ⇒ Q � �0, �

b2

y0�.

x0xa2 �

y0yb2 � 1

x0xa2 �

y0yb2 �

x02

a2 �y0

2

b2

yy0 � y0

2

b2 �x0x � x0

2

a2

y � y0 �b2x0

a2y0�x � x0�M�x0, y0�:

y� �b2xa2y

β

βα

Q

F1(c, 0)

F2(−c, 0)

(x0, y0)M

y

x

x2

a2 �y2

b2 � 1, a2 � b2 � c2, MF2 � MF1 � 2a

By the Law of Cosines,

Let

Slopes: MF1: y0

x0 � c; QF1:

�b2

y0c; QF2:

b2

y0c

z � MF1.MF2 � MF1 � 2a.

cos � ��MF1�2 � f 2 � d2

2 f �MF1�cos � �

��F2�2 f 2 � d2

2 f �MF2� ,

d2 � �MF1�2 � f 2 � 2 f �MF1�cos �.

�F1Q�2 � �MF1�2 � f 2 � 2 f �MF1�cos �

d2 � �MF2�2 � f 2 � 2 f �MF2�cos �

�F2Q�2 � �MF2�2 � �MQ�2 � 2�MF2��MQ�cos �

464 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

5. (a) In

In

(b)

Let

Then and

(c)

y2 �x3

�2a � x�

�x2 � y2�x � 2ay2

r3 cos � � 2a r2 sin2 �

r cos � � 2a sin2 �

r � 2a tan � sin �

x � 2at2

1 � t2, y � 2at3

1 � t2.sin2 � �t2

1 � t2

t � tan �, � < t < .

y � r sin � � �2a tan � sin ��sin � � 2a tan � sin2 �, ��

2< � <

1

t1 + t2

x � r cos � � �2a tan � sin ��cos � � 2a sin2 �

� 2a tan � sin �

� 2a sin2 �cos �

� 2a� 1cos �

� cos �� r � OP � AB � OB � OA � 2a�sec � � cos ��

�OAC, cos � �OA2a

⇒ OA � 2a cos �.

�OCB, cos � �2aOB

⇒ OB � 2a sec �.

6. (a)

(b) Disk:

(c) Disk:

�a�ba2c a2c�a2 � c2 � a4 arcsin�c

a�� � 2�b2 � 2� �abe � arcsin�e�

�4�ba2 �a

0 �a4 � c2x2 dx �

2�ba2c cx�a4 � c2x2 � a4 arcsin�cx

a2��a

0

S � 2�2�� �a

0 ba�a2 � x2 ��a4 � �a2 � b2�x2

a�a2 � x2 � dx

V � 2� �a

0 b2

a2 �a2 � x2� dx �2�b2

a2 �a

0 �a2 � x2� dx �

2�b2

a2 a2x �13

x3�a

0�

43

�ab2

� 2�a2 ��ab2

c ln�c � a

e �2

� 2�a2 � ��b2

e � ln�1 � e1 � e�

�2�ab2c

�b2c�b2 � c2 � b4 lncb � b�b2 � c2 � b4 ln�b2��

�4�ab2 �b

0 �b4 � c2y2 dy �

2�ab2c cy�b4 � c2y2 � b4 lncy � �b4 � c2y2�

b

0

S � 4� �b

0 ab�b2 � y2��b4 � �a2 � b2�y2

b�b2 � y2 � dy

�43

�a2bV � 2� �b

0 a2

b2 �b2 � y2� dy �2�a2

b2 �b

0 �b2 � y2� dy �

2�a2

b2 b2y �13

y3�b

0

A � 4 �a

0 ba�a2 � x2 dx �

4ba �1

2�x�a2 � x2 � a2 arcsin�xa��

a

0� �ab

Problem Solving for Chapter 10 465

(c)

1 20

π2 (d) for

Thus, and are tangent lines to curve atthe origin.

y � �xy � x

� ��

4,

3�

4.r��� � 0

(b)

r � cos 2� sec �

r cos � � cos 2�

r cos ��sin2 � � cos2 �� � cos2 � � sin2 �

r cos � sin2 � � sin2 � � cos2 � � r cos3 �

sin2 ��1 � r cos �� � cos2 ��1 � r cos ��

r2 sin2 � � r2 cos2 ��1 � r cos �1 � r cos ��7. (a)

Thus, y2 � x2�1 � x1 � x�.

1 � x1 � x

1 � �1 � t2

1 � t2�1 � �1 � t2

1 � t2��

2t2

2� t2

y2 �t2�1 � t2�2

�1 � t2�2 , x2 ��1 � t2�2

�1 � t2�2

(e)

��5 � 12

, ±�5 � 1

2��2 � �5�

�3 � �5

�1 � �5�

�5 � 1

2

t4 � 4t2 � 1 � 0 ⇒ t2 � �2 ± �5 ⇒ x �1 � ��2 ± �5�1 � ��2 ± �5� �

3 � �5

�1 ± �5

y��t� ��1 � t2��1 � 3t2� � �t � t3��2t�

�1 � t2�2 �1 � 4t2 � t4

�1 � t2�2 � 0

8.

θa − y

a2ay − y2

x � a arccos�a � ya � � �2ay � y2, 0 ≤ y ≤ 2a

� a�arccos�a � ya � �

�2ay � y2

a � � a�arccos�a � y

a � � sin�arccos�a � ya ���

x � a�� � sin ��

� � arccos�a � ya �

y � a�1 � cos �� ⇒ cos � �a � y

a9. (a)

(b) is on

the curve whenever is on the curve.

(c)

Thus,

On ���, ��, s � 2�.

s � �a

0dt � a.

x��t� � cos � t2

2, y��t� � sin

�t2

2, x��t�2 � y��t�2 � 1

�x, y�

��x, �y� � ���t

0cos

�u2

2 du, ��t

0sin

�u2

2 du�

Generated by Mathematica

466 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

10. For

Hence, the curve has length greater that

(Harmonic series) � .

>2��

12

�14

�16

�18

� . . .�

�2� �1 �

13

�15

�17

� . . .�

S �2�

�2

3��

25�

�2

7�� . . .

y �2�

, �23�

, 2

5�,

�27�

, . . .

t ��

2,

32

, 5�

2,

7�

2, . . . 11.

Line segment

Area �12

ab

yb

�xa

� 1

ay � bx � ab

r�a sin � � b cos �� � ab

r �ab

a sin � � b cos �, 0 ≤ � ≤

2

12. (a)

(b)

(c) Differentiating,d

d��tan �� � sec2 �.

⇒ tan � � ��

0sec2 � d�

tan � �h1 ⇒ Area �

12

�1�tan �

�12�

0sec2 � d� α

1

x = 1r = secθ

Area � ��

0

12

r2 d� 13. Let be on the graph.

r2 � 2 cos 2�

r2 � 2�2 cos2 � �1�

r2 � 4 cos2 � � 2

r2�r2 � 4 cos2 � � 2� � 0

r 4 � 2r2 � 1 � 4r2 cos2 � � 1

�r2 � 1�2 � 4r2 cos2 � � 1

�r2 � 1 � 2r cos ��r2 � 1 � 2r cos � � 1

�r, ��

14. If a dog is located at in the first quadrant, then its neighbor is at

and

The slope joining these points is

slope of tangent line at

Finally, � ≥�

4.r �

d�2

e����4����,

r��

4� �d�2

⇒ r � Ce���4 �d�2

⇒ C �d�2

e��4

r � Ce��

r � e���C1

ln r � �� � C1

drr

� �d�

⇒ drd�

� �r

dydx

dydrdxdr

drd�

sin � � r cos �

drd�

cos � � r sin ��

sin � � cos �sin � � cos �

�r, ��.r cos � � r sin ��r sin � � r cos �

�sin � � cos �sin � � cos �

�x2, y2� � ��r sin �, r cos ��.�x1, y1� � �r cos �, r sin ��

�r, � ��

2�:�r, ��

Problem Solving for Chapter 10 467

15. (a) The first plane makes an angle of with the positive axis, and is 150 miles from P:

Similarly for the second plane,

(b)

� �sin 45�190 � 450t� � sin 70�150 � 375t��2�1�2 � ��cos 45��190 � 450t� � cos 70�150 � 375t��2

d � ��x2 � x1�2 � � y2 � y1�2

� sin 45 �190 � 450t�.

y2 � sin 135 �190 � 450t�

� cos 45 ��190 � 450t�

x2 � cos 135 �190 � 450t�

y1 � sin 70 �150 � 375t�

x1 � cos 70 �150 � 375t�

x-70 (c)

The minimum distance is 7.59 miles when 0.4145.t �

00

1

280

17.

produce “bells”; produce “hearts”.n � �1, �2, �3, �4, �5n � 1, 2, 3, 4, 5

6−6

−4

4

n = 3

6−6

−4

4

n = −1

6−6

−4

4

n = −5

6−6

−4

4

n = 4

6−6

−4

4

n = 0

6−6

−4

4

n = −4

6−6

−4

4

n = 5

6−6

−4

4

n = 1

6−6

−4

4

n = −3

6−6

−4

4

n = 2

6−6

−4

4

n = −2

16. The curve is produced over the interval 0 ≤ � ≤ 10�.