chapter 10 conics, parametric equations, and polar coordinates · 362 chapter 10 conics, parametric...
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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
2π
�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
=θ
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
2π
�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
2π
�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
2π
�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< � <
�
2θ
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�.