C H A P T E R 9 Conics, Parametric Equations, and Polar Coordinates
Section 9.1
Conics and Calculus . . . . . . . . . . . . . . . . . . . . 177
Section 9.2
Plane Curves and Parametric Equations . . . . . . . . . . 188
Section 9.3
Parametric Equations and Calculus . . . . . . . . . . . . 192
Section 9.4
Polar Coordinates and Polar Graphs . . . . . . . . . . . . 198
Section 9.5
Area and Arc Length in Polar Coordinates . . . . . . . . 205
Section 9.6
Polar Equations of Conics and Kepler’s Laws . . . . . . . 210
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 222
C H A P T E R 9 Conics, Parametric Equations, and Polar Coordinates Sectio Sec tion n 9.1
Conics Conics and Calcul Calculus us
Solutions to Odd-Numbered Exercises
1. y 2
3. x x
4 x
2 y
Vertex: 0, 0
Vertex: 3, 2
p
p
1 > 0
x2 9
y 2 4
7.
1
Center: 0, 0 Ellipse Matches (f)
9. y 2
6 x
Focus:
Directrix: x
< 0
y 2 16
x2 1
1
Hyperbola Center: 0, 0 Vertical transverse axis. Matches (c)
x 4 2 x 3
11. x x
Vertex: 0, 0
3
0
3 2
4
8
x 4 4 x 1
3
Vertex: 3, 2
y
Focus: 3.25, 2 (0, 0)
12
y 22 0 y 22
y
8
3 2 ,
1
2
2
Opens downward Matches graph (e).
Opens to the right Matches graph (h).
5.
32
x
4
Directrix: x
4
2.75
(− 3, 2)
2
4
x
−8
−6
−4
−2 −2
8
−4
13. y2
y 2
4 y
4 y
4 x
0
4
4 x
15. x2
4
y 22 41 x x 1
4 x
4 y
4
x 2
4 x
4
4 y
0 4
4
x x 22 4 1 y 2
Vertex: 1, 2
Vertex: 2, 2
y
6
Focus: 0, 2 Directrix: x
2
y
Focus: 2, 1 Directrix: y
4
4
(− 2, 2)
3
(− 1, 2)
x
−6 x
2
2 2
4
−4
−2
2
−2
6
−4
177
178
Chapter 9
17. y 2 y2
x y
Conics, Parametric Equations, and Polar Polar Coordinates Coordinates
19. y 2
0
1
4 x
4
0
y 2
4 x
x 41 x
1
y 4 x 4 2
y 12 4 14 x 14 1 1 Vertex: 4 , 2 1 Focus: 0, 2
4 1
Vertex: 1, 0
2
4
Focus: 0, 0 −5
2
Directrix: x
1
Directrix: x
2
2
−6
6
−3
−4
y 22 4 2 x x 3
21. y 2
4 y
25. x2
8 x
y
4
4
0
y
20
x
0
c, 4
9a
3b
c, 11
Solving Solving this system, we have have a Therefore, y
5 2 14 3 x 3 x
x x 12 9 a2
25, b2
y
52
25
9, c2
x2
46 y
0
x2
24 y
96
29. x2
4 y 2
4
3 or 5 x
2
16a
4b
5 14 3, b 3 ,
14 x
3 y
c
9
x2 4
c.
a2
3.
Foci:
0.
1
y 2 1
y
2
4, b2
1
1, c2
(0, 0)
3
−1
3, 0 ± 3,
x
1
−2
3 2
9 x2
4 y 2
36 x
24 y
36
4
x x
x 8
4
4
0
x2 4 x 4 4 y2 6 y 9 36 36 36 9 x
16
8
22
4
Vertices: 1, 10, 1, 0
4
Vertices: ± 2, 0
33.
12
(1, 5)
Foci: 1, 9, 1, 1
e
y
Center: 1, 5
4 5
k
Center: 0, 0
e
31.
2
27. Since the axis of the parabola parabola is vertical, vertical, the form of the equation is y ax2 bx c. Now, Now, substituting the values of the given coordinates into this equation, we obtain
3
x x h2 4 p y
23.
a2
9, b2
4, c2
y 32 9
36
1
5 y
Center: 2, 3 Foci:
2, 3 ± 5
6
Vertices: 2, 6, 2, 0 e
(− 2, 3) 2
5 3
x 6
4
2
2
Section 9.1
12 x2
35.
12 x2
20 y 2
12 x
20 y 2
x
1 4
x 122
5, b2
Center:
1 , 1 2
1 2
±
Vertices:
Foci:
3, c2
0
2 y
1
37
60
y
12
3
3
20
x2
3 x
1
a2
3 x
2 y 2
9 4
2, c 2
4 y
2 y
y
0.25
1
y 12 y
57
2
12 x
±
37
Vertices:
2 y
y
y
1
2
x
1
4
3
−3
−3
39. Center: 0, 0 Focus: 2, 0 Vertex: 3, 0 Horizontal major axis a
x2 9
3, c
y 2 5
2
b
⇒
41. Vertices: 3, 1, 3, 9 Minor axis length: 6 Vertical major axis Center: 3, 5
5
a
4, b
x 32
1
9
3
y
52
16
43. Center: 0, 0 Horizontal major axis Points on ellipse: 3, 1, 4, 0
Since the major axis is horizontal,
x2 a2
y 2 b2
1.
Substituting the values of the coordinates of the given points into this equation, we have
9 1 2
a
2
b
1, and
16 a2
The solution to this system is a2
1.
16, b2
Therefore, x2 16
y 2 167
1,
x2 16
7 y 2 16
1.
167.
12
9 4
2
1
1
1 4
3 x
1 7 2 4
3 x
x
12 x 8
1 ±
−2
−3
(Graph each of these separately.)
(Graph each of these separately.) 1
1 4
1 7 , 1 , , 1 2 2
20
12 x 12 x2 20
2
2, 1
12 x 12 x2 20 57
1 ±
0
Solve for y: 2 y 2
12 x
12
2
Solve for y: 20 y 2
1
32, 1
32
Foci:
2 y
4, b2
Center:
5, 1
2 y 2
4
2
±
x 322
2, 1
1 2
37
5 a2
x2
37.
40 y
179
Conics and Calculus
7
2
4 x2
2
4
180
45.
Chapter 9
y 2 1 a
x2 4
47.
1
1, b
Conics, Parametric Equations, and Polar Coordinates
2, c
5
x
12
4 a
2, b
y 22
1 1, c
1
5
Center: 0, 0
Center: 1, 2
Vertices: 0, ± 1
Vertices: 1, 2, 3, 2
Foci: 0, ± 5
Foci: 1
Asymptotes: y
±
1 x 2
Asymptotes: y
y
2 ±
1 x 2
1
y
4
1 x
2
1 x
4
5, 2
±
2
2
1
2
3
2
4
2 4 5
4
9 x2
49.
9 x2
y
4 x
2
4
x
6 y
18
y 2
6 y
9
18
22
1 a
1, b
3, c
y 32
9
0 36
9 y 2
x2 2 x 1
9
9 y 2
54 y
80
0
6 y
9
80
32
0
9 y
10
3
±
1 x 3
1
81
0
1
Degenerate hyperbola is two lines intersecting at 1, 3.
10, 3
Asymptotes: y
y
Vertices: 1, 3, 3, 3 ±
2 x
x 12
1
Center: 2, 3 Foci: 2
x2
51.
36 x
y
3 x
3 ±
2
x
4
y
2
2 2
x
2
2
4
4
6
2
6
4
6
9 y 2
53.
9 y 2
2
x
6 y
2 x
9
x2
y 32 2 a
62
2 x
1
62
x 12
18
0
±
±
9
y
3 x2
a 7
2
2 x
2, b
1
6 x
2 y 2
12 y
27
0
6 y
9
27
1
y 32 6
6, c 10
3
18
−5
2
2 x
62
x2
2 x 9
19
3 ±
1 x2 3
(Graph each curve separately.)
10, 3
Solve for y: 2 y 2
81
−7
6 y
y 2 x
19
9
32
y
12
7
Vertices: 1, 3, 3, 3 ±
1
Center: 1, 3 Foci: 1
x
y 32
18
1
2 5
4
−7
6 y
81
2 y 2
x 12
Solve for y: 9 y2
−5
Vertices: 1, 3 Foci: 1, 3
1
1
2, b 3 2, c 2 5
Center: 1, 3
3 x2
55.
54 y
3 x2
6 x
27
3 x2
6 x 2
9
3 ±
(Graph each curve separately.)
3 x2
18
2 x 2
3
Section 9.1
57. Vertices: ± 1, 0 Asymptotes: y ± 3 x Horizontal transverse axis Center: 0, 0 a
1, ±
b a
±
x2 Therefore, 1
b
±3 ⇒
y 2 9
3
3
Therefore, the equation is of the form y2 9
1.
181
59. Vertices: 2, ± 3 Point on graph: 0, 5 Vertical transverse axis Center: 2, 0 a
b 1
Conics and Calculus
x 22
1.
b2
Substituting the coordinates of the point 0, 5, we have 25 9
4 b2
1
b2
or
y2 9
Therefore, the equation is
61. Center: 0, 0 Vertex: 0, 2 Focus: 0, 4 Vertical transverse axis a
2, c
4, b2
y2 Therefore, 4
2 x, y 4 3 Horizontal transverse axis Asymptotes: y
c2
x2 12
1.
a2
a
At x
2
2 x 9
1,
6: y
At 6, 3 : y
2 yy 0,
3, y 3
or 2 x At 6, 3 : y
A
4 y 2
6 x
16 y
2 x 3
1.
0
1, C 4
3
3
±2
At 6, 3 : y
9
y 22
±
4
3
or 9 x At 6, 3 : y
0
or 9 x
2 3
1.
2 3
2 3 y
3
9
60
0
x
6
60
0
4 y 2
16 y
9 2 3
2 3 y
x 6
0 69. y2 A
AC 4 > 0
b a
(b) From part (a) we know that the slopes of the normal lines must be 9 2 3 .
x 6
±
2. Therefore,
6
3
9
3
2
3 3 y
21
y
9 3
2 3 x 9
3
x 9 y
±6
3 3 y
or 2 x 67. x2
Slopes of asymptotes:
±
x 22 94
3
9
y
Center: 3, 2
12
x 32
x2 9
9 . 4
63. Vertices: 0, 2, 6, 2
Thus, b
65. (a)
4 y
4 x
71. 4 x2
0
0, C 1
A
Parabola
15
4 y
2
C 4
Circle
Ellipse 73. 9 x2 A
9 y 2
C 9
Circle
36 x
6 y
34
0
75.
3 x2 A
2 y2
6 x
3
6
6 x
4 y
5
0
3, C
Hyperbola
3 x2
2, AC
< 0
2 y2
0
182
Chapter 9
Conics, Parametric Equations, and Polar Coordinates 79. (a) A hyperbola is the set of all points x, y for which the absolute value of the difference between the distances from two distance fixed points (foci) is constant.
77. (a) A parabola is the set of all points x, y that are
equidistant from a fixed line (directrix) and a fixed point (focus) not on the line. (b) x
h2
4 p y
k or y
k 2
4 p x
h
(b)
x h2 a2
(c) See Theorem 9.2. (c) y
83. y
81. Assume that the vertex is at the origin. x2
32 4 p1 9 4
h or y
1 or
y k 2
k ±
a2 a x b
x
h2
b2
1
h
ax2
The equation of the tangent line is
p
The pipe is located
9 4
ax02
Let y
0. Then:
meters from the vertex.
2
2ax0 x
ax02
2ax0 x
Therefore,
x0 2
ax0
3
Focus
2ax0 x
y
y
2
(− 3, 1)
k 2
b2
b x a
k ±
y 2ax
4 py
y
(3, 1)
x
or y 2ax 0 x ax 02.
x0
2ax02
is the x-intercept.
y
1
x
−3
−2
−1
1
2
3
( x0, ax02 )
y = ax 2
x
x0
( 2 , 0)
85. (a) Consider the parabola x2 4 py. Let m0 be the slope of the one tangent line at x1, y1 and therefore, the second at x2, y2. From the derivative given in Exercise 32 we have: m0 1
m0
1 x or x1 2 p 1
1 x or x2 2 p 2
m0 is the slope of
1
2 pm0 2 p
m0
Substituting these values of x into the equation x2 4 py, we have the coordinates of the points of tangency 2 pm0, pm02 and 2 pm0, pm02 and the equations of the tangent lines are
y pm02 m0 x 2 pm0 and
y
p m02
1
m0
x
2 p . m0
The point of intersection of these lines is
pm02 1 , p and is on the directrix, y m0
p.
y
x 2 = 4py 2 p p − , 2 m0 m0
(
) (2 pm0, pm02) x
y = − p
( p(mm − 1) , − p) 0
0
—CONTINUED—
Section 9.1
Conics and Calculus
85. —CONTINUED—
(b) x2
4 x
4 y
8
0
x 22 4 y 2 x
4
4
dy dx
0
dy dx
1 x 2
At 2, 5, dydx
1
At 3, 4 , dydx 5
2.
Tangent line at 2, 5: y
Tangent line at 3, 4 : y
5
5
Since m1m2
2 12
87. y dy dx
x x
1
3
⇒
x1 x12
1
5 x 2
1 x 2
x
1 2
y
0
m
1
4 y
1
1
0.
0.
5 4
12 , 0 lies on this line.
2 x1. Also, m
1
1
1 2 x1 x1 1
x1
1
2 x1
2
x1
1
2 x1
2 1
01 x0 1
Thus, x0
y
1 4
1
1
1
3
2 3
3
2 3 3
2
2 ±
1 . 1
x1
1
22 41 2 21
2 ±
2 3
2
1
1 ±
3
3.
3 3 2 3
y
2
1
x0
3
y1 x1
0
2 x1
Choosing the positive value for x1, we have x1
2 x
2 x
the lines are perpendicular.
1 1
m
⇒
2
y1 x1
x12
0 and the point of intersection
At x1, y1 on the mountain, m
2
1
2
2 x
x1
4 2 x 1,
1
2.
2 x
5
2 x
Point of intersection:
Directrix: y
1. Vertex 2, 1
1
(− 1, 1)
2 3
−2
1
−1
1
−1
x0
1
−2
1
x0
2 3 3
x0.
The closest the receiver can be to the hill is 2 33
1 0.155.
( x1, y1) ( x0, 0)
x
183
184
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
89. Parabola Vertex: 0, 4 x2
4 p y
4
42
4 p0
4
p
1
Circle Center: 0, k Radius: 8 y
x2
y
k 2
42
0 k 2 64
x
x
2
y
−6
y 4
4
− 4 −2
2
−2
k 2
6
4
8
−6
x2
−8
y 4 3
2
y
Since the y-value is positive when x
4
A
2
x3 12
64 12
0
2 16 2 4 x
16 4
x2 4
4
3 3 3
91. (a) Assume that y
64
48
3
(Center is on the negative y-axis.)
3
±
4
64
4
3 64 x2.
0, we have y
4
2
dx
16 3
2
1 x 64 2 2 48
x
2
64 arcsin
32 arcsin
1 2
x 8
4
0
15.536 square feet
ax2.
y
(− 6 0, 20 )
20 (b) f x
a602
⇒
1 2 x , f x 180
a
2 360
1 x 90
0
93. x2
2 1 x 902 90 2
x
1 60 11,700 90
1 1800 13 90
20 13
10 2 13
4 py, p
2
2
dx
2 90
902 ln x
902 ln 60
902 ln 60
60 2 9 ln
90 ln
⇒
y
15
902 x2 dx
x
−60 − 45 −30 −15
0
5
60
902 x2
60
(formula 26) 0
11,700 902 ln 90
30 13
902 ln 90
30 13 90
13 128.4 m 3
1 1 3 , , 1, , 2 4 2 2
95.
y
As p increases, the graph becomes wider. 1
y
p =
1 4
24
p = p =
3 2
1 2
x
−16
−8
8
2
3
p = 1 p = 2
16
(60, 20) 20
1 2 x 180
10
1
1 180
1 x 90
60
S 2
64 x2
4 3 64 x
4 3 x
k
−4
x2 4 4
17
16
15
14
13
12
11
4
10
5
9
6
8
7
7
8
6
9
5
10
4
11
3
12
2
13
1
14
15
16
17
x
15 30 45 60
Section 9.1
97. a
5 ,b 2
2, c
5 2
2
3 2
22
The tacks should be placed 1.5 feet from the center. The string should be 2a
e
c a
P
2a
99. A
5 feet long.
y
c
x
c
a
P
e
c a
A P2 A P2
A A
P P
35.34au 35.34au
0.59au 0.9672 0.59au
y 2 52
1
103.
x2 102
P
A
2
A
P
P
2
A A
A
P
2
P P
105. 16 x2
9 y 2
96 x
36 y
18 yy 96
y 18 y 2
5 x 2
10 y
At 8, 3: y
32 x
2 yy 0 52 y
(a, 0)
P
2 x 102
A
a
101. e
185
Conics and Calculus
8 12
36
y 0 when x
2 3
At x 3
3, y
2 or
32 x 96 32 x
18 y
96 36
undefined when y
2.
6.
Endpoints of major axis: 3, 2, 3, 6
2
x 8. It 2 25 will cross the y-axis when x 0 and y 3 8 3 3 . The equation of the tangent line is y
3. y is
0
y
4 y
36 y 0
x
36
3
At y
2, x
0 or
6.
Endpoints of minor axis: 0, 2, 6, 2 Note: Equation of ellipse is
2
107. (a) A
4
0
1 4 2
x
2
dx
x 4
2
V 2
(b) Disk:
0
y
y
1 y 2
2
S 22 y
1 4 2
—CONTINUED—
4 y
1 4 4 x
x
1 4 2
4 arcsin
dx
2
y 22
2
x 2
9
0
x
2 or, A
1 3 x 3
2
0
ab 21
2
8 3
2
x
2 4
x
16 3 x2
0
2
x
x 32
1
dx
2
x2
16
2 3
2
4 x
16
3 x2
4 y
3 x 16 3 x2 16 arcsin
3 x
4
2
0
2 9 9
4 3 21.48
16
1
186
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
107. —CONTINUED—
2
V 2 x 4
(c) Shell:
2
x
2
dx
0
x
x
1 x 2
2 1
2 1
2 3 8
y
1
y
2
4 y 2 1 y 2
a
x2 25
1 3 y 2 1 y 2
8 1
3 y 2 dy
1 3 y 2
1
0
4 6 3
7, b
2100
49 25 2 6, e
2
C 47
1
0
0
16 3
r 2
100
ab a10
10 a
⇒
a
20
Hence, the length of the major axis is 2 a
1, we have
Area ellipse
1 e2 sin2 d
5, c
3 ln 2 3 34.69
111. Area circle
2
2
0
y2 49
1
1 3 y2 dy 1 y 2
3 y 1 3 y 2 ln 3 y
0
For
2
109. From Example 5, C 4a
2 3 2
x
1 y 2
0
0
2 4 3
2 y
1
S 22
4 x212 dx
2 x
c a
40.
2 6 . 7
24 sin2 d 49
281.3558 37.9614 113. The transverse axis is horizontal since 2, 2 and 10, 2 are the foci (see definition of hyperbola).
115. 2a c
Center: 6, 2 c
4, 2a
6, b2
c2
a2
10
6
⇒
a
b
⇒
5
11
7
Therefore, the equation is
x
9
62
y 22 7
1.
117. Time for sound of bullet hitting target to reach x, y:
Time for sound of rifle to reach x, y:
Since the times are the same, we have: 4c2 vm2
4c x vmvs
c2
x c2 y 2
y
vs ( x, y)
x c2 y 2 vs
2c vm
x c
2
y
2
1 x2 c vs2vm2
y
2
x c2 y 2 2 s
v
c v 2
c2
vm2 2 x vs2
2 m
y
vs
x c2 y 2 vs2
2
vm2 x vs2c vsvm
2
1
y
y
x c
2
vs
x
2
2c vm
y 2 2 2 vs vm
vs2 vm2
1 c2
2
(− c, 0) rifle
(c, 0) target
x
Section 9.1
119. The point x, y lies on the line between 0, 10 and 10, 0. Thus, y x236 y264 1. Using substitution, we have: x2 36
16 x2 7 x2
64
910
10 x2
180 x
x
x. The point
10
Conics and Calculus
187
also lies on the hyperbola
1
2 576
1476
x
0 180 ±
1802 47 1476 27
192 2 14
180 ±
90 ±
96 2
7
Choosing the positive value for x we have: x
90
x2 a2
2 y 2 b2
1
⇒
2 y 2 b2
2 y 2 b2
1
⇒
2 y 2 b2
1
⇒
2
x
1
2a2a2 b2 2a2 b2
⇒
x
±
2a a2 b2 2a2 b2
x2 2
a
2 y2 b2
2
b
x2 a2
x2
1
y2
x2 2
a
b2
1 2a2c 2 a2 2a2 b2
b4
22a2
b2
2 y 2 b2
⇒
y
⇒
x2 a2
At
b2
1
a
a2
b2
b2
1
1 a2
a2
b2
±
2ac 2a2 b2
b2
±
x2 2 , c a2
x2 2
2
There are four points of intersection: x2 a2
160 96 2 3.462 7
6.538 and y
7
121.
1
96 2
2a2
b2
b2
2 2a2 b2
2ac b2 ,± , 2a2 b2 2 2a2 b2
2 y2 b2
1
⇒
2 x a2
4 yy 0 b2
⇒
y e
2 y 2 b2
1
⇒
2 x c2
4 yy 0 b2
⇒
y h
2ac b2 ,± 2a2 b2 2 2a2 b2
b2 x 2a2 y
b2 x 2c 2 y
2ac b2 , the slopes of the tangent lines are: , 2a2 b2 2 2a2 b2 b
y e
2a2
2
2ac 2a2 b2
2 2
b
2 2
a
2
b
c a
b2
and
y h
2c 2
2ac 2a2 b2
2 2
b
2
a2
b2
a c
Since the slopes are negative reciprocals, the tangent lines are perpendicular. Similarly, the curves are perpendicular at the other three points of intersection.
123. False. See the definition of a parabola.
127. False. y2 lines.
x
2
2 x
2 y
0 yields two intersecting
125. True
129. True
188
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
Section 9.2 1. x
t , y
Plane Curves and Parametric Equations
1
t
(b)
(a)
t
0
1
2
3
y
4 1
x
0
2
1
3
2
x −1
y
1
1
0
2
3
1
2
3
−1
−2
(c)
2 −3
3
−1
(d) x2
t
y
1
−4
x2, x ≥
0
3t 1
5. x
t 1
y
2t 1
y
t 2
y
2 t 2
y
2
y
x 12
x
y
2 x 23
2 x
3 y
x
1
3
5
1
7. x
y
t 3
3. x
1
t 3 implies t x13 1
0
y
4
y
1 4 x
2
3
2
1
1
13. x
2t
y
t 2
y
2
3
x
2
2
4
x
4
2 2
9. x
t, t ≥ 0
y
y
x2
11. x
y
t 2
2, x
≥
0 y
y
t 1 t t 1 x
1
x
x 2
4
2
x 4 2
y y
3
8
2
4 4
1 x 2
3
4
5
2
6
x
4
−1 −2
x
2
15. x
et , x
y
e3t 1
y
3
x
>
y
0
1, x
2
5 4 >
0
3 2 1 x −2
−1
1 −1
2
3
4
4
8
12
Section 9.2
17. x
sec
y
cos
0
≤
xy
y
x
19. x
2
,
3 cos , y
<
2
≤
x2
y 2
3 sin
x
4 sin 2
y
2 cos 2
x2 16
sin2 2
y 2 4
cos2 2
y 2 4
1
21.
9.
y
1 4
1 x
2
1, y
≥
Squaring both equations and adding, we have
<
Plane Curves and Parametric Equations
≤
x2 16
x
1
4
2
2
4
4
2
y
3
4 −6
2
6
1 x
1
2
3
−4
2 3
23.
x
y
1
x 42 4
y 12 1
x 42 4
y 12 1
4
2 cos
cos2
sin2
1
25.
sin
y
1
x 42 4
y 12 16
x 42 4
2
y 12 16
4
8
cos2
sin2
1
−5
x
4 sec
29. x
t 3
y
3 tan
y
3 ln t
x2 16
sec2
y
3 x 3 ln
31.
ln x
2
x2 16
y 2 9
et
y
e3t
1 x
3 et y
tan2 −1
x
et
2
4 sin
10
−2
−4
y 9
2 cos
3
−1
27.
x
3 y
5
1 −2
6
1 x
y
1 x3
x > 0 y > 0
−9
9 3
−6
−1
5
−1
189
190
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
33. By eliminating the parameters in (a) – (d), we get y 2 x domains. These curves are all smooth except for (b).
(a) x
t , y
2t 1
1. They differ from each other in orientation and in restricted (b) x
y
cos
1 ≤
3
x
≤
1 ≤
1
2 cos 1 y
≤
3
dx dy 0 when 0, ± , ± 2 , . . . . d d
2
y
1 x −2
y
−1
1
3
2
−1
2 1 x −2
−1
1
2
−1
(c) x
et
y
x > 0
2et 1
(d) x
y > 1
e t
y
x > 0
y
2et 1
y > 1
y
4
4
3
3
2
2
1
1 x
−1
1
2
x
3
−1
1
35. The curves are identical on 0 < < . They are both smooth. Represent y 37. (a)
4
2
21
3
x2
(b) The orientation of the second curve is reversed.
4
(c) The orientation will be reversed. −6
6
−6
(d) Many answers possible. For example, x 1 y 1 2t , and x 1 t , x 1 2t . −4
39.
x1
t x2
x1
y
y1
t y2
y1
y
y
y1
y
y1
41.
x
t
y2 x2
x
h
y
k b sin
cos
sin
1
h
a
y1
x x1 y2 x2 x1
y1 x1
m x
−4
x
x x1 x2 x1
6
y
y1
k
b
x h2
x x1
2
a
y k 2 b2
a cos
x1
43. From Exercise 39 we have
45. From Exercise 40 we have
x
5t
x
2
4 cos
y
2t .
y
1
4 sin .
Solution not unique
Solution not unique
47. From Exercise 41 we have a
5, c
4
⇒
x
5 cos
y
3 sin .
b
Center: 0, 0 Solution not unique
3
t ,
Section 9.2
51. y
49. From Exercise 42 we have a
4, c
5
⇒
b
3
3 x
Plane Curves and Parametric Equations
2
53. y
Example
x3
Example
x
4 sec
x
t ,
y
3t 2
x
y
3 tan .
x
t 3,
y
3t 11
x
y
t 3
3 t,
y
t
x
tan t ,
y
tan3 t
sin
59. x
3 cos3
2 cos
y
3 sin3
Center: 0, 0 Solution not unique 55. x
2 sin
y
21
57. x
cos
y
3
2
1
3
t ,
5
5
4
−6
−2
−2
16
6
7
−1
−1
−4
Not smooth at x, y ± 3, 0 and 1 0, ± 3, or 2 n .
Not smooth at 2n
61. x
2 cot
y
2 sin2
63. See definition on page 665.
4
−6
6
−4
Smooth everywhere 65. A plane curve C , represented by x f t , y gt , is smooth if f and g are continuous and not simultaneously 0. See page 670.
67. x
4 cos
y
2 sin 2
69. x
cos sin
y
sin cos
Matches (d)
Matches (b)
71. When the circle has rolled radians, we know that the center is at a , a.
sin sin180 cos Therefore, x
C b
cos180
BD or BD b sin b
a
b
t 2
⇒
x
≥
0
x
t 2
⇒
y
≥
0
θ
b cos .
The graph of the parametric equations is only a portion of the line y
C
A
B
73. False
y
P
AP or AP b cos b
a b sin and y
x
191
x.
a D
x
192
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
75. (a) 100 mihr
1005280 3600
440 ftsec 3
(d) We need to find the angle (and time t ) such that
440 x v0 cos t cos t 3
x
y h v0 sin t 16t 2
y
(b)
4403 sin
3
4403 cos 400 3 4403 sin 16 t
t
t 2 10.
From the first equation t 1200440 cos . Substituting into the second equation,
t 16t 2
30
10 3
120 16 sec 44 120 tan 1. 16 44
2 440 1200 1200 16 sin 3 440 cos 440 cos 2
7 400 tan 0
400 0
2
2
It is not a home run—when x 400, y (c)
≤
20.
400 tan
2
60
We now solve the quadratic for tan : tan 120 44 2
16 0
2
tan 0.35185
400 0
⇒
400 tan 7 16
120 44
2
19.4
Yes, it’s a home run when x 400, y > 10.
Section 9.3 1.
dy dx
dydt dxdt
Parametric Equations and Calculus 4
2t
2
t
3.
dy dydt 2 cos t sin t 1 dx dxdt 2 sin t cos t
Note: x 5. x 2t , y 3t 1
y
1
⇒
y 1 x and
7. x t 1, y t 2 3t
dy dy dt 3 dx dx dt 2
dy 2t 3 1 when t 1. 1 dx
d 2 y 0 Line dx2
d 2 y 2 concave upwards dx2
9. x 2 cos , y 2 sin
dy 2 cos cot 1 when . dx 2 sin 4 csc2 csc3 d 2 y 2 when . 2 2 sin dx 2 4 concave downward
dy 1 dt
11. x 2 sec , y 1 2 tan dy 2 sec2 dx sec tan
2 sec 2 csc 4 when . tan 6
d 2 y 2 csc cot 2 dx sec tan 2 cot3
concave downward
6 3
when
. 6
0
Section 9.3
13. x cos3 , y sin3
15. x 2 cot , y 2 sin2
dy 3 sin2 cos dx 3 cos2 sin tan
1
dy 4 sin cos 2 sin3 cos dx 2 csc2
when
. 4
At
d 2 y sec2 1 2 2 4 3 cos sin 3 cos sin dx
193
Parametric Equations and Calculus
dy 3 2 3 3 , , and . dx 3 8 3 2
2
,
y
Tangent line:
3 3 3 2 x 2 8 3
3 3 x 8 y 18 0
sec4 csc 4 2 when . 3 3 4
At 0, 2,
concave upward
dy , and 2 dx
0.
Tangent line: y 2 0
At 2 3,
3 dy 1 , , and . 2 6 8 dx
y
Tangent line:
3 x 8 y 17. x 2t , y t 2 1, t 2
(a)
10 0
19. x t 2 t 2, y t 3 3t , t 1
(a)
10
5
8
−1 −6
3 1 x 2 3 2 8
6
−4
−3
(b) At t 2, x, y 4, 3, and
(b) At t 1, x, y 4, 2, and
dx dy dy 2, 4, 2 dt dt dx
dx dy dy 3, 0, 0 dt dt dx
(c)
dy 2. At 4, 3, y 3 2 x 4 dx y 2 x 5
(d)
(c)
dy 0. At 4, 2, y 2 0 x 4 dx y 2
(d)
10
5
(4, 2)
(4, 3) −1 −5
8
5
−4
21. x 2 sin 2t , y 3 sin t crosses itself at the origin, x, y 0, 0.
At this point, t 0 or t . dy 3 cos t dx 4 cos 2t
At t 0:
dy 3 3 and y x. Tangent Line 4 4 dx
At t ,
3 dy 3 x Tangent Line and y dx 4 4
−3
194
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
23. x cos sin , y sin cos
Horizontal tangents:
dy sin 0 when 0, , 2 , 3 , . . . . d
Points: 1, 2n 1 , 1, 2n where n is an integer. Points shown: 1, 0, 1, , 1, 2
3 5 dx cos 0 when , , ,. . .. d 2 2 2
Vertical tangents: Points:
1n 12n
2
Points shown:
5 1 , , 1 2
1 , 1n 1
2 , 1,
3 , 2
25. x 1 t , y t 2
Horizontal tangents:
27. x 1 t , y t 3 3t dy 2t 0 when t 0. dt
Horizontal tangents:
Point: 1, 0
dy 3t 2 3 0 when t dt
± 1.
Points: 0, 2, 2, 2 dx 1 0; none dt
Vertical tangents:
dx 1 0; none dt
Vertical tangents:
3
3
(2, 2)
−4 −2
5
4
(1, 0)
(0, − 2)
−1
−3
29. x 3 cos , y 3 sin
Horizontal tangents:
3 dy 3 cos 0 when , . d 2 2
31. x 4 2 cos , y 1 sin
Horizontal tangents:
Points: 0, 3, 0, 3 dx 3 sin 0 when 0, . d
Vertical tangents:
3 dy cos 0 when , . d 2 2
Points: 4, 0, 4, 2 dx 2 sin 0 when x 0, . d
Vertical tangents:
Points: 3, 0, 3, 0
Points: 6, 1, 2, 1 2
4
(0, 3) (4, 0)
0 −6
6
(− 3, 0)
(2,
9
(6,
− 1)
35. x t 2, y 2t , 0
≤
(3, 0)
− 1)
(4,
− 2)
(0, − 3) −4
−4
33. x sec , y tan
Horizontal tangents:
dy sec2 0; none d
Vertical tangents:
dx sec tan 0 when x 0, . d
−4
6
2
dy dt
2
4t 2 4 4t 2 1
t 2 1 dt
0
4
(1, 0)
2
2
s2
(− 1, 0)
≤
dx dy dx 2t , 2, dt dt dt
Points: 1, 0, 1, 0
−6
t
t t 2 1 ln t t2 1
2 5 ln 2 5 5.916
2
0
Section 9.3
37. x et cos t , y et sin t , 0
2
t ≤
≤
s
dx dt
0
2
dy dt
2e
2
2t
dt 2
0
dt
et 1 dt
0
2
2et
0
21 e
1 1 9 dt 4t 2
0
2
2
1
S
1.12
2
dx 3a cos2 sin , d
dy 3a sin2 cos d
1 6
1 u2 du
0
1 ln 37 12 3
t
0
6 6 37 3.249
dt
dx dy a1 cos , a sin d d
9a2 cos4 sin2
9a2 sin4 cos2 d
S 2
a2 1 cos 2
a2 sin2 d
0
2
sin cos c os2 sin2 d
12a
2 2a
0
1 cos d
0
2
6
43. x a sin , y a1 cos ,
0
0
1 36t dt t
1 ln 1 u2 u u 1 u2 12
2
S 4
1
6
u 6 t, du
41. x a cos3 , y a sin3 ,
195
1 dx dy 3 , dt 2 t dt
39. x t, y 3t 1,
dx dy et sin t cos t , et cos t sin t dt dt 2
Parametric Equations and Calculus
0
2
sin 2 d 3a cos 2
6a
0
6a
2 2a
0
sin
1 cos
4 2a 1
d
cos
0
8a
45. x 90 cos 30t , y 90 sin 30 t 16t 2
(a)
(d) y 0
35
⇒
90 sin t 16t 2
x 90 cos t
0
240
x
0
(b) Range: 219.2 ft
⇒
t
90 sin 16
902 902 cos sin sin 2 16 32
902 2 cos 2 0 32
45
⇒
By the First Derivative Test, 45
4
maximizes the range. (c)
dx dy 90 cos 30 , 90 sin 30 32t . dt dt y 0 for t s
45 . 16
dx 90 cos , dt
45 16
90 cos 30 2 90 sin 30 32t 2 dt
s
0
230.8 ft
dy 90 90 sin 32 90 sin 32 sin 90 sin dt 16
9016sin
90 cos 2 90 sin 2 dt
0
9016sin
0
9016sin
90 dt 90t
0
902 sin 16
ds 902 cos 0 d 16
⇒
2
By the First Derivative Test, 90 maximizes the arc length.
196
Chapter 9
47. (a)
Conics, Parametric Equations, and Polar Coordinates
x t sin t
x 2t sin2t
y 1 cos t
y 1 cos2t
0
t ≤ 2
≤
0
≤
t ≤
3
3
1
1
y 1 cos 2 t 1
The time required for the particle to traverse the same path is t 4 .
3
−
−1
−1
49. x t , y 2t ,
dx dy 1, 2 dt dt
51. x 4 cos , y 4 sin ,
4
0
2
0
4
32 5
0
2
32
2
cos d 32 sin
0
4
32
0
4
4 dt 2 5 t 1
(b) S 2
4 cos 4 sin 2 4 cos 2 d
S 2
t dt
0
2 5 t 2
dx dy 4 sin , 4 cos d d
4
4 dt 4 5 (a) S 2 2t 1
(c) x 2 t sin 2 t
3
−
(b) The average speed of the particle on the second path is twice the average speed of a particle on the first path.
0
t dt
0
5 t
4
2
0
16 5
53. x a cos3 , y a sin3 ,
dx dy 3a cos2 sin , 3a sin2 cos d d
2
S 4
0
57. One possible answer is the graph given by
y
4
2 x −2
2
4
−2
−4
61. x r cos , y r sin y
r sin r sin r cos d 2
2
2
2
0
2 r 2
θ
sin d
0
x
2 r 2 cos
0
2 r 21 cos
a
dx dt
2
12 2 a 5
dy dt
2
dt
See Theorem 9.8, page 678.
x t , y t .
−4
59. s
b
See Theorem 9.7, page 675.
2
12 a2 sin5 5 0
sin4 cos d
0
dy dy dt 55. dx dxdt
S 2
2
a2 cos4 sin2 9a2 sin4 cos2 d 12a2 a sin3 9
Section 9.3 63. x t , y 4 t , 0 A
4
4 t
0
x
3 16
3 y 32
x, y
≤
1 2 t
4
t
2
0
t
4
163 323 4t t 2 34 3 16 32 t t dt t t 64 3 1 2 8 t t t 2 3
t 12 dt
4t
1 2
0
dt
3 4 t dt 64 2 t 2
0
1 2
1
4 t t
4
4
≤
dt
197
Parametric Equations and Calculus
1
3 32
4
2
4 t dt
0
4
16t
0
0
4
0
8t 12
1 2
4
3 2
4 22 8 t t 5 5 0
34, 85
65. x 3 cos , y 3 sin ,
dx 3 sin d
0
3 sin 2 3 sin d
V 2
2
0
sin3 d
54
2
0
1 cos2 sin d
54
2
54 cos
cos3 0 36 3 2
67. x 2 sin2
y
2
y 2 sin2 tan dx 4 sin cos d A
π
x
2
8
sin4 d
0
sin3
3 2 cos 3 sin cos 4 8 8 0
−1
1 −1
−2
3 2
71. 6 a2 is area of cardioid (f).
69. ab is area of ellipse (d).
75. (a) x
−2
2
2 sin2 tan 4 sin cos d 8
0
1
0≤θ< 2
1 t 2 2t , y , 20 2 1 t 1 t 2
1 t 2 1 t 2
8 3 ab
is area of hourglass (a).
2
≤
t ≤ 20
The graph is the circle x 2 y 2 1, except the point 1, 0. Verify: x2 y 2
73.
2
2t 1 t 2
2
−3
1 t 22 1 2t 2 t 4 4t 2 1 1 t 22 1 t 22
(b) As t increases from 20 to 0, the speed increases, and as t increases from 0 to 20, the speed decreases. 77. False
d g t d y dt f t f t g t g t f t dx2 f t f t 3 2
3
−2
198
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
Section 9.4 1.
Polar Coordinates and Polar Graphs
4, 2 x
y
4 sin 2 4 cos 2
x, y
3.
3
5. 2, 2.36
4 sin 3
4 cos 3
0
x
4
y
0, 4
4,
x, y
2 cos2.36
y
2 sin2.36 0.996
x, y
2 3
π
2
2
1.004
1.004, 0.996 π
2
2, 2 3
π
(4, 36 )
2
x
( 2, 2.36 )
π
(
− 4, −
π
3
0
)
1
0 1
2
3
0 1
7. r ,
5, 3 4
x, y
3.5355, 3.5355
9. r ,
3.5, 2.5
11. x, y
x, y
2.804, 2.095
r
1 4 −1
1
−1
2
3
2, 5 4
y
2
−2
(2.804,
x −2
5 , , 2, , 4 4 4
−1
1
−3
2
x
3 2
−4
±
1, 1
tan 1
y y
(− 3.54, 3.54)
1
− 2.095)
−3
(1, 1)
−1
1
x
1
13. x, y r
tan
±
3, 4
2
y
9 16 ± 5
5
4 3
(− 3, 4)
4 3
2.214, 5.356, 5, 2.214, 5, 5.356
2 1 x −4
15. x, y
r ,
3, 2 3.606, 0.588
19. (a) x, y
y
4, 3.5 4
−3
−2
−1
1
17. x, y
52 , 43
r ,
2.833, 0.490
(b) r ,
4, 3.5
π
2
(4, 3.5) 0
3
1
2
(4, 3.5) 1 x
1
2
3
4
Section 9.4 21. x2
2
y
a2
23.
π
2
r a
Polar Coordinates and Polar Graphs
y
4
π
2
r sin 4 r 4 csc a
0 0 2
25.
3 x
2
0
3r cos r sin 2
0
y
r 3 cos sin
y 2
27.
9 x
r 2 sin2 9r cos
2
r
4
r 2
9 cos sin2
r 9 csc2 cos
3 cos sin
π
2
π
2
0 1
0 1
29.
x2
9
2
9
y
3
4
5
6
7
2
31.
r 3 r 2
2
33.
r sin r 2
y
x2 2
x2
1
x2
y
y
y
2
1 2
2
r sin
tan r tan
y 2
y
r
x2 tan
1 4
y
2
y x y x
x2 y 2 arctan
0 y
x
2
1
1
y
2
2 π
1 2
π
x 1 2
2 π
π − π
x 1 2
35.
r 3 sec
x
x
3
3
0
− 2 π
37. r 3
y
r cos 3
1 2
0
3
≤
4 cos
6
< 2 − 12
6
2
−6
1
x 1
39. r 2
0
≤
sin
2
41. r
4
< 2 5
−4
2 1 cos
Traced out once on < <
5
5
− 10
−5 −2
199
200
Chapter 9
43. r 2 cos
0
Conics, Parametric Equations, and Polar Coordinates
3 2
2
< 4
≤
45. r 2
−3
0
3
4 sin 2
≤
<
2
2
−3
3
−2
−2
r 2
2r h cos k sin
r 2
2hr cos
2
2hx
2ky
0
y 2 2ky k 2
0
h2
x2 x2
x2 2hx h2
y
2
2hx
k 2
ky
h2
r 2 dy dx
2
Center: h, k
k r sin
51. 2, 0.5, 7, 1.2 2
4
2
20
16 cos
d 22
2 242 cos 3 6
2 cos 3 sin 1 3 cos 2 2 sin
2 cos 3 sin 1 6 cos2 2 sin 3
−8
4
2
Tangent line: y
⇒
3 y
(c) At
57. (a), (b) r 3 sin
5
5
−4
−1
3 2 3, 3
⇒
3 4 3, 94 3 3 3 4
x, y
9 4
y
cos
r , 3,
dy , 3 dx
−4
(c) At
227 cos0.5
4
dy , 0. 2 dx dy 2 At 2, , . dx 3 3 dy 0. At 1, , dx 2
3 sin 3 sin
At 5,
Tangent line: y
72
53 28 cos 0.7 5.6
55. (a), (b) r 31
3 sin
2 5 4.5 2
3 cos sin cos 2 3 cos cos sin 2
r ,
k 2
k 2
k 2
4, 2 3 , 2, 6 d
53.
y
x h2 y
49.
Radius: h2
r 2h cos k sin
47.
3
x
3 x
9 2
1.732.
dy , 2 dx
x, y
0, 3
1 x x
1.0.
3
0
1.2
Section 9.4 59.
r 1
dy 1 d
61.
sin
2 sin
cos 0, sin
1 2
Horizontal tangents: dx 1 d
0
⇒
3 5 , , , 2 2 6 6
2, 3 2 , 12, 6 , 12, 5 6
3 cos 0
3 , 2 2
5, 2 , 1, 3 2
Horizontal:
sin sin cos cos
sin
r 2 csc 3 dy 2 csc 3 cos 2 csc cot sin d
sin cos cos sin
cos 1
201
Polar Coordinates and Polar Graphs
sin2 sin2 1
2 sin2 sin 1
2 sin 1sin 1
sin 1, sin
1 2
⇒
0
7 11 , , 2 6 6
32, 7 6 , 32, 11 6
Vertical tangents:
63. r 4 sin cos2
65. r 2 csc 5 10
2
−3
3 −12
12
−6
−2
Horizontal tangents:
2 , 3, 3 2
Horizontal tangents: 7,
0, 0, 1.4142, 0.7854 , 1.4142, 2.3562 67.
r 3 sin
69. r 21
π
sin
π
2
2 2
r
3r sin
y
2
3 y
3 2
Circle r
3 2
Cardioid 0
x 2 x 2
y
2
Symmetric to y-axis, 0
9 4
1
2
3
32
Center: 0,
Tangent at the pole: 0 71. r 2 cos3
Rose curve with three petals
π
2
Symmetric to the polar axis Relative extrema: 2, 0,
2,
2 , 2, 3 3
0 2
0
6
r
2
0
4
2
3 2
Tangents at the pole:
2
2 3
5 6
0
2
0
5 , , 6 2 6
2
2
1
2
3
202
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
73. r 3 sin 2
Rose curve with four petals
π
Symmetric to the polar axis, Relative extrema:
± 3,
, 4
Tangents at the pole: 0,
± 3,
2
, and pole 2
5 4
0
3
2
, 3 2 give the same tangents. 75. r 5
77. r 41
π
2
y
2
π
2
Cardioid
Circle radius: 5 x2
cos
25 0
0 2
79. r 3
4
2 cos
2
6
81.
π
r 3 csc r sin 3
Symmetric to polar axis 0 1
y 0
3
2
2 3
2
3
4
5
3
Horizontal line
2
0 1
83. r 2
π
2
Spiral of Archimedes Symmetric to
2
0
4
2
3 4
5 4
3 2
r
0
2
3 2
2
5 2
3
0 1
Tangent at the pole: 0 85. r 2
10
2
Limaçon
r
6
π
2
4
4 cos2
Lemniscate
π
Symmetric to the polar axis,
, and pole 2
2
Relative extrema: ± 2, 0 0
0
r
±2
±
6
4
2
0
Tangents at the pole:
1
3 , 4 4
2
3
2
Section 9.4
87. Since
89. r
r 2
sec 2
1 , cos
2
Hyperbolic spiral r ⇒ as ⇒ 0
the graph has polar axis symmetry and the lengths at the pole are
Polar Coordinates and Polar Graphs
, . 3 3
r
2 2 2 sin 2 sin ⇒ r r sin y
y
2 sin
x = −1 4
Furthermore,
−6
r ⇒ as ⇒
2
6
lim
→ 0
2 sin 2 cos lim 2 0 1 →
−4
3
r ⇒ as ⇒ . 2 Also, r 2
1 2 cos
y = 2
r 2 r cos
r x
−3
3
−1
rx
r
2 x
r
2 x 1
x
.
Thus, r ⇒ ± as x ⇒ 1.
91. The rectangular coordinate system consists of all points of the form x, y where x is the directed distance from the y-axis to the point, and y is the directed distance from the x-axis to the point. Every point has a unique representation.
The polar coordinate system uses r , to designate the location of a point. r is the directed distance to the origin and is the angle the point makes with the positive x-axis, measured clockwise.
Point do not have a unique polar representation.
97. r 31
95. r 2 sin circle
93. r a circle
Matches (c)
b line
cos
Cardioid Matches (a)
99. r 4 sin
(a) 0
≤
≤
2
(b)
π
2
0 1
2
≤ ≤ 2
(c)
≤ ≤ 2 2
π
π
2
2
0 1
2
0 1
2
203
204
Chapter 9
Conics, Parametric Equations, and Polar Coordinates π 2
101. Let the curve r f be rotated by to form the curve r g . If r 1, 1 is a point on r f , then r 1, 1 is on r g . That is, g 1
Letting 1 g
g 1
r 1
(r , θ + φ )
1.
f
, or 1
1
f
, we see that
(r , θ )
.
f
φ θ 0
103. r 2
sin
(a) r 2
sin
4
2
2 2
sin cos
(b) r 2
cos
2
cos
4 4
−6 −6
6
6
−4 −4
(c) r 2
sin
2
(d) r 2
sin
cos 4
4
−6
−6
6
6
−4
−4
105. (a) r 1
sin
(b) r 1
π
2
sin
4
π
2
Rotate the graph of r 1
0 1
107. tan
2
r 21 cos 2 sin dr d
At , tan is undefined
⇒
. 2
At
3
−6
0 1
r 2 cos 3 dr d 6 sin 3
, tan 0 6
⇒
2
3
−3
sin
through the angle 4.
109. tan
−3
3
−2
0.
2
Section 9.5
6 61 1 cos
r
111.
6 r 1 cos dr 6 sin d 1 cos 2
tan
2 , tan 3
At
1
cos 1
⇒
Area and Arc Length in Polar Coordinates
dr 6 sin d 1 cos 2
7
ψ θ
cos sin
1
−8
7
−3
12
3
3.
2
, 60 3
113. True
Section 9.5
115. True
Area and Arc Length in Polar Coordinates
1. (a) r 8 sin
(b) A
8 sin 2
1 2 2
π 2
2
d
0
2
64
sin2 d
0
2
32
1 cos 2 d
0
0 2
A
3. A
7. A
42
1 2 2
4
6
2 cos 3 d 2
6 1 sin 6 6 3 0
2
5. A
2
1
sin 2 2 2 0
16
16
0
1 2
2
32
2
sin d
2
9. A
2 3 1 2 cos sin 2 2 4 2
3 2
1 2
2
4 2
0
4 1 1 sin 4 2 4 0
2
1 2
cos 2 d
8
2
1 2 cos d
2 3
3 4 sin sin 2
2 3
2 3 3 2
2
−1
4
−2
11. The area inside the outer loop is
1 2 2
2 3 2
2
3
1 2 cos d
0
4 sin sin 2
2 3 0
4 3 3 . 2
−1
4
From the result of Exercise 9, the area between the loops is A
4 3 3 2
2 3 3 2
−2
3 3.
205
206
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
13. r 1
cos
15. r 1
cos
r 1
cos
r 1
sin
Solving simultaneously, 1
cos 1
Solving simultaneously,
cos
1
2 cos 0
3 , . 2 2
Replacing r by r and by in the first equation and solving, 1 cos 1 cos , cos 1, 0. Both curves pass through the pole, 0, , and 0, 0, respectively.
2 , 1, 32 , 0, 0
Points of intersection: 1,
cos 1
cos
sin
tan
1
Replacing r by r and by in the first equation and solving, 1 cos 1 sin , sin cos 2, which has no solution. Both curves pass through the pole, 0, , and 0, 2, respectively.
19. r
5 sin
r 3 sin
2
2
2
,
3 2 2 7 , , , 0, 0 4 2 4
2
π 2
r 2
Solving simultaneously, 4
3 7 , . 4 4
Points of intersection:
17. r 4
sin
5 sin 3 sin sin
1 2
Solving simultaneously, we have 0
2
1
2, 4.
Points of intersection:
2, 4, 2, 4
5 , . 6 6
Both curves pass through the pole, 0, arcsin 4 5, and 0, 0, respectively. Points of intersection:
32, 6 , 32, 56 , 0, 0
21. r 4 sin 2
π 2
r 2 r 4 sin 2 is the equation of a rose curve with four petals and is symmetric to the polar axis, 2, and the pole. Also, r 2 is the equation of a circle of radius 2 centered at the pole. Solving simultaneously,
4 sin 2 2 2
5 , 6 6
5 , . 12 12
Therefore, the points of intersection for one petal are 2, 12 and 2, 5 12. By symmetry, the other points of intersection are 2, 7 12, 2, 11 12, 2, 13 12, 2, 17 12, 2, 19 12, and 2, 23 12.
0 1
3
Section 9.5 23. r 2
3 cos
25. r cos
sec θ 2
r = 4
r 2
sec 2
r
Area and Arc Length in Polar Coordinates
3 sin
Points of intersection:
8
−4
0, 0, 0.935, 0.363 , 0.535, 1.006 −4
The graphs reach the pole at different times ( values).
r = 2 + 3 c os θ
The graph of r 2 3 cos is a limaçon with an inner loop b > a and is symmetric to the polar axis. The graph of r sec 2 is the vertical line x 12. Therefore, there are four points of intersection. Solving simultaneously, 2
6 cos 4 cos 1
r = cos θ
−4
5
sec 2
3 cos
2
1
−5
r = 2
−
3 sin θ
0
10
2 ±
cos
6
arccos
arccos
Points of intersection: 0.581,
10
2
6
2
10
6
± 2.607
1.376
2.6068.
, 2.581, ± 1.376
27. From Exercise 21, the points of intersection for one petal are 2, 12 and 2, 5 12. The area within one petal is
12
1 2
0
12
12 1 8 sin 4 4 0
Total area
29. A
4
1 2
4
12
1 2
2 2
4 sin 2 d
4
5 12
d (by symmetry of the petal)
−6
6
12
0
2
2 d
5 12
sin22 d 2
16
5 12
1 2
2
4 sin 2 d
43
3
2
5 12
12
16 4 3 3
4 3. 3
−4
4 4 3 3 3
2 2
6
3 2 sin d
0
2
2 11 12 cos sin2
0
−9
9
11 24 −6
31. A
2
1 2
6 2
4 sin d
0
6
1 1 sin2 2 4
16
8 2 3 3
6
−3
2 2
2
4
2 d
6
2 4 3 3 3
5
−6
0
1 2
6
33. A
2
1 2
a1 cos 2 d
0
a2 4
3 sin 2 a 2 2 sin 2 4 4 0
a2
3a 2 a 2 5a 2 2 4 4
207
208
Chapter 9
35. A
a 2 8
a 2 8
a 2 8
a 2 8
Conics, Parametric Equations, and Polar Coordinates
1 2
a1 cos 2 d
π 2
2
a2 2
2
a 3 2 2
3 2
2 cos
cos 2 d 2 0
sin 2 a2 3 2 sin 2 2 4 2 2
3 2 4
2a
a
a2 2 2
37. (a) r a cos2
(b)
4
a=6
a=4
r 3
ar 2 cos2 −6
x 2 y 232 ax 2
6
−4
(c) A
2
1 4 2
6 cos
2
10
4 cos d 2
2
0
40
4
cos d
10
0
2
2
2
1
2 cos 2
1
0
2
1 cos 2 2 d
0
2 cos 4 3 1 d 10 sin 2 sin 4 2 2 8 0
15 2
39. r a cosn
For n
For n
1:
r a cos A
a 2
r a cos 2
2
a 2 4
A
a
4
a cos 2 2 d
0
a
3:
6
1 2
0
For n
r a cos 3 A
8
a 2 2
π 2
2
For n
2:
1 2
0
4:
r a cos 4
6
a cos 3 2 d
0
a 2 4
A
π 2
16
1 2
8
a cos 4 2 d
0
a 2 2
π 2
a
0
In general, the area of the region enclosed by r a cosn for n a 22 if n is even.
a
0
1, 2, 3, . . . is a 24 if n is odd and is
Section 9.5 41. r a
43. r 1
r 0 s
Area and Arc Length in Polar Coordinates
sin
r cos 2
a2 02 d a
0
2 0
2 a
s
2
3 2
1 sin 2
cos 2 d
2
(circumference of circle of radius a)
2 2
3 2
1 sin d
2
2 2
3 2
≤
2
≤
1 sin
2
45. r 2 , 0
cos
4 2 1
sin
4 2 2
0
1 47. r , ≤ ≤ 2
d
3 2
2
8
49. r sin3 cos , 0
≤
≤
1 4
0.5
−1 − 0.5
−1
2
0.5
2
−1
Length 4.16
53. r ea
r aea
6 cos2 36 sin2 d 6 cos sin 3
S 2
6 sin
Length 4.39
Length 0.71
51. r 6 cos r
−1
− 0.5
2
S 2
ea cos ea 2
2
72
2 1
a2
e2a cos d
0
2
36 sin 2
0
36
2 1
a2
a2 a 2 1 e 4a2 1
8
2a
sin 2
4
6 cos2 2 64 sin2 2 d 4 cos 2 sin 1
S 2
0
4
32
cos 2 sin c os2 2 4 sin2 2 d 21.87
0
57.
Area
1 2
f d 2
1 2
Arc length
f 2 f 2 d
59. (a) is correct: s 33.124.
r 2 d
r2
dr 2 d d
2
e2a 2a cos sin 4a2 1
55. r 4 cos 2 r
aea 2 d
2
sin cos d
0
0
0
2
0
209
210
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
61. Revolve r a about the line r b sec where b > a > 0.`
π
2
f
a
2a
f
0
a
2
b a cos a2 02 d
S 2
a
b
0
0
2 a b a sin
2 a2 b
63. False. f
1 and g
2 0
4 2ab
1
have the same graphs.
65. In parametric form,
b
s
dx dt
a
2
dy dt
Using instead of t , we have x
2
dt .
r cos f cos and y
r sin f sin . Thus,
dx dy f cos f sin and f sin f cos . d d
It follows that
2
dx d
Therefore, s
dy d
2
f 2 f 2.
f 2 f 2 d
Section 9.6
1. r
1
Polar Equations of Conics and Kepler’s Laws
2e e cos
3. r
2 , parabola 1 cos
(a) e
1, r
(b) e
0.5, r
(c) e
1.5, r
1
1 0.5 cos 2
1
3 1.5 cos 2
2 , ellipse cos
6 , hyperbola 3 cos
1
2e e sin 2 , parabola 1 sin
(a) e
1, r
(b) e
0.5, r
(c) e
1.5, r
e = 1.0
4
4
1
1 0.5 sin 2
1
3 1.5 sin 2
e = 0.5
e = 1.5
e = 0.5
−9
−4
9
8
e = 1.5 e = 1.0 −4
−8
2 , ellipse sin
6 , hyperbola 3 sin
Section 9.6
5. r
1
Polar Equations of Conics and Kepler’s Laws
4 e sin
(a)
5
(b)
e = 0.1
− 30
5 − 30
30
30
e = 0.25 e = 0.5 e = 0.75 e = 0.9
− 40
− 40
The conic is a parabola.
The conic is an ellipse. As e → 1, the ellipse becomes more elliptical, and as e → 0 , it becomes more circular. (c)
e = 1.1
80
e = 1 e = 2 − 90
90
− 40
The conic is a hyperbola. As e → 1, the hyperbolas opens more slowly, and as e → , they open more rapidly. 7. Parabola; Matches (c)
13. r
9. Hyperbola; Matches (a)
1
1
15. r
sin
Parabola since e
1
1 3 Vertex: , 2 2
6 2 cos 1
17. r 2
Ellipse since e
1 < 1 2
Vertices: 2, 0, 6,
sin
4
4 2 sin 1
2 12 sin
Ellipse since e
1 < 1 2
π
0 1
r
3 12 cos
π
2
11. Ellipse; Matches (b)
2
2
Vertices:
43, 2 , 4, 32
π
2 0 1
3
0 1
19. r
1
5 2 cos 1
Hyperbola since e Vertices: 5, 0,
5
2 cos
2 > 1
5 , 3
21. r
2
3 6 sin 1
Vertices:
38, 2 ,
π
π
2
2
0 4
6
8
Hyperbola since e
32 3 sin
3 > 1
3 3 , 4 2
0 1
3
4
211
212
Chapter 9
23.
Conics, Parametric Equations, and Polar Coordinates
25.
Ellipse
1
−2
Parabola
2
2 −2
1
−2
−2
1
27. r
1
sin
4
2
Rotate the graph of r
cos
6
Rotate the graph of
1
1
6
29. r
r
sin
counterclockwise through the angle
4
.
2
6 cos
clockwise through the angle
−2
10
−8
4
−6
−3
31. Change to
4
5
: r 5
3 cos
4
33. Parabola
.
e
r
35. Ellipse
r
1 , y 2
1, d 1
ed 1 e sin
1
r
1
e
r
2, x 1
1,
d 1
ed e cos 1
1 cos
39. Parabola
1, d 1
ed e cos 1
Vertex: 1,
2 2 cos
e
2
1, d 2, r
2 1 sin
12 12 sin
1 2 sin 43. Hyperbola
Vertices: 2, 0, 8,
1, x
37. Hyperbola
41. Ellipse
e
.
6
5
2
e
3 16 , d 5 3 1 1
ed e cos
165 35 cos
16 5 3 cos
45. Ellipse if 0
32 , 9, 32
Vertices: 1, e
r
5 9 , d 4 5 ed 1 e sin
1 4
94 54 sin
9 5 sin
if e
<
e
<
1, parabola
1, hyperbola if e
>
1.
Section 9.6 47. (a) Hyperbola e
(b) Ellipse e
(c) Parabola e
1 2
2
1
>
1
<
r 2
3, b
r 2
53. A
1
259 cos
2
4 ,b 5
3
9 1625 cos 2
2
1 2
0
16
1
4, e
213
3
5 3
5, e
5, c
1
4, c
49. a
(d) Rotated hyperbola e
51. a
Polar Equations of Conics and Kepler ’s Laws
9
0
2 3 d 2 cos
1 d 10.88 2 cos 2
55. Vertices: 126,000, 0 , 4119, a
126,000 4119 2
65,059.5, c
65,059.5
4119
60,940.5, e
When 60, r
92.957
106 mi, e
1 e2a 1
e cos
Aphelion distance: a1
40,627 84,000 , d 4119 43,373 40,627
11,004.49 miles.
5.900
r
e 91,404,618 mi
109 km, e
1 e 2 a 1
e cos
5.537 10 9 1 0.2481 cos
Perihelion distance: a1
e 94,509,382 mi
0.2481
Aphelion distance: a1
e
e
4.436
7.364
10 9 km
10 9 km
5.537 109 1 0.2481 cos
(a) A
1 2
9
1 2
248
(b)
59. a
0.0167
92,931,075.2223 1 0.0167 cos
Perihelion distance: a1
61. r
345,996,000 15,004.49. 23,059.5
Distance between the surface of the earth and the satellite is r 4000
r
c a
ed 411984,000 43,373 345,996,000 1 e cos 1 40,627 43,373 cos 43,373 40,627 cos
r
57. a
d 9.341 10 km 10 d 1 5.537 0.2481 cos 21.867 yr 5.537 10 1 0.2481 cos d 10 d 9.341 10 1 5.537 0.2481 cos 5.537 109 1 0.2481 cos
0
9
1 2
0
1 2
2
2
9
18
2
9
2
0
2
9
2
18
0.8995 rad
In part (a) the ray swept through a smaller angle to generate the same area since the length of the ray is longer than in part (b). (c) r
5.537 10 90.2481 sin 1 0.2481 cos 2
9
s
0
2 5.537 109 1 0.2481 cos
2.559 109 km 1.17 21.867 yr s
0.899
2
109 km
108 kmyr
2 5.537 109 1 0.2481 cos
4.119 109 km 1.88 21.867 yr
1
d 2.559
109 sin 0.2481 cos 2
1.3737297
108 kmyr
1
109 sin 2 d 4.119 0.2481 cos 2
1.3737297
109 km
214
Chapter 9
63. r 1
Conics, Parametric Equations, and Polar Coordinates
ed ed and r 2 1 sin 1 sin
Points of intersection: ed , 0, ed ,
r 1:
dy dx
1 1
At ed , 0,
r 2:
dy dx
1 1
ed cos ed cos sin 1 sin 2 sin ed ed cos sin cos 1 sin 2 sin
dy dy 1. At ed , , 1. dx dx
ed ed cos cos sin 1 sin 2 sin ed cos ed sin cos 1 sin 2 sin
At ed , 0,
dy dy 1. At ed , , 1. dx dx
Therefore, at ed , 0 we have m1m2 11 1, and at ed , we have m1m2 1 1 1. The curves intersect at right angles.
Review Exercises for Chapter 9 1. Matches (d) - ellipse
3. Matches (a) - parabola
16 x 2 16 y 2 16 x 24 y 3 0
5.
x 2 x
12
3 4
y
1 3 9 3 1 9 y 2 y 4 2 16 16 4 16 2
x
x
1
2
y
1
2
1 1 , 2
2
Circle Center:
1
3 4
12, 34
Radius: 1 3 x 2 2 y 2 24 x 12 y 24 0
7.
3 x 2 8 x 16 2 y 2 6 y 9 24 48 18
x 4 2 2
y
3 2
3
3 x 2 2 y 2 12 x 12 y 29 0
9.
3 x 2 4 x 4 2 y 2 6 y 9 29 12 18
x
1
Hyperbola
Ellipse
Center: 4, 3
Center: 2, 3
Vertices: 4
±
2, 3
Asymptotes: y 3
±
Vertices:
3 x 4 2
y 3 2 2 2 1 13 12
2, 3
±
2
y
x
1
y
1
2
1 6
2
4
3 4
2 x
6
4
2
(2, − 3)
3
2
Rev iew Ex erc ises for Cha pter 9 13. Vertices: 3, 0, 7, 0
11. Vertex: 0, 2
15. Vertices: ± 4, 0
Directrix: x 3
Foci: 0, 0, 4, 0
Foci: ± 6, 0
Parabola opens to the right
Horizontal major axis
Center: 0, 0
p 3
Center: 2, 0
Horizontal transverse axis
a 5, c 2, b 21
a 4, c 6, b 36 16 2 5
x 22
x2 y 2 1 16 20
y 2 2 4 3 x
0
y 2 4 y 12 x 4 0
25
17.
215
y 2 1 21
5 x 2 y 2 1, a 3, b 2, c 5, e 9 4 3
19. y x 2 has a slope of 1. The perpendicular slope is 1. y x 2 2 x 2
By Example 5 of Section 9.1,
dy 1 5 2 x 2 1 when x and y . dx 2 4
2
C 12
5 sin2 d 15.87. 9
1
0
y
Perpendicular line:
5 1 1 x 4 2
4 x 4 y 7 0 21. (a) V ab Length 12 16 192 ft 3
(b) F 262.4
3
4 3
3 y 9 y 2 dy
3
8 3 9 3 9 62.4 3 2 2 2 2
8 62.4 3 3
3
3
y 9 y 2 dy
3
y 8 3 1 62.4 y 9 y 2 9 arcsin 9 y 2 3 2 3 2 3 3
3
9 y 2 dy 3
3
8 27 7057.274 62.4 3 2
3
(c) You want 4 of the total area of 12 covered. Find h so that
h
2
y
x =
4 9 y 2 dy 3 3
0
h
9 y 2 dy
0
9 arcsin 3
h
y 1 y 9 y 2 9 arcsin 2 3
h
h 9 h 2
9 8
9 . 4
9 − y 2
Area of filled tank above x-axis is 3π.
4
2
h
9 8
0
4 3
x −
2
2 −
2
−
4
Area of filled tank below x-axis is 6π.
By Newton’s Method, h 1.212. Therefore, the total height of the water is 1.212 3 4.212 ft. (d) Area of ends 212 24 Area of sides PerimeterLength
2
16
1
0
256
7 sin 2 d 16 from Example 5 of Section 9.1 16
122 1 167 sin 0 4 1 167 sin 8 2 1 167 sin 4 4 1 167 sin 38 1 167 sin 2 353.65
2
Total area 24 353.65 429.05
2
2
2
2
216
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
23. x 1 4t , y 2 3t t
x 1 4
⇒
y 2 3
y
x 1 4
4
2 1
3 11 y x 4 4
x −
1
1
2
3
5
1
−
4 y 3 x 11 0
−
2
Line 25. x 6 cos , y 6 sin
6 6 x
2
y
27. x 2 sec , y 3 tan
y
2
1
x 22 sec2 1 tan2 1 y 32
4
x 22 y 32 1
2
x2 y2 36
y
x −
Circle
4
−
2
2 −
2
−
4
4
8
Hyperbola
4 2 x −
29. x 3 3 2t 3 5t
31.
x
2 −
2
−
4
4
8
y 4 2 3 2 1 16 9
y 2 2 6t 2 4t
Let
(other answers possible)
4
x 3 2 16
cos 2 and
y 4 2 9
sin 2 .
Then x 3 4 cos and y 4 3 sin .
35. (a) x 2 cot , y 4 sin cos , 0
33. x cos 3 5 cos y sin 3 5 sin
<
<
4
5 − 12
−7
12
8 −4
(b) 4 x2 y 4 4 cot2 4 sin cos
−5
16 csc2 sin cos
16
82 cot
8 x
cos sin
37. x 1 4t y 2 3t
(a)
dy 3 dx 4
(b) t
x 1 4
y
(c) 5
No horizontal tangents
3 3 x 11 y 2 x 1 4 4
4
2 1 x
1
2
3
5
Rev iew Ex erc ises for Cha pter 9
217
39. x 1 t y 2t 3 dy 2 2t 2 dx 1t 2
(a)
(b) t
No horizontal tangents t 0
1 x
y
(c)
y
6
2 3 x
4
2
x
4
41. x y
1
2
y 2 5 sin
1 t 2 2t
(a)
2t
2 dy t 2t 2 t 12t 1 2 (a) 0 when t 1. 2 dx t 2t 2 2 2t 1 2 Point of horizontal tangency: (b) 2t 1
1 x
⇒
t
13 , 1
1 1 1 2 x
3 5 cos dy 2.5 cot 0 when , . dx 2 sin 2 2
Points of horizontal tangency: 3, 7, 3, 3
2
(b)
x 3 2 4
y
2 2
25
y
(c)
8
4
1
1 1 x x 2
12 1 2 x
x
x
4
4 x 2 4 x 2 1 x 2 4 x1 x 5 x 1 x 1 (c)
y
3 2
x −
2
−
1
2 −
1
−
2
3
45. x cos 3 y 4 sin 3
(a)
dy 12 sin 2 cos 4 sin 4 tan 0 when 0, . 2 3 cos sin cos dx
But,
dy dx 0 at 0, . Hence no points of horizontal tangency. dt dt
(b) x 23
y 23 1 4
y
(c) 4
x
4
4
43. x 3 2 cos
2t 1
y
2
2
2
4
4
8
4
1
218
Chapter 9
Conics, Parametric Equations, and Polar Coordinates 49. x r cos sin
47. x cot
y r sin cos
y sin 2 2 sin cos
(a), (c)
dx r cos d
2
−3
3
dy r sin d
−2
2 cos 2 2 sin 2 d
s r
0
dx dy dy 1 4, 1, and d dx 6 d 4
(b) At
,
r
d
0
51. x, y 4, 4
7
4
3 4 2, 4
r 3 cos
−
1
−
2
−
3
−
4
−
5
2
3
4
(4,
−
5
4)
r 21 cos
55.
r 2 2r 1 cos
r 2 3r cos
x 2 y 2 2 ± x 2 y 2 2 x
x 2 y 2 3 x
x 2 y 2 2 x 2 4 x 2 y 2
x 2 y 2 3 x 0
r 2 cos 2 cos 2 sin 2
57.
0
x
7 r , 4 2, , 4
53.
1 1
1 2 r 2
y
r 4 2 4 2 4 2
r 2 2
59.
r 4 r 2 cos2 r 2 sin 2
r 4 cos 2 sec
4 2 cos 2 1
x 2 y 2 2 x 2 y 2
cos1
r cos 8 cos 2 4 x 8
x2 2
x y 2
4
x 3 xy 2 4 x 2 4 y 2 y 2 x 2
44 x x
y 63. x 2 y 2 a 2 arctan x
61. x 2 y 2 2 ax 2 y r 4 a r 2 cos 2 r sin
2
r 2 a 2 2
2
r a cos sin
65. r 4
67. r sec
π
2
Circle of radius 4 Symmetric to polar axis, and pole
cos
π
2
r cos 1, x 1
Centered at the pole 2,
1
0 2
6
Vertical line
0 1
Rev iew Ex erc ises for Cha pter 9 69. r 21 cos
71. r 4 3 cos
Cardioid
Limaçon
Symmetric to polar axis
Symmetric to polar axis
π
π
2
2
0
0
1
0
r
4
2
2
2 3
0
3 3
2
1
0
r
1
73. r 3 cos 2
4
3 5 2
2
2 3 11 2
4
π
2
Rose curve with four petals
Symmetric to polar axis,
2
, and pole
2 ,
Relative extrema: 3, 0, 3, Tangents at the pole:
0 4
3, ,
3, 32
3 4 4 ,
75. r 2 4 sin 2 2 r
±2
π
2
sin 2
Rose curve with four petals Symmetric to the polar axis, Relative extrema:
0
, and pole 2 3 ± 2, , ± 2, 4 4
Tangents at the pole: 0,
77. r
2
2
3 cos 4
79. r 4 cos 2 sec
Strophoid
Graph of r 3 sec rotated through an angle of 4 5
Symmetric to the polar axis r ⇒ as ⇒ r ⇒ as ⇒
2
2
8
−1
4 −1
−6
6
−4
7
219
220
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
81. r 1 2 cos
(a) The graph has polar symmetry and the tangents at the pole are
(b)
3
, . 3
dy 2 sin 2 1 2 cos cos 2 sin cos 1 2 cos sin dx
Horizontal tangents: 4 cos 2 cos 2 0, cos When cos
1
33
±
8
, r 1 2
3 33 1 33 , arccos 4 8
1 32 8
1
±
3 33 , 4
0.686, 0.568
3 33 1 33 , arccos 4 8 3 33 1 33 , arccos 4 8
1 33 8
1 ±
0.686, 0.568
2.186, 2.206
3 33 1 33 , arccos 4 8
2.186, 2.206.
Vertical tangents: 1 sin 4 cos 1 0, sin 0, cos , 4
12,
0, ,
± arccos
(c)
14,
± arccos
1,
0, 3,
1 0.5, ± 1.318 4 2.5
−5
1
− 2.5
83. Circle: r 3 sin dy 3 cos sin 3 sin cos sin 2 dy tan 2 at , 3 2 2 dx 3 cos cos 3 sin sin cos sin 6 dx
Limaçon: r 4 5 sin
3 dy 5 cos sin 4 5 sin cos dy at , 5 cos cos 4 5 sin sin dx 6 dx 9 Let be the angle between the curves: tan
3 39 1 13
Therefore, arctan
2 3 . 3
2 3 49.1. 3
33 8
Rev iew Ex erc ises is es for Cha pter pt er 9
85. r 1 cos , r 1 cos
The points 1, 2 and 1, 3 2 are the two points of intersection (other than the pole). The slope of the graph of r 1 cos is m1
dy r sin r cos cos 1 cos sin cos sin 2 . cos r sin sin 1 cos cos sin sin cos dx r
At 1, 2, m1 1 1 1 and at 1, 3 2, m1 11 1. The slope of the graph of r 1 cos is m2
dy cos 1 cos sin 2 . sin 1 cos sin cos cos dx
At 1, 2, m2 1 1 1 and at 1, 3 2, m 2 11 1. In both cases, m 1 1m 2 and we conclude that the graphs are orthogonal at 1, 2 and 1, 3 2.
89. r sin cos 2
87. r 2 cos A 2
1 2
2 cos 2 d 14.14
0
9 2
A 2
1 2
3
0.10 −3
2
sin cos cos 2 2 d
0
32
6
0.5
−3
− 0.5
0.5
− 0.1
91. r 2 4 sin 2
1 A 2 2
93. r 4 cos , r 2
2
1 A 2 2
4 4 sin 2 d
0
0
1 4 d 2
2
−3
−3
3
6
−2
−3
95. s 2
a 21 cos 2
a 2 sin 2 d
0
2 2 a
1 cos d 2 2 a
0
97. r
0
2 ,e1 1 sin
Parabola
sin
1 cos
d 4 2 a1 cos 12
99. r
0
π
π
2
2
0 2
0 2
4
6
8
8a
6 2 2 ,e 3 2 cos 1 23cos 3
Ellipse
4 cos 2 d 4.91 3
3
2
3
221
C H A P T E R 9 Conics, Parametric Equations, and Polar Coordinates
Section 9.1
Conics and Calculus . . . . . . . . . . . . . . . . . . . . 424
Section 9.2
Plane Curves and Parametric Equations . . . . . . . . . . 434
Section 9.3
Parametric Equations and Calculus . . . . . . . . . . . . 439
Section 9.4
Polar Coordinates and Polar Graphs . . . . . . . . . . . . 444
Section 9.5
Area and Arc Length in Polar Coordinates . . . . . . . . .452
Section 9.6
Polar Equations of Conics and Kepler’s Laws . . . . . . . 458
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469 469
C H A P T E R 9 Conics, Parametric Equations, and Polar Coordinates Sectio Sec tion n 9.1
Conics Conics and Calcul Calculus us
Solutions to E ven-Numbered Exercises
2. x2
8 y
4.
Vertex: 0, 0 p
x2 9
y2 9
22 16
y 12
4
1
Center: 2, 1 Ellipse Matches (b)
2 > 0
Opens upward Matches graph (a).
6.
x x
1
8.
Circle radius 3. Matches (g)
x x
22
9
y2 4
1
Hyperbola Center: 2, 0 Horizontal transverse axis. Matches (d)
10. x2
12. x x
8 y
0
x2
y 4 2 y
8 y
x x (0, 0) x
−8
Focus: 0, 2
12
y
Vertex: 0, 0 Directrix: y
−4
8
4
2
0
12
4 2 y
2
Vertex: 1, 2
y
Focus: 1, 4
−4
2
Directrix: y
x −8
0
−4
4
−8
−4
− 12
−8
8
(1, − 2)
− 12
14. y 2
6 y
y 2
8 x
6 y
25
9
16. y 2
0
8 x
25
4 y
y 2
9
y 32 4 2 x x 2
8 x
4 y
8
Focus: 0, 2
Directrix: x
4
Directrix: x
0
x − 20
− 16
4
0
8 x
12
− 12
−8
(− 2,
−4
4
2
Vertex: 2, 2
y
Focus: 4, 3
12
y 22 4 2 x x
Vertex: 2, 3
y
4 2
4
x −6
−4
−2
4
(2,
− 3)
−4 −8
− 12
424
−6 −8
− 2)
6
Section 9.1
18.
1
y
6 y 6 y
x
x
6 x
10
2
x
42
x
42 5
4 2 y 3
1
6
6 x
2
8 x
16
10
20. x2
2 x
8 y
9
x2
2 x
1
8 y
0 9
1
x 12 4 2 y 1
6 y 3
42
10
42
8 x
425
Conics and Calculus
Vertex: 1, 1 5
3
2
Focus: 1, 3
Vertex: 4, 3 5
Directrix: y
4
−8
10
1
Focus: 4, 6 1
− 10 −2
19 6
Directrix: y
10
−4
22. x2
2 x
x
12
4 2 y
8 y
15
0
2
24. Vertex: 0, 2
y 22 42 x 0 y 2
26.
4
y
0
y x2
4 x
x 22 4 x x2
8 x
7 y 2
2
x 14 a2
y 10
y
70 1
4
10, c2
−6
−2
16 x2
34.
16 x2
4
14 25 y2
4 x
4
2
x 42 8 y
0
22
2
1, b
16
y 42 14
1
1 2 ,c 4
3 4
8 y
e
150 y
25 y2
x 22 58
2 ±
3 2
2 2 ,c 5
a2
b2
Center: 2, 3 Foci:
2
±
Vertices:
6 y
279
0
y 32 25
279
10
1
64
e
c a
3 5
±
9 40
10 4
, 3
x −5
, 4
2
0 225 y
1 x −1
1
−2 −3 −4
3 10 , 3 20
2
y
−4
−3
−2
−1 −1 −2
(− 2,
− 4)
−4
3
2
3
4
−1
5 a2, , b2 8
0
Vertices: 1, 4, 3, 4
64 x
Foci:
7
Center: 2, 4
14
8 or p
−3
6
± 14, 0
2
−6
2
8 x
1 a
−4
Foci: ± 2, 0
0
x
4
Center: 0, 0 Vertices:
x
2
2
14, b2
e
32.
6
4
Vertex: 4, 0
2
28. From Example 2: 4 p
x2
30. 5 x2
4 y
(2, −3)
−5
426
Chapter 9
36 x2
36.
36 x2
9 y 2
4 x 3
Conics, Parametric Equations, and Polar Coordinates
4 9
48 x
9 y 2
x 232 14 a2
1, b2
Center:
1 2 ,c 4
36 y
43
4 y
4
43
y
22
1
0
9
1
16
36
3 4
2 ,2 3
3
2 2 Vertices: , 3, ,1 3 3 Foci:
2 ,2 3
±
2
3
Solve for y: 9 y 2
4 y
4
2
36 x
y 22 y
2 x2
38.
50 x2
50 x2
x
y
2
1 36 x2 3
6.4 y
25 y 2
120 x
160 y
12 x 5
36 25
10, b2
Center:
Foci:
±
5 a2
2
4.8 x
652
Vertices:
36
−4
2
−1
3.12 78
32 y 5
48 x
7
(Graph each of these separately.)
0
0
256 25
78
72
256 250
1
5
±
5
6 16 , 5 5
Solve for y: y2
25 y 2
10
6 16 , 5 5
6 16 , 5 5
y 165 2
5, c2
43
9
36 x2 48 x 7
48 x
7
10
±
6.4 y
10.24
2
2 x
y 3.22 7.12 y
40. Vertices: 0, 2, 4, 2
3.2
−7
±
4.8 x
4 x
7.12
3.12
5
10.24
−1
2 x2
4 x
2 x2
(Graph each of these separately.) 42 Foci: 0, ± 5
1 2 Horizontal major axis
Major axis length: 14
Center: 2, 2
Center: 0, 0
Eccentricity:
a
2, c
x 22 4
1
⇒
b
y 22 3
Vertical major axis
3
c
1
x2 24
5, a
7
y 2 49
1
⇒
b
24
Section 9.1
44. Center: 1, 2
46.
Vertical major axis
a
Points on ellipse: 1, 6, 3, 2
x
1
y 2
4
16
1
5, b
3, c
a2 b2 34
y
10 8 6 4 2
Vertices: ± 5, 0
(1, 6) 6
± 34, 0
Foci:
1.
y
2
y2 9
Center: 0, 0
From the sketch, we can see that h 1, k 2, a 4, b 2 2
x2 25
427
Conics and Calculus
x
(3, 2)
Asymptotes: y
(1, 2)
3 ± x 5
x −4
−2
−8
− 4 −2
4
8 10
−4 −6 −8
4
−10
−2
48.
y 1 2 122 a
12, b
x 4 2
52
5, c
50. y 2
1
y 2
a 2 b2 13
Center: 4, 1 Vertices: 4, 11, 4, 13 Foci: 4, 14, 4, 12 12 Asymptotes: y 1 ± x 5
9 x2
9 x2 y 2 36
a
6, b
36 x
72
0
4 x
4
72
1
x 22 4
2, c
36
36
a2 b2 2 10
Center: 2, 0
4
Vertices: 2, 6, 2, 6 Foci: 2, 2 10, 2, 2 10
y
20
Asymptotes: y
± 3 x
2
y
5 x
1
−5
2
7
6
8 5 x −2
− 20
−1
4
5
−5
52. 9 x2 6 x 9 4 y2 2 y 1
9 x
32
y 12 14 a
1 ,b 2
1 ,c 3
4 y
12
x 32 19
78
81 4 1
9 x 2
1
13
y
6 a
2
Foci:
3,
3,
1
1 , 2
±
Asymptotes: y
3,
1 13 6
1
±
3 2
3 x 2
x −3
−1
1 ,b 3
Vertices:
−1
Foci:
3
9
54 x
10 y
y 2 10 y
55
25
1, c
y
52
1
0
55
1
1
81
25
10 3
10
Center: 3, 5
1
2
x 32 19
3
6 x
y
1
Center: 3, 1 Vertices:
9 x 2
54.
10 , 5 3
3 ±
3 ±
1 ,5 3
−8
2 0
Solve for y: y2
10 y
y
25
9 x2
54 x
55
52
9 x2
54 x
80
y
5
±
9 x2 54 x
(Graph each curve separately.)
25
80
428
Chapter 9
56.
3 y 2 3 y 2
4 y
Conics, Parametric Equations, and Polar Coordinates
4
2
x
1 1, b
x 2 6 x
y 22 a
6 x
12 y
0
9
0
1
x 32
3
58. Vertices: 0, ± 3
12
9
Asymptotes: y
3
Vertical transverse axis a
3, c 2
6
3
Slopes of asymptotes:
Center: 3, 2
Thus, b
Vertices: 3, 1, 3, 3
± 3 x
−4
±
a b
±3
1. Therefore,
10
Foci: 3, 0, 3, 4
y 2 9
−4
x2 1
1.
Solve for y: 3 y 2
4 y
y
4
2
6 x
12
x2
6 x 3
12
x
22
y
2
±
x2
6 x 3
12
(Graph each curve separately.) 60. Vertices: 2, ± 3
62. Center: 0, 0
Foci: 2, ± 5
Vertex: 3, 0
Vertical transverse axis
Focus: 5, 0
Center: 2, 0 a
3, c
5, b2 2
y Therefore, 9
c2
a2
x 2
Horizontal transverse axis
16
a
2
16
1.
66. (a) ±
3, c
3 x 4
y 2 4
c
y2 16
1.
2 x2
x2 2
1, y 2
y
4 x 2 y
2 x y
At x
4: y
a2
b2
3 ± and b 4
y 2 36
2
64 and b
36.
A
2
4 x
At 4, 6: y
4, C
AC < 0
Hyperbola
1
3
0
70. 25 x2 A
10 x
200 y
25, C 0
Parabola
6
4, 2 yy 4 x
4
±2
6
4 x 3
16
0,
4 3
4 or 4 x
4 x 3
±
3 y
4 or 4 x
2
0
2
3 y
0
(b) From part (a) we know that the slopes of the normal lines must be 34.
1.
6
At 4, 6: y
y
± 6, y
At 4, 6: y
Solving these equations, we have a Therefore, the equation is x2 64
At 4, 6: y
3 a 4
100 2
68. 4 x2
a2
10
b Slopes of asymptotes: ± a c2
c2
Horizontal transverse axis Center: 0, 0 since asymptotes intersect at the origin.
5, b2
x2 Therefore, 9
64. Focus: 10, 0
Asymptotes: y
119 0
6
3 x 4
4 or 3 x
4 y
36
0
6
3 x 4
4 or 3 x
4 y
36
0
4 y
0
72. y2 A
x
0, C 1
Parabola
5
Section 9.1
2 x2
74.
2 x2 A
y
2
2 xy
3 y
3 y
0
y
2
76.
2 xy
9 x2 A
2, C 1, AC > 0
78. (a) An ellipse is the set of all points x, y, the sum of whose distance from two distinct fixed points (foci) is constant.
x h2
y
2
a
k 2
2
b
1 or
x h2 2
b
y k 2 a2
54 x
81
36
54 x
16 y
61
0
4 y2
4 y
x2
4 py
82
4 p
1600 3 x2
y
4
1600 y 3
2 100
x
⇒
±
128 3
± 6.53
a2 b2 0
<
e
<
1
1
For e 1, the ellipse is elongated.
21 . p
x
4 x
4
4 y
0
4
dy dx
0
dy dx
1 x 2
meters.
x
)
1
At 0, 0, the slope is 1: y x. At 6, 3, the slope is 2: y 2 x 9. Solving for x,
(
3 0, 100
)
(
3 8, 100
)
1 100
x −4
4
2 x
9
3 x 9
2 100
−8
For e 0, the ellipse is nearly circular.
2 x
4 100
(
c ,c a
(b) x2
6400 y 3
5 100
3 −8, 100
Therefore, for distinct tangent lines, the slopes are unequal and the lines intersect.
p
80. e
y
y
x
y
3.
3
8
Point of intersection: 3, 3 86. The focus of x2 d
8 y
42 y is 0, 2. The distance from a point on the parabola, x, x28, and the focus, 0, 2, is
x 02
8 2 . x2
2
Since d is minimized when d 2 is minimized, it is sufficient to minimize the function f x f x
x3 16
4
9, C 4, AC > 0
1003
(b) The deflection is 1 cm when
f x
84. (a) Without loss of generality, place the coordinate system so that the equation of the parabola is x2 4 py and, hence,
82. Assume that the vertex is at the origin.
(a)
4 y2
9 x2
Ellipse
Ellipse
(b)
429
Conics and Calculus
x
2
2 x
x2 8
2
4
2
2 .
3
2
x
3
x 16
16 1 x2
x.
x −3
−2
−1
1
−2
0
⇒
x
2
( x, x8 )
1
0 implies that x x
x 2 = 8y
(0, 2)
8 2 4 x
y
0.
This is a minimum by the First Derivative Test. Hence, the closest point to the focus is the vertex, 0, 0.
2
3
430
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
88. (a) C 0.0853 t 2
(b)
90.
0.2917 t 263.3559
x
1 2 y 4
320
1 x y 2
0
1
s
x 2 1
18 0
y2 4
4
dC 0.1706 t 0.2971 (c) dt
1
y
1 4 20 4
2 5
5 4 3 2 1
1 y 4 4
2
3
4
2
4 ln y
4 ln 4
ln2
x
1
0
y
−1
y 2 dy 4
1 2
4
4 y2 dy
0
4 y 2
20
4
0
4 ln 2
5 5.916
5
−1
The consumption of fruits is increasing at a rate of 0.1706 pounds/year.
92. x2
20 y
94. A
h
2
4 py dy
0
x2 20
y
x 10
y
h
4 p y12 dy 0
r
S 2 x
1
0
x 10
2
2 2 32 100 x 10 3
r
dx r
0
x 100 x2 2 dx 10 0
100 r 232 15
Vertex
0
8 ph32 3
e
c a
0.0167
c 149,570, 000
98.
Focus
h
1000
96. (a) At the vertices we notice that the string is horizontal and has a length of 2 a.
(b) The thumbtacks are located at the foci and the length of string is the constant sum of the distances from the foci.
2 32 y 3
p 4
c 2,497,819
Focus Vertex
Least distance: a
Greatest distance: a
100. e
A A
P P
122,000 122,000
121,881 0.9367 130,119
4000 4000
x2 a2
119 4000 119 4000
x2 a2 x2 a2
y 2 a2b2a2
1
1
1
y 2 a2
c2a2
y2 a21 e2
As e → 0, 1 x2 a2
c
1
a2a2
y 2
x2 a2
y 2 b2
102.
c
147,072,181 km
152,067,819 km
e2 → 1 and we have
1 or the circle x2
y
2
a2.
Section 9.1
104.
x2 4.52
y 2 2.52
1
x2
4.52 1
Conics and Calculus
y
y 2 2.5 2
5 ft x
x
9 2 ± 2.5 5
y
3 ft
2
9 ft
V Area of bottomLength V
106. 9 x2
4.52.5 16 2
90
4 y 2
18 x
0.5
0
24 y
8 yy 36
9 2.52 5
16
144 1 y 2.52 5 2 36 x
Area of topLength
36
y
2
y
2
dy (Recall: Area of ellipse is ab.) y 2.5
2.52 arcsin
0.5
0
72 0.5 6 5
90
2.52 arcsin
1 318.5 ft3 5
0
24 y 0
8 y 24 y 18 x 36
y
18 x 8 y
36 24
y 0 when x 2. y undefined when y At x 2, y 0 or 6.
3.
Endpoints of major axis: 2, 0, 2, 6 At y
3, x
0 or
4.
Endpoints of minor axis: 0, 3, 4, 3 Note: Equation of ellipse is
108. (a) A
4
4
0
3 16 4
x
2
x
22
4
dx
V 2
0
y
y
1 y 2
4
S 22
0
8 7 3
3 16 4
x
9
3 x 16 2 4
(b) Disk:
y 32
3 16 4
9 16 16
2
x
x
1
16 arcsin
dx
2
x 4
4
0
9 16 x 8
12
1 3 x 3
4
0
48
2
x
3 x
4 16
2
1
x
2
9 x2 1616 x2
1616 x2 9 x2 dx 1616 x2
7 x 256 7 x2 256 arcsin
—CONTINUED—
7 x
16
0
4
0
4
4
3 16 4
48 7 8 7 3
2
x
4 256 7 x2 3 256 7 x2 dx dx 2 4 4 16 x 0
256 arcsin
7 4
138.93
431
432
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
108. —CONTINUED—
4
(c) Shell:
V 4 x 0
x
4 9 3
3
S 22
0
3
4
0
110. (a)
4 81 9
16 9 2 7 8 9 7
x2 a2
2 x a2
4 9 3
At P, y
3
1 2
2 16 3
y
1
0
64
16 y 2 99 y 2
99
2
y
99
81 ln3 7
2
y
2
16 y 2
dy
12
81 7 y 2
3
0
81 ln 9 168.53
(b) Slope of line through c, 0 and x0, y0: m1 Slope of line through c, 0 and x0, y0: m2
y0 x0
x0
xb2 ya2
b2 x0 a2 y0
m.
x0
1
b2 y0c
x0
m1 m 1 m1m
1
b2 x0 a2 y0 b2 x0 a2 y0
c y0
b2 x0c a2 y0c
arctan
c
a2 y02 b2 x0 x0 c a2 y0 x0 c b2 x0 y0
a2b2 b2 x0c x0 y0c2 a2 y0c
b2a2 x0c y0c x0c a2
b2 y0c
arctan
x0
c y0
x0
a2 y02 b2 x02 b2 x0c a2 x0 y0 a2cy0 b2 x0 y0
b2 y0c
b2 x0 a2 y0 b2 x0 a2 y0
c
a2 y02 b2 x0 x0 c a2 y0 x0 c b2 x0 y0
a2b2 b2 x0c x0 y0a2 b2 a2cy0
b2a2 x0c y0c x0c a2
b2 y0c
arctan
b2 y0c
, the tangent line to an ellipse at a point P makes equal angles with the lines through P and the foci.
Since
y0
y0
tan
4
7 y 2 dy
a2 y02 b2 x02 x0 y0a2 b2
2
1
m2 m 1 m2m
2 3 2
x
2
y
y0
2 yy 0 b2 y
(c) tan
dx
7 y 81 7 y 2 81 ln 7 y
3 712
y 2 b2
3 9
y
2
x
4 y
x
1 x 2
3 16 4
c
c
Section 9.1
112. (a) e
c a
a2 b2
a
x h2
a2
x h2
a2
x 22
(b)
4
y
⇒
k 2
b2
ea2 a2 b2. Hence,
y k 2 a21 e2
y 32 41 e2
114. The transverse axis is vertical since 3, 0 and 3, 3 are the foci.
Center:
1
c
1.
3,
3 , 2a 2
3 2
2, b2
a2
5 4
32 54
1.
c2
Therefore, the equation is
1
y 32 2 1
7
−3
433
Conics and Calculus
x
9 −1
(c) As e approaches 0, the ellipse approaches a circle.
116. Center: 0, 0
Horizontal transverse axis Foci: ± c, 0 Vertices: ± a, 0 The difference of the distances from any point on the hyperbola is constant. At a vertex, this constant difference is
a c
c
a
2a.
Now, for any point x, y on the hyperbola, the difference of the distances between x, y and the two foci must also be 2a.
x c2 y
02
x c2 y 02
2a
x c2 y 2
2a
x c2 y 2
4a2
4a2
4a x
a2
a x
a4
a2 x2
a c
4 xc
xc
x2c2 x c 2
2
x2 a2
Since a2
b2
c2, we have x2a2
0.001186,000 , a
118. c
150, 2a
b
1502 932 13,851 x2 932
When y x2
y 2 13,851
2a2cx a 2
2 2
a y
c2
2
x c2 y 2 4a x
2
c2
c2
2cx
a
c2
y
y
752 13,851
x 110.3 miles.
x c2 y 2
c2
y
y
2
( x, y)
2
(c, 0)
a2
1
1.
x2 a2
120.
y
2
75, we have
932 1
2
2
2 x a2
1
y
(− c, 0) (− a, 0) (a, 0)
y 2
y 2b2
93,
y2 b2
1
b2 x 2 yy 0 or y 2 b a2 y y0
b2 x0 x a2 y0
x0
a2 y0 y
a2 y02
b2 x0 x
b2 x02
b2 x02
a2 y02
b2 x0 x
a2 y0 y
a2b2
b2 x 0 x
a2 y 0 y
y0 y b2
1
x0 x a2
x
434
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
Ax2 Cy2 Dx Ey F 0
122.
4 4 2 2
(Assume A 0 and C 0; see (b) below)
D E A x2 x C y 2 y F A C
D D2 E E 2 D 2 E 2 A x2 x C y2 y F R 2 2 A A C C 4 A 4C 2
D A C
x
y
2
E C A
(a) If A C , we have
D 2 A
x
(b) If C 0, we have
2
R AC
y
E 2 R A 2C
A x
(c) If AC > 0, we have
C y
2
2
F Ey
D 2 . 4 A
If A 0, we have
which is the standard equation of a circle.
x D2 A y E 2C R R A C
D 2 A
E 2C
2
F Dx
E 2 . 4C
2
These are the equations of parabolas.
1
(d) If AC < 0, we have
x D2 A y E 2C R R A C 2
2
which is the equation of an ellipse.
±1
which is the equation of a hyperbola. 124. True
Section 9.2
≤
x
≤
y 2 sin 2 ≤
4
(a)
128. True
Plane Curves and Parametric Equations
2. x 4 cos2
0
126. False. The y4 term should be y2.
2
y
0
4
≤
2
4
2
(c)
3
5
−1
x
0
2
4
2
0
y
2
2
0
2
2
(b)
−3
y
(d) 3 2 1 x
1
−2
2
3
x cos2 4 y 2 sin2 4
5
x y2 1 4 4 x 4 y2, 2
≤
y
≤
2
(e) The graph would be oriented in the opposite direction.
Section 9.2 4. x 3 2t
6. x 2t 2
y 2 3t y 2 3
435
Plane Curves and Parametric Equations
y t 4 1
3 2 x
x 2
y
2 y 3 x 13 0
2
1
x2 1, x 4
≥
0
For t < 0, the orientation is right to left.
y
For t > 0, the orientation is left to right. y
6
6 5
4
4 2
3 2 x
2
4
6
1
8
x −1 −1
1
2
3
4
5
6
8. x t 2 t , y t 2 t
Subtracting the second equation from the first, we have x y 2t or y
x y2 4
t
x y 2
t
2
1
0
1
2
x
2
0
0
2
6
y
6
2
0
0
2
y
4
x y 2
3 2
Since the discriminant is
x
B 4 AC 22 411 2
−1
0,
2
3
4
−1
the graph is a rotated parabola.
4 10. x t, t ≥ 0
12. x 1
y 3 t ≥
0 x 1
y
3
y
2 1
y t 2
−1
1
2
x y 2 1 y 3
1 1 implies t t x 1
1 x 1
x −2
14. x t 1
y t 1
y 3 x 4, x
−3
1 t
y
5
1
4 3
y
3
−1
2
1 −2
1 x
−3
−2
x
2 1
2
3
4
5
−3
16. x et , x
>
0 3
y e2t 1 y
x2
1 1 2 1, x x
18.
y
2 >
1
0
x tan2 y sec2
4
sec2 tan2 1
3
x −3
−2
−1
1 −1
3
y
y x 1
2 1
−2 −3
x
≥
0
x
1
2
3
4
436
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
x 2 cos
20.
x cos
22.
y 6 sin
2 6 x
2
y
y 2 sin 2
2
x 4 2 cos
24.
y 1 2 sin
y 4 sin cos
cos2 sin2 1
x 42 4 cos2
1 x2 sin2
x2 y 2 1 ellipse 4 36
y
y 12 4 sin2
1 x2
x 42 y 12 4
± 4 x
3
3
y
−1 −2
4
7
2
2
x −6
−4
4
6
−5
−3
−2
26. x sec
x cos3
28.
2
−3
3
y2 tan2
−2
x23 cos2
2
y23 sin2
−2
30. x ln 2t
− 1.5
32. x e2t
3
4
y et
y t 2 t
1.5
y sin3
y tan x2 sec2
e x 2
−2
3
−1
1 e 2 x e 2 x y r 4
y 2 x
−1
y > 0
3
−1
y x, x
>
0
34. By eliminating the parameters in (a) – (d), we get x2 y 2 4. They differ from each other in orientation and in restricted domains. These curves are all smooth. 4t 2 1 1 1 (b) x 4 2 y (a) x 2 cos , y 2 sin t t t
y
x
≥
0, x 2
y 0
3 y
1
2 x
−3
−1
1
3
1 x −1
−3
1
3
−1
−2
(c) x t x
≥
y 4 t y
0
≥
(d) x 4 e2t 2
0
< x
≤
y et y > 0
0
y
y
3
3
2
1
1
x
1
2
3
x −3
−2
−1
Section 9.2
Plane Curves and Parametric Equations
36. The orientations are reversed. The graphs are the same. They are both smooth.
38. The set of points x, y 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 h r cos
40.
x h a sec
42.
y k r sin
cos
x h r
x h sec a
sin
y k r
y k tan b
cos2 sin2
x h2 y
y k b tan
x h2 r 2
y k 2 r 2
x
1
a2
h2
y k 2 b2
1
k 2 r 2
44. From Exercise 39 we have
46. From Exercise 40 we have
x 1 4t
x 3 3 cos
y 4 6t .
y 1 3 sin .
Solution not unique
48. From Exercise 41 we have a 5, c 3
⇒
b4
x 4 5 cos y 2 4 sin .
Solution not unique
Center: 4, 2 Solution not unique
50. From Exercise 42 we have a 1, c 2
⇒
b 3
52. y
2
54. y x2
x 1
Example
Example
x 3 tan x t , y
y sec .
Center: 0, 0 Solution not unique The transverse axis is vertical, therefore, x and y are interchanged. 56. x
t 1
x t , y
y 1 cos
y t 2
x t 3,
y t 6
60. x 2 sin
y 2 4 cos
6
x t ,
2 t 1
58. x 2 4 sin
sin
−6
2
y 2 cos
9
6
−9
4
9 5
− −2
Not smooth at x 2n 1
−3
0
Smooth everywhere
437
438
Chapter 9
62. x y
Conics, Parametric Equations, and Polar Coordinates
3t
2
64. Each point x, y in the plane is determined by the plane curve x f t , y gt . For each t , plot x, y. As t increases, the curve is traced out in a specific direction called the orientation of the curve.
1 t 3 3t 2
−3
3
1 t 3 −2
Smooth everywhere 66. (a) Matches (ii) because 1
x
≤
≤
0 and 1
≤
y
≤
(b) Matches (i) because x y 22 1 for all y.
2.
68. x cos3
70. x cot
y 2 sin2
y 4 sin cos
Matches (a)
Matches (c)
72. Let the circle of radius 1 be centered at C . A is the point of tangency on the line OC. OA 2, AC 1, OC 3. P x, y is the point on the curve being traced out as the angle changes AB AP . AB 2 and AP ⇒ 2 . Form the right triangle CDP. The angle OCE 2 and
DCP
2 3 sin 2
2 3 2 . sin3 3 cos 2 cos 3 3 sin sin 3 2
x OE Ex
y EC CD 3 sin
y
3
C 2
α
x
1
cos 3
Hence, x 3 cos cos 3 , y 3 sin sin 3 .
74. False. Let x t 2 and y t . Then x y 2 and y is not a function of x.
76. (a) x v0 cos t y h v0 sin t 16t 2 t
x v0 cos
⇒
y h v0 sin
(b) y 5 x 0.005 x 2 h tan x h 5, tan
0.005 v02
1
⇒
2
16 sec2 2 x v02
16 sec2 2 x v02
(c)
80
4
, and
16 sec2 4 16 2 2 v0 v02 32 6400 0.005
x x 16 v0 cos v0 cos
y h tan x
⇒
v0 80.
Hence, x 80 cos45t y 5 80 sin45t 16t 2.
P = (x, y)
θ
A
1
D
0
250
−5
(d) Maximum height: y 55 at x 100 Range: 204.88
B E
x
Section 9.3
Section 9.3 2.
dy dx
6. x
dydt
dxdt
d 2 y dx 2
1
13t 23
3
12 t
3t 23
4.
3 t 12 t
dy dx
8. x
when t 1.
6 t 6
6 concave upwards
dyd 12e 2 2e dxd
t 2
3t 2, y
dy dx
d 2 y dx2
2 2t 3
1 3 2 e 4
1 4e3 2
2t
2 when t 0. 3
222t
3
2t 3
4
2t 32
4
9
concave downward 10. x
cos , y dy dx
d 2 y dx2
12. x
t , y dy dx
cot
3
dy is undefined when 0. dx
d 2 y 3 csc2 3 is undefined when 0. 3 sin sin dx2
t 1
14. x
1 2 t 1
dy dx
1 2 t
d 2 y dx2
3 sin
3 cos sin
t t 1
2 when t 2.
1
1 t 132
sin , y
1
2
d y dx2
t 1 2 t t 12 t 1 t 1 1 2 t
when t 2.
1
cos
sin 0 when . cos
1 cos cos sin2 1 cos 2 1 cos 1
1 cos
2
1 when . 4
concave downward
concave downward
16. x
2
dy dx
3 cos , y
3
Tangent line: x At 2, 5,
18. x
t 1, y
(a)
dy is undefined. dx
2
, and
1 1, t 1 t
4
5
−3
1
Tangent line: y
4
2 sin
2 cos 2 cot 3 sin 3
At 1, 3, 0, and
At
−4
dy dx
(b) At t 1, x, y
0.
dx dy 1, dt dt
5
dy 3 3 7 , 2 , , and dx 2 6
2 3 . 3
(c)
dy dx
1.
0, 2, and
1,
dy dx
At 0, 2, y
2 y
y
2
2 3 x 3
4
3 3 2
(d)
4
−3
1
Tangent line:
2 3 x
439
Parametric Equations and Calculus
t , y 3t 1 dy dx
Parametric Equations and Calculus
3 y
4 3
3
5
0 −4
1 x x
2
0
when t
0.
440
Chapter 9
20. x
4 cos , y
Conics, Parametric Equations, and Polar Coordinates
(a)
3 4
3 sin ,
(b) At
4
−6
dx dt
6
(d)
−4
dy dx
(c)
3 4 3 . At , , 4 2 2
y
3
t 2
t , y
t 3
At this point, t dy dx
At t
24. x
3 x 4
3 x 4
4
2
42, 32, and
dy dt
3 2 dy , 2 dx
3 4
4
−6
6
3 2
−4
3t 1 crosses itself at the point x, y
2 2,
2, 1.
or t 2.
1
3t 2 3 2t 1
dy dx
1,
At t 2,
2 y
22. x
3 , x, y 4
0 and y
dy 9 dt 3
3 and y
21
cos
2 , y
1. Tangent Line
1
3 x
2 or y
dy 2 sin 0 when 0, d
Horizontal tangents:
Points: 4n , 0, 22n
3 x
5. Tangent Line
± , ± 2 ,
. . ..
1 , 4 where n is an integer.
Points shown: 0, 0, 2 , 4, 4 , 0 dx 2 d
Vertical tangents:
26. x
t 1, y
t 2
Point:
1 9 , 2 4
Vertical tangents:
30. x
cos , y
28. x
dy 2t 3 dt
0 when t
3 . 2
Vertical tangents:
t 2
t 2, y
t 3
3t
dy 3t 2 dt
3
0 when t
± 1.
Points: 2, 2, 4, 2
dx 1 dt
Vertical tangents:
0; none Point:
dy 3 5 7 , , . 4 cos 2 0 when , 4 4 4 4 d
22, 2,
Horizontal tangents:
2 sin 2
Horizontal tangents: Points:
0; none
3t
Horizontal tangents:
2 2
dx d
Points: 1, 0, 1, 0
, 2 ,
sin
2 2
22, 2
,2 ,
0 when 0, .
74, 118
dx 2t 1 dt
1 0 when t . 2
Section 9.3 32. x
4 cos2 , y
2 sin
34. x
3 dy 2 cos 0 when , . d 2 2
Horizontal tangents:
Since dxd 0 at 2 and 3 2, exclude them. dx d
Vertical tangents:
cos sin 0 when
8
441
Parametric Equations and Calculus
cos2 , y
cos
dy d
Horizontal tangents:
sin
0 when x
0, .
Since dxd 0 at these values, exclude them. dx d
Vertical tangents:
0, .
2
cos sin 0 when
3 . 2 2
,
Point: 4, 0
Exclude 0, . Point: 0, 0
36. x
t 2
1, y
4t 3
3, 1
t ≤ 0
≤
dx dy dx 2t , 12t 2, dt dt dt s
2
0
4t 2 144t 4 dt
1
dy dt
38. x
36t 232 54
4t 2
144t 4
0
2t 1
1 2
36t 2 dt
s
1 3732 4.149
1 2
54
t , y
t 5 10
dy t 4 1 dx , 1, dt 6t 3 dt 2
2
S
1
1
t 4 2
1 2t 4
1 2t 4
42. x
t 4 2
1
2
1
t 4 2
t 10
5
44. x
1 2t 4
2
1
cos sin , y
1 1 ln 2 3
dt
2
dt
sin cos ,
2 cos2 2 sin2 d
0
d
2 2
2
0
4a
0
0 2
1
dt 2
2
t 2
2
dt
0
1 1
2 2
dx cos d
dt
0
1 ln 3 2
a sin ,
dx d
0.549
a sin ,
a2 sin2 a2 cos2 d
t 2
1 2
0
2
t
1 2
1
dy sin d S
1 2
≤
2
S 4
779 240
dy dt
a cos , y
2
1 dt 2t 4
1 6t 3
2
dx dt
t 1 1 ln t 1 2
2
2
t
≤
1 t
t 2, 0
0
40. x
1
1
ln 1
1 1 2t dx dy , dt 1 t 2 dt 2 1 t 2
0
0
2
1
1
arcsin t , y
2
d 4a
0
2 a
dy a cos d
442
Chapter 9
46. x
Conics, Parametric Equations, and Polar Coordinates
4t , y 1 t 3
(a) x 3
4t 2
y 3
1
t 3
(b)
4 xy
1 dy dt
4
−6
t 38t 4t 23t 2 1 t 32
4t 2 t 3 1 t 32
6
Points: 0, 0,
3 0 when t 0 or t 2.
2 4 3 4 4 3 1.6799, 2.1165 , 3 3
−4
1
(c) s
2
0
1
8
0
48. x
3
4t 2 t 3 1 t 32
1
2
16 t 8 1 t 34
dt 2
0
1
4t 6
4t 5
4t 3
4t 2
1 dt
dt 6.557
4 sin
4
dy 4 cos d
sin ,
2
t 8 4t 6 4t 5 4t 3 4t 2 1 t 32
3 cos , y
dx d
41 2t 3 1 t 32
−6
6
2
s
9 sin2 16 cos2 d 22.1
−4
0
50. x
t , y
4
dx dy 1, dt dt
2t ,
52. x
2
2
(a) S 2 4
2t 1
2 5 4 t t
0
t 1
(b) S 2
a cos , y
8 5
S 2
dx d
2
4 dt 5 t 2
b sin ,
2
1
0
t 1, 1
2
2
2
54. x
≤
t ≤ 2, y-axis
1 dt
dx dy t 2, 1 dt dt
4 dt
0
13 t , y 3
0
a sin ,
4 5
13 4 t t 3
9
2
2 sin2 b2 cos2 d b sin a
(a) S 4
0
2
4
ab sin
0
1
a2
b2
a2
cos2 d
e 2 b 2a a bb arcsin a a b e a a b ca: eccentricity
2ab
2
2
e cos 1
2
2
—CONTINUED—
2
2
e
2 b2
2
e sin 1 e2 cos2 d
ab
e
0
2
2
0
2
e2 cos2 arcsine cos 2
4ab
e 1 e
arcsin ab e
e
9
1732 232 23.48
dy b cos d
2
arcsine
2
x4 132
1
Section 9.3
443
Parametric Equations and Calculus
54. —CONTINUED—
2
2 sin2 b2 cos2 d a cos a
(b) S 4
0
2
2 a cos b
4
0
2a 2 c sin b c
2a c b2 c
2 a 2
a
2
2
b
b2 ln c
ln
a
b2 c 2
a2 b2 b
c 2 sin2 d
2
2 ab2
2 c cos b
c2 sin2 b2 ln c sin b2
c2
2
4a c 0
c2 sin2 d
c 2 sin2
0
b2 ln b
2 a 2
ln11 b2
e
56. (a) 0
e e
58. One possible answer is the graph given by
(b) 4
t , y t .
x
y 4 3 2 1
x −4
−3
−2
1
2
3
4
−2 −3 −4
b
dx dt
2
dx dt
2
60. (a) S 2 gt a
b
(b) S 2 f t a
dy dt
2
dy dt
2
62. Let y be a continuous function of x on a ≤ x ≤ b. Suppose that x f t , y gt , and f t 1 a, f t 2 b. Then using integration by substitution, dx f t dt and
dt
dt
b
y dx
a
64. x
4 t , y t,
0
A
t
4
Let u
x
1 2 4
x, y
66. x
t
1
, 0
2 4 t
t
≤
≤
2
dt
4 u2 du
0
t 2
gt f t dt .
t 1
4
1 u 4 2
u2
4 arcsin
u 2
2
0
4 t , then du 1 2 4 t dt and t 4 u2. 1
0
4 t t
4
0
y
dx dt
1 t 2 2 4
1 2 4
1
2 4
t
t
dt
0
t 1 dt 4 4 4 t
dt
1 t dt 2 4
1 2 32 t 2 3
0
1 4
38 , 38
cos , y
3 sin ,
dx d
sin
0
3 sin 2 sin d
V 2
2
0
sin3 d
18
2
18 cos
cos3 0 3 2
12
0
4
28
3
8 3 t
0
4 t
4
8 3
444
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
dx 2 csc2 d
68. x 2 cot , y 2 sin2 ,
0
0
2 sin 2 csc d 8 2
A 2
2
2
70.
3 2 8 a
0 < x
≤
0
2
12
144 x2 x
4
2
72. 2 a2 is area of deltoid (c).
is area of asteroid (b).
76. (a) y 12 ln
d 8
144 x2
74. 2 ab is area of teardrop (e).
(b) x 12 sech
12
t t , y t 12 tanh , 0 12 12
t
≤
60
60
0
12 0
0
12
Same as the graph in (a), but has the advantage of showing the position of the object and any given time t .
0
(c)
1 sech2t 12 dy t sinh secht 12 tant 12 dx 12
Tangent line: y t 0 12 tanh
y
24
t 0 t 0 t x 12 sech 0 sinh 12 12 12
(0, y0)
16 12
y t 0 sinh
t 0 x 12
8
( x, y)
4 x
y-intercept: 0, t 0
2
Distance between 0, t 0 and x, y: d
12 sech
t 0 12
2
12
tanh
t 0 12
4
6
10
12
2
12
d 12 for any t ≥ 0.
78. False. Both dxdt and dydt are zero when t 0. By eliminating the parameter, we have y x 23 which does not have a horizontal tangent at the origin.
Section 9.4 2.
2,
Polar Coordinates and Polar Graphs
4
7
4.
7 2 sin 4
7 2 x 2 cos 4 y
2
x 3 cos 1.57 0.0024
x
0
y
0
y 3 sin 1.57 3
x, y
0.0024,
3
π
2
0, 0
(− 0.0024, 3)
π
π
2
2
2,
6. 3, 1.57
x, y
x, y 2, 2
(−
0, 76 7 0 cos 6 7 0 sin 6
2)
(0, 0) 0 1
2
0 1
0 1
2
Section 9.4
8. r , 2,
x, y
11 6
1.7321,
10. r , 8.25, 1.3
x, y
1
12. x, y 0, 5 r
2.2069, 7.9494 8
2
6
±5
tan undefined
y y
(−1.7321, 1)
(2.2069, 7.9494)
3 3 , , 5, , 2 2 2 y
2 x −2
x −2
5,
4 1
−1
1
−1
1
−2
2
x
2
−2
−1
1
2
3
−1
−4 −1
−2
−6 −8
−3
−2
−4
(0, −5)
−5
14. x, y 4, 2 r
±
y x
16 4 ± 2 5
1
2
3
4
5
−1
2 1 tan 4 2
−2
(4, −2) −3
0.464
−4
2 5, 0.464 , 2 5, 2.678
−5
16. x, y 3 2, 3 2
r ,
18. x, y 0, 5
r ,
6, 0.785
5, 1.571
20. (a) Moving horizontally, the x-coordinate changes. Moving vertically, the y-coordinate changes.
(b) Both r and values change. (c) In polar mode, horizontal (or vertical) changes result in changes in both r and .
22. x2 y 2 2ax 0
24.
π
2
2
r 2ar cos 0 r r 2a cos 0
r cos r sin
2
0
a
0
2a
2
28.
4
4
6
x2 y 22 9 x2 y 2
0
r 22 9r 2 cos2 r 2 sin2
0
π
2
2
0 4
12
r 2 9 cos 2
8 csc 2
π
2
8
r 2r 2 9cos 2 0
r 2 4 sec csc
π
r 10 sec
xy 4
26.
x 10 r cos 10
r 2a cos
445
Polar Coordinates and Polar Graphs
0 1
2
2
446
Chapter 9
30.
r 2
Conics, Parametric Equations, and Polar Coordinates
32.
r 5 cos
r 2 4
34.
5 6
r 2 5r cos
x2 y 2 4
2
x y 5 x
y
x
1
3 y x 3
25 25 y2 4 4
x2 5 x
5 2
2
y2
52
2
y
3 3
x −1 −1
x
y
y
1
5 6
tan tan
2
4
2
3 2
1
1 x −2 − 1
1
2
3
4
x
6
−2
−1
1
−2
2
−1
−3 −4
38. r 51 2 sin
r 2 csc
36.
−2
r sin 2
0
≤
40. r 4 3 cos
< 2
y 2
0
≤
< 2
3 −10
y 2 0
6
10
−4
y
10
3 −18
−6
1
x −1
42. r
1
2
2 4 3 sin
Traced out once on 0
44. r 3 sin ≤
≤ 2
0
≤
1 46. r 2 .
52
Graph as
< 4 4
r 1
3
−6
6
1
1 , r 2 .
It is traced out once on 0, . 1.5
−3
3 −4 −1 −2
2
− 1.5
Section Section 9.4
Polar Coordina Coordinates tes and Polar Graphs Graphs
48. (a) The rectangular rectangular coordinates coordinates of r 1, 1 are r 1 cos 1, r 1 sin 1. The rectangular coordinates coordinates of r 2, 2 are r 2 cos 2, r 2 sin 2. d 2 x x2 x12 y2 y12
r 2 cos 2 r 1 cos 12 r 2 sin 2 r 1 sin 12
r 22 cos2 2 2r 1r 2 cos 1 cos 2 r 12 cos2 1 r 22 sin2 22 2r 1r 2 sin 1 sin 2 r 12 sin2 1
r 22 cos2 2 sin2 2 r 12 cos2 1 sin2 1 2 r 1r 2cos 1 cos 2 sin 1 sin 2
r 12 r 22 2r 1r 2 cos 1 2
d r r1 2 r 22 2r 1r 2 cos 1 2
(b) If 1 2, the points lie on the same line passing through the origin. In this case, d r r1 2 r 22 2r 1r 2 cos0
r 1 r 22 r 1
r 2
(c) If 1 2 90, then cos 1 2 0 and d r Pythagorean Theorem! r1 2 r 22, the Pythagorean (d) Many answers are possible. For example, consider the two points r 1, 1 1, 0 and r 2, 2 2, 2. d
1 22 212 cos 0
2
5
Using r 1, 1 1, and r 2, 2 2, 5 2, d
12 22 2 12 cos
5 5. 5. 2
You always obtain the same distance.
50.
10, 76 , 3, d
52. 4, 2.5, 12, 1
7 10 102 32 2103 cos 6
109 109 60 cos
6
d 42 122 2412 cos2.5 1
160 160 96 cos 1.5 12.3
109 109 30 3 7.6
54. r 21 sin
56. (a), (a), (b) (b) r 3 2 cos
2 cos sin 2 cos 1 sin dy 2 cos cos 2 sin 1 sin dx
At 2, 0,
dy 1. dx
3 At 4, , 2
At 3,
7 dy , is undefined. dx 6 dy 0. dx
4
−8
4
−4
r , 1, 0
⇒
x x, y 1, 0
Tangent line: x 1 (c) At 0,
dy does not exist (vertical tangent). dx
447
448
Chapter 9
Conics, Parametric Parametric Equations, and Polar Coordinates Coordinates
58. (a), (a), (b) (b) r 4
r a sin
60.
dy a sin cos a cos sin d
6
−8
8
4
at r , 4,
⇒
dx a sin2 a cos2 a1 2 sin2 0 d
x x, y 2 2, 2, 2 2
Tangent line: y 2 2 1 x 2 2
sin
y x 4 2
dy , 4 dx
3 , , 2 2
0,
−6
(c) At
2a sin cos 0
±
1
2
,
3 5 7 , , , 4 4 4 4
2 2 2 3 Vertical: , , , 2 4 2 4
Horizontal: 0, 0, a,
1.
a
62.
r a sin cos2
a
64. r 3 cos 2 sec sec
dy a sin cos3 2a sin2 cos a cos3 sin d
2asin cos3 sin3 cos
2a sin cos cos2 sin2 0
2
−2
4
−2
2
0, tan 1, Horizontal:
3 , 4 4
Horizontal tangents: 2.133,
± 0.4352
42 , 4 , 42 , 34 , 0, 0 a
a
66. r 2 cos3 2
r 3 cos
68.
π
2
2
cos r 3r cos
2
x2 y2 3 x −3
3
3 x x 2
2
0
−2
Horizontal tangents:
Circle: r
1.894, 0.776 , 1.755, 2.594 , 1.998, 1.442 Center:
3 2
32, 0
Tangent at pole:
70. r 31 cos
1
9 y 4 2
2
π
2
Cardioid Symmetric to polar axis since r is a function of cos . 0 1
0
3
2
2 3
r
0
3 2
3
9 2
6
2
4
Section Section 9.4
72. r sin5
449
Polar Coordina Coordinates tes and Polar Graphs Graphs
74. r 3 cos 2
Rose curve with five petals Symmetric to
Rose curve with four petals
2
, and pole 2
Symmetric to the polar axis,
Relative extrema occur when
Relative extrema: 3, 0, 3,
3 5 7 9 dr 5 cos5 0 at , , , , . d 10 10 10 10 10 Tangents at the pole: 0,
Tangents at the pole:
2 3 4 , , , 5 5 5 5
, 3, , 2
3,
3 2
3 , 4 4
5 7 and given th the sa same tan tangen gents. ts. 4 4
π π
2
2
0
0
1
2
76. r 2
Circle Circle radius radius:: 2 2
80. r 5 4 sin
78. r 1 sin
Limaçon
Cardioid
2
x y 4
Symmetric to
π
2
2
π
2
0
r
0
1
1
2
9
6
7
0
6
2
5
3
1
2 π
2
0 2
r
82.
6 2 sin 3 cos
2r sin sin 3r cos cos 6
84. r
Hyperbolic spiral
2 y 3 x 6 Line π
2
1
4
2
3 4
5 4
3 2
r
4
2
4 3
1
4 5
2 3
2
0 1
1
0
4
450
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
86. r 2 4 sin
π
2
Lemniscate Symmetric to the polar axis,
, and pole 2 0
Relative extrema: ± 2, 2
0
r
0
6 ±
2
2
±2
2
5 6
2
0
±
Tangent at the pole: 0
88. Since
90. r 2 cos 2 sec
r 2 csc 2
1 , sin
Strophoid r ⇒ as ⇒
the graphs has symmetry with respect to 2. Furthermore,
r ⇒ as ⇒
r ⇒ as ⇒ 0
2
r cos 4 cos2 2
r r 1 Also, r 2 2 2 y sin sin
r
r 2 cos 2 sec 22 cos2 1 sec
r ⇒ as ⇒ .
ry 2 y r
2
x 4 cos2 2
lim
→ ± 2
2 y . y 1
4 cos2 2
x = −2
2
2
Thus, r ⇒ ± as y ⇒ 1. −3
3
4
−2
−4
4
y = 1 −2
92. x r cos , y r sin x2 y2 r 2, tan
y x
94. Slope of tangent line to graph of r f at r , is dy f cos f sin . dx f sin f cos
If f 0 and f 0, then is tangent at the pole.
96. r 4 cos 2
98. r 2 sec
Rose curve
Line
Matches (b)
Matches (d)
Section 9.4
Polar Coordinates and Polar Graphs
100. r 61 cos
(a) 0, r 61 cos
(c)
9
2
6 1
2 cos cos sin sin 2 2
r 6 1 cos −9
15
−9
(b)
, r 6 1 4
cos
61 sin 15
4
12 − 12
12
−3
−9
The graph of r 61 cos is rotated through the angle 2.
15
−6
The graph of r 61 cos is rotated through the angle 4.
102. (a) sin
2
2
sin cos
cos
(b) sin sin cos cos sin sin
f cos
cos sin
r f sin
(c) sin
2
r f sin
2
f sin
3 3 3 sin cos cos sin 2 2 2
cos
r f sin
3 2
f cos
104. r 2 sin 2 4 sin cos
(a) r 4 sin
cos 6 6
(b) r 4 sin
2
−3
−3
−2
sin cos
−2
2 2 cos 3 3
(d) r 4 sin cos 4 sin cos 2
−3
3
3
−2 −2
4
3
2
−3
2
3
(c) r 4 sin
cos 2 2
451
452
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
106. By Theorem 9.11, the slope of the tangent line through A and P is
π 2
f cos f sin f sin f cos
Radial line
Polar curve r = f (θ)
P = (r , θ)
This is equal to tan
ψ
Tangent line
tan tan sin cos tan . 1 tan tan cos sin tan
θ 0
A
Equating the expressions and cross-multiplying, you obtain
f cos f sin cos sin tan
sin cos tan f sin f cos
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 cos2 sin2
f
tan
108. tan
At
r 31 cos dr d 3 sin
3 1 , tan 4
tan cos2 sin2
f r . f dr d
110. tan
22 2 2.
2
2
At
sin 3 , tan 6 2 cos 3
arctan
−8
−6
2
6
−4
−5
r 5 undefined dr d 0
⇒
. 2
6
−9
9
−6
114. True
Section 9.5
116. True
Area and Arc Length in Polar Coordinates
2. (a) r 3 cos
(b) A
2
2
3 cos
2
1 2
2
d
0
2
9
cos2 d
0
0 2
A
32
4
2
9 4
3 2
.
4
5
112. tan
23 0.7137 40.89
2 2 1.041 59.64 2
arctan
4 sin 2 r dr d 8 cos 2
9 2
2
1 cos 2 d
0
9 sin 2 2 2 2 0
9 4
Section 9.5
4. A
2
1 2
0
36
1
0
8. A
18 4
1 2
6. A
cos 4 d 2
2
2
cos 5 d
0
10
1 1 sin10 2 10
0
20
9 2
2
2
1 sin d
10
1 2
0
32
sin2 2 d
36
sin 4 4 4 0
18
2
0
4
6 sin 2 2 d
4
4
453
Area and Arc Length in Polar Coordinates
10. A
2 1 2 cos sin 2 4 0
2
1 2
2
4 6 sin 2 d
arcsin 2 3
2
3 8 4
2
16 48 sin 36 sin d
arcsin 2 3
2
16
48 sin 36
arcsin 2 3
1
cos 2 2
d
2
34 48 cos 9 sin 2
arcsin 2 3
1.7635
2
−8
8
−12
12. Four times the area in Exercise 11, A
2
1 2
4 3 3 . More specifically, we see that the area inside the outer loop is
2
2
21 2 sin 2 d
6
4 16 sin 16 sin2 d 8 6 3.
6
6
The area inside the inner loop is 2
1 2
3 2
21 2 sin 2 d
7 6
4 6 3.
Thus, the area between the loops is 8 6 3 14. r 31
sin
r 31
sin
−4 −1
4 6 3 4 12 3. 16. r 2
sin
3 cos
r cos
Solving simultaneously, 31
4
31
Solving simultaneously,
sin
2 sin 0
0, . Replacing r by r and by in the first equation and solving, 31 sin 31 sin , sin 1, 2. Both curves pass through the pole, 0, 3 2, and 0, 2, respectively. Points of intersection: 3, 0, 3, , 0, 0
2
3 cos cos
1 2 5 . , 3 3 Both curves pass through the pole, (0, arccos 23), and 0, 2, respectively. cos
Points of intersection:
12, 3 , 12, 53 , 0, 0
454
Chapter 9
18. r 1
Conics, Parametric Equations, and Polar Coordinates
20.
cos
r 3 cos
π
2
r 2
Solving simultaneously, 1
4
Line of slope 1 passing through the pole and a circle of radius 2 centered at the pole.
cos 3 cos cos
1 2
0 1
3
Points of intersection:
2, 4 ,
5 , . 3 3
2,
4
Both curves pass through the pole, 0, , and 0, 2, respectively. Points of intersection:
22. r 3
32, 3 , 32, 53 , 0, 0
sin
Points of intersection:
π
2
r 2 csc
0 1
2
17 3 2
172
, arcsin
3
17 3 2
,
172 3,
, arcsin
3.56, 0.596 , 3.56, 2.545 The graph of r 3 sin is a limaçon symmetric to 2, and the graph of r 2 csc is the horizontal line y 2. Therefore, there are two points of intersection. Solving simultaneously, 3
sin 2 csc
sin2 3 sin 2
sin
0 2
arcsin
24. r 31 r
17
3 ±
17 3 2
0.596.
cos
r =
6 1 cos
The graph of r 31 r 61
cos is a cardioid with polar axis symmetry. The graph of
5
− 10
5
cos
is a parabola with focus at the pole, vertex 3, , and polar axis symmetry. Therefore, there are two points of intersection. Solving simultaneously, 31
6 1 − cos θ
cos
1
6 cos
±
2
r = 3(1 − cos θ )
1 cos 2 2 cos 1
arccos 1 Points of intersection:
2 .
3 2, arccos 1 2 4.243, 1.998 , 3 2, 2 arccos 1 2 4.243, 4.285
−5
Section 9.5
26. r 4 sin r 21
28. A
2
1 2
4
sin
91
2
sin d
0
2
Points of intersection: 0, 0, 4, 2
455
Area and Arc Length in Polar Coordinates
1 sin 2 d
18
0
9 3 8 2
(from Exercise 14)
7
The graphs reach the pole at different times ( values). 6
−7
7
r = 4 sin θ
−7 −6
6
−2
30. r 5
r = 2 (1 + sin θ )
3 sin and r 5
32. A
3 cos intersect at
4 and 5 4. A
1 2 2
4
30
2
9 4 592 4
30
2 2
9 4
4 cos
4
4
−4
8
12
−8
34. Area Area of r 2a cos Area of sector twice area between r 2a cos and the lines , . 3 2 A
a 2
2 a 2 3 2 a 2 3 2 a 2 3
a2
2
1 2
36. r a cos , r a sin
tan 1, 4
2
2a 2
θ
=
2a
2a 2
1 2
2
2
4
a cos 2 d
0
a2
1
0
1
cos 2 d
3
2
A
2a cos 2 d
3
2
π
2
3
sin 2 2 2 3
3 2 3 4
2 a 2
6
3 3a 2
cos 2 d 2
1 1 a 2 4 2 1 2 sin 2 a 2 2 2
1 2 a 4
0
1 2 a 8
π
3
2
a
2a
r = a sin θ
0
0
a θ
4
π
a
π = − 3
2
sin2
59 30 2 50.251 2
−12
6
4 cos 2 4 sin d
2
r = a cos θ
6
2 sin 2 d
−4
2
6
5 4 59 9 30 cos sin 2 2 4 4
6
1 2
5 3 sin 2 d
2
3 sin 2 d
2
5 4
592 54
2
1 2
2
3 3
456
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
38. By symmetry, A1
4 cos2 1 d 2a2
3
1 2 a 2 2
1 5 2 a 2 6 5 a 2 12
2a 2
2
a
12
2a
cos 2 d
4
6
a 2 3 2 2
a2 1
3 2
a2
4 1
2
2a sin d θ =
A2
5π 6
A6
1 cos 2 d
5 a 2 12
2 sin 2
5 6
12
a
2
3 3 2
a
2
θ = −
π
3
a2 d
6
4
1 cos 2 d a 2
6
a 2 2 sin 2
0
a 2 12
6
a2
3 3 2
a 2 12
a2
5 3 12 2
4
A7
2a sin 2 a2 d
6
a2
4
6
A6 A7 A4
[Note: A1
a2
r sec 2 cos ,
40.
r cos 1 x
1
2 cos2
2
4
4 sin2 1 d a 2 sin 2
6
a2
1 12
1
y 2
A
2
< < 2 2
r 2 cos2 1 r 2
y
1
2
x2
x 2
y 2
x
0
−1
x 21 x 1 x
4
sec 2 cos 2 d
0
4
2
x 2 x3
1 2
3
area of circle of radius a]
x 2 y 2 x x 2 y 2 2 x 2 y 2 x
sec2 4 4 cos2 d
4
0
sec2 4
21
θ =
π
6
0
r = 2 a cos
4
2a sin 2 d 2
4
2a
r = 2
3 12 2
π
A1 a
A4
A5
A7
a
A3
3 θ =
2
5 6
π
2a
r = 2a sin θ
6
1 2 2
θ =
2
0
6
4
π
0
2
2a cos 2 2a sin 2 d
a 2 4
1 2 2
a 2 sin 2
6
A6
4
5 6
5 a 2 12
1 2
6
A5
3
A4
A3
2a cos 2 a2 d
6
6 a2 sin 2 2 3
A4.
6
a2 2
and A3
1 2
A2
A1
A2
4
cos 2 d tan 2 sin 2
0
2
2
θ
Section 9.5
44. r 81
42. r 2a cos r s
2a sin
Area and Arc Length in Polar Coordinates
r
8
cos , 0
≤
≤ 2
sin
2
2a cos 2 2a sin 2 d
s
2
457
81 cos 2 8 sin 2 d
2
0
2
2
2a d 2
2
2
2 a
1 2 cos cos2 sin2 d
16
0
16 2
1 cos d
0
16 2
1 cos
0
16 2
0
46. r sec , 0 ≤ ≤
3
sin
1 cos
1 1
cos
cos
d
d
32 2 1
cos
0
64
50. r 2 sin2 cos , 0 ≤ ≤
48. r e , 0 ≤ ≤ 10
2
3
−2 −3
4
− 25
4
5
−5
−2
−3
Length 31.31
Length 1.73 exact 3
Length 7.78
52. r a cos r
a sin
2
a cos cos a2 cos a2 sin2 d
S 2
0
2 a2
2
cos2 d a2
0
a2
54. r a1 r
2
1 cos 2 d
0
sin 2 2
2
0
2a2 2
cos
a sin
a1
S 2
21 cos sin a
cos 2
a2 sin2 d 2 a2
0
sin 1
cos 2
2 cos d
0
2 2 a2
1 cos 32 sin d
0
56. r
4 2 a2 1 5
cos 52
0
32 a2 5
58. The curves might intersect for different values of :
r 1
See page 696.
S 2
0
2 sin
1 d 42.32
458
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
2 60. (a) S 2 f sin f
2 d
f
2 (b) S 2 f cos f
2 d
f
62. r 8 cos , 0
(a) A
1 2
≤ ≤
2
r d
0
(Area circle
1 2
2
64 cos d 32
0
1
0
r2 42
cos 2 sin 2 d 16 2 2
16
0
16 ) 1
(b)
A
0.2
0.4
0.6
0.8
1.0
1.2
1.4
6.32
12.14
17.06
20.80
23.27
24.60
25.08
(c), (d) For 4 of area 4 12.57: 0.42 1
For 2 of area 8 25.13: 1.57 2 3
For 4 of area 12 37.70 : 2.73 (e) No, it does not depend on the radius.
64. False. f 0 and g intersection.
Section 9.6 2. r
sin 2 have only one point of
Polar Equations of Conics and Kepler’s Laws
2e 1 e cos
4. r
2 , parabola 1 cos
(a) e
1, r
(b) e
0.5, r
(c) e
1.5, r
1
1 0.5 cos 2
1
3 1.5 cos 2
2 , ellipse cos
6 , hyperbola 3 cos
2e 1 e sin
1, r
(b) e
0.5, r
(c) e
1.5, r
4
e = 1.5
2 , parabola 1 sin
(a) e
1
1 0.5 sin 2
1
3 1.5 sin 2
2 , ellipse sin
6 , hyperbola 3 sin
9
e=1
−9
e = 1.5
3
e=1 −9
9
e = 0.5 −4
6. r
1
e = 0.5
4 0.4 cos (b) r
(a) Because e 0.4 < 1, the conic is an ellipse with vertical directrix to the left of the pole. (c)
−3
9
7
− 10
10 −8
8
1
4 0.4 cos
The ellipse is shifted to the left. The vertical directrix is to the right of the pole 4 r . 1 0.4 sin The ellipse has a horizontal directrix below the pole.
−7
8. Ellipse; Matches (f)
−5
10. Parabola; Matches (e)
12. Hyperbola; Matches (d)
Section 9.6
14. r
6 1 cos
Parabola since e
16. r
5 5
3 sin
1
Ellipse since e
Vertex: 3, 0 π
2
Vertices:
Polar Equations of Conics and Kepler’s Laws
3 5
1 1
18. r 3
35sin
r <
1
5 5 3 , , , 8 2 2 2
3 1
2 cos
6 2 cos
2 23 cos
Ellipse since e
π
2
0 4
8
Vertices: 6, 0,
0 1
6
2
2 < 1 3
65,
π
2
0 1
20. r
6
3
7 sin
Hyperbola since e Vertices:
2
1
7 3
22. r
73sin >
1
Vertices:
3 3 3 , , , 5 2 2 2
3
4
4 2 cos
Hyperbola since e
1.
2
2 > 1
43, 0,
4,
2 π
2
0
0
2
24.
2
Hyperbola
4
−6
6
26.
−6
6
−4
−4
6
28. r
1
Hyperbola
4
cos
30. r
3
Rotate the graph of r
3
7 sin 2 3
Rotate graph of r
6 1 cos
counterclockwise through the angle
6
6
3
Clockwise through angle of 2 3.
3
.
4
6 −6
− 14
6
10
−4
− 10
. 7 sin
5
459
460
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
32. Change to
2
: r
6
1
34. Parabola
sin
6
e
e
r
3 , y 4
2,
d 2
1
ed e sin
1
234 34 sin
4
e
3 , x 2
r
6 3 sin
40. Parabola
Vertex: 5, 1,
d 1
1
ed e cos
1
32 32 cos
2
3 3 cos
46. r
e
r
ed 1 e sin
1, d 10 ed 1 e cos 1
10 cos
Vertices: 2, 0, 10, 0
1 , d 8 3
1
e
r
2 , 4, 32
r
1 sin
44. Hyperbola
Vertices: 2,
1, d 1
38. Hyperbola
42. Ellipse
e
ed 1 e sin 1
r
36. Ellipse
1, y
83 13 sin
8 3 sin
3 10 , d 2 3 ed 1 e cos
5 1
32 cos
10 2 3 cos
4 is a parabola with horizontal directrix above the pole. 1 sin
(a) Parabola with vertical directrix to left pole.
(b) Parabola with horizontal directrix below pole.
(c) Parabola with vertical directrix to right of pole.
(d) Parabola (b) rotated counterclockwise 4.
x 2 a2
48. (a)
x 2b 2
y
y 2 b2 2
a2
x 2 a2
(b)
1
a 2b 2
x 2b 2
y
y 2 b2
2
a2
1
a 2b 2
b 2r 2 cos2 a 2r 2 sin 2 a 2b 2 2
2
2
2 2
2
2
b r cos a r sin a b r 2 b 2 cos 2 a 21
r 2
cos 2
a 2b 2
a2
a 2b 2 b 2 a 2 cos 2 a 2 2
a 2b 2
a 2
r 2 a 2
cos 2 b 2
1
b ca 2 cos 2 1
a 2b 2 2 2 c cos 2
2
b e 2 cos 2
r 2 b 2 cos2 a 2 1 r 2 a 2 r 2
cos 2 a 2
cos 2
a 2b2
b 2
a 2b 2
a 2b 2 2 2 a c cos 2
b
1
2
2
e cos2
1
b2 c a 2 cos 2 2
Rev iew Ex erc ises for Cha pter 9
50. a 4, c 5, b 3, e r 2
5 4
52. a 2, b 1, c 3, e
9 1 2516 cos 2
1 54. A 2 2
2
2
r 2
2 2 d 4 3 2 sin
3 2
1 1 34 cos 2
2
1 d 3.37 3 2 sin 2 2
ed 1 e cos
56. (a) r
(b) The perihelion distance is a c a ea a1 e.
When 0, r c a ea a a1 e.
When , r
Therefore,
1 e2a 1e
a1 e.
The aphelion distance is a c a ea a1 e. a1 e
ed 1e
When 0, r
1 e2a 1e
a1 e.
a1 e1 e ed a1 e 2 ed .
Thus, r
58. a 1.427
1 e2a
. 1 e cos
109 km
60. a 36.0
e 0.0543 r
1
r
1.422792505 109 e 2 a 1 e cos 1 0.0543 cos
Aphelion distance: a1 e 1.5044861
1
10 6 mi, e 0.206
e 2 a 34.472 10 6 1 e cos 1 0.206 cos
Perihelion distance: a1 e 28.582
Perihelion distance: a1 e 1.3495139
Aphelion distance: a1 e 43.416
109 km
109 km
r a sin b cos
62.
r 2 ar sin br cos x 2 y 2 ay bx x 2 y 2 bx ay 0 represents a circle.
Review Exercises for Chapter 9 2. Matches (b) - hyperbola
4. Matches (c) - hyperbola
6. y 2 12 y 8 x 20 0 y
y 2 12 y 36 8 x 20 36
y 6
2
461
16
42 x 2
12
Parabola Vertex: 2, 6 x −4
8
12
10 6 mi
10 6 mi
462
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
8. 4 x 2 y 2 16 x 15 0
y
4 x 2 4 x 4 y 2 15 16
x 2 2 14
1
(2, 0)
y 2 1 1
x
1
2
3
−1
Ellipse Center: 2, 0
−2
Vertices: 2, ± 1
4 x 2 4 y 2 4 x 8 y 11 0
10.
y
4
1 4 x x 4 y 2 2 y 1 11 1 4 4 2
x 12 2 2
y 1 2
2
3
1
1
x −2
1
Hyperbola
−1
1 2, 1 Vertices: 2 Asymptotes: 1
−2
Center:
1 ,1 2
3
4
±
y
x
±
1 2
12. Vertex: 4, 2
Focus: 4, 0 Parabola opens downward p 2
x 42 4 2 y
2
x 2 8 x 8 y 0 14. Center: 0, 0
16. Foci: 0, ± 8
Solution points: 1, 2, 2, 0
Asymptotes: y
Substituting the values of the coordinates of the given points into
Center: 0, 0
x 2 y 2 1, b2 a2
b
2
a
2
a y x 4 x asymptote → a 4b b
1, 4b 2 1.
Solving the system, we have a2
18.
x 2 16 3 y 2 and b 2 4, 1. 3 4 16
21 x 2 y 2 1, a 5, b 2, c 21, e 4 25 5 By Example 5 of Section 9.1,
2
C 20
1
0
Vertical transverse axis c8
we obtain the system
1 4
± 4 x
21 sin2 d 23.01. 25
b2 c2 a2 64 4b2 ⇒
b2
64 17
⇒
a2
y2 x2 1 102417 6417
1024 17
⇒
17b2 64
Rev iew Exe rcise s for Cha pter 9
20. y
1 2 x 200
(a) x 2 200 y
y
1 2 x 200
y
1 x 100
1 y 2
(b)
x2 450 y
Focus: 0, 50
1
x2 10,000
100
S 2
1
x
0
22. (a) A 4
a
0
b
0
b
b 2 y 2 dy
a
ab 0
0
b
b 4
c 2 y 2 dy
0
2 a 2
ab 2
c
a
(c) Disk: V 2
0 a
S 22
0
a
0
ln
ca e
2
2 a 2
b2 2 2 b 2 a x 2 dx 2 2 a a
b 2
e
a 2
x 2
dx
0
2 b 2 2 1 a x x 3 2 a 3
2
24. x t 4, y t 2
x 3 3
9
7 6
0
4 ab 2 3
0
arcsin ab e
2 b 2 2
2
x 32
y
a
e
26. x 3 3 cos , y 2 5 sin
y x 42
a
dx
cx 2 b a c x dx 2 cx a 4 c 2 x 2 a 4 arcsin 2 a c a 2
0
e e
a
4
b
ln11
a 4 a 2 b 2 x 2 b a 2 x 2 a a a 2 x 2
Parabola
4 a 2b 3
a b 2 c a c a 2 c 2 a 4 arcsin a 2c a
⇒
0
t x 4
2 a cy b 4 c 2 y 2 b 4 ln cy b 4 c 2 y 2 b 2c
2 a 2 b c b 2 c 2 b 4 ln cb b b 2 c 2 b 4 lnb 2 b 2c
b
4 b a2
2 a 2 2 1 b y y 3 2 b 3
y 2 5
2
y 22 25
1
1
28. x 5 sin3 , y 5 cos3
5 x
2 3
y 23
5
1
x23 y23 523 y
Ellipse
5
x2 dx 38,294.49 10,000
b 4 a 2 b 2 y 2 a b 2 y 2 dy b b b 2 y 2
0
4 a b2
a2 2 2 a 2 2 b y dy b2 b2
b
S 4
b x 4b 1 a 2 x 2 dx x a 2 x 2 a 2 arcsin a a 2 a
(b) Disk: V 2
6 4
4
y
3
2
2 1 x −1
1
2
463
3
4
5
6
7
7 6 5 4 3 2 1
x −6
−2 −3
2
−4 −6
x −2 −1
−4
1 2 3 4 5 6
7 8
4
6
464
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
30. x h 2 y k 2 r 2
x 5 2 y
32. a 4, c 5, b2 c2 a2 9,
3 2 2 2 4 Let
y 2 x 2 1 16 9
y 2 x 2 sec 2 and tan 2 . 16 9
Then x 3 tan and y 4 sec .
sin
34. x a b cos t b cos
ab t b
ab t b
y a b sin t b
(b) a 3, b 1
(a) a 2, b 1
(c) a 4, b 1
x cos t cos t 2 cos t
x 2 cos t cos 2t
x 3 cos t cos 3t
y sin t sin t 0
y 2 sin t sin 2t
y 3 sin t sin 3t
4
2
4
y = 0 −2 ≤ x ≤ 2 −3
−6
3
6
−6
6
−4
−2
−4
(e) a 3, b 2
(d) a 10, b 1 x 9 cos t cos 9t
(f) a 4, b 3
x cos t 2 cos
y 9 sin t sin 9t y sin t 2 sin
10
t 2
x cos t 3 cos
t 2
y sin t 3 sin
4 − 15
6
−6
6
− 10
−4
−4
36. x t u r cos r sin
38. x t 4
r cos sin
y t 2
y v w r sin r cos
(a)
r sin cos
dy 2t 2t 0 when t 0. 1 dx
Point of horizontal tangency: 4, 0
y
(b) t x 4 y x 4 2
(c) r θ
t
r θ
θ
v
u
y
6
w
5
( x, y) x
t 3
4
15
−6
t 3
4 3 2 1 x
1
2
3
4
5
6
465
Rev iew Ex erc ises for Cha pter 9
40. x
1 t
42. x 2t 1
y t 2
y
2t dy (a) 2t 3 1t 2 dx
(a)
No horizontal tangents
t 0
1 2
t 2t t 2 2t 2 2t 2 dy 2 dx
1 (b) t x
1 t 0 when t 1. t 2t 2 2
Point of horizontal tangency: 1, 1
1 y 2 x
(b) t
(c)
x 1 2
y
1
y
4
3
x 122 2 x
(c)
12
4
x 3 x
y
2
2
1 x −2
2
4
x −2
−1
1
2
46. x e t
44. x 6 cos
y et
y 6 sin
(a)
3 6 cos dy cot 0 when , . 6 sin 2 2 dx
(a)
Points of horizontal tangency: 0, 6, 0, 6 (b)
(c)
y
x 6
2
No horizontal tangents (b) t ln x
2
y 6
et 1 1 dy 2t 2 t dx e e x
1
1 y eln x e ln1 x , x > 0 x (c)
4
y
3
2 x −4
−2
2
2
4
−2
1
−4
x
1
y 6 sin
y 2 cos
dx 6 sin d
8
−8
dy 6 cos d
8
−4
3 dx (b) At , 1.134, 2 , 6 d 2
3
50. x 6 cos
48. x 2 sin
(a), (c)
2
dy dy 0.441 0.5, and dt dx
s
36 sin 2 36 cos 2 d 6
0
(one-half circumference of circle)
0
6
1
466
Chapter 9
Conics, Parametric Equations, and Polar Coordinates
52. x, y 1, 3
y
(−1, 3)
r 1 2 3 2 10
2
r ,
arctan 3 1.89 108.43
1
10, 1.89, 10, 5.03
x −3
−2
−1
1
2
3
−1 −2 −3
r 10
54.
r
56.
r 2 100
1 2 cos
2r r cos 1
x 2 y 2 100
x 2 y 2 x 1 2 ±
4 x 2 y 2 x 1 2 3 x 2 4 y 2 2 x 1 0
58.
r 4 sec
3
4
60.
cos 3
3 4
tan 1
4
12 cos 32 sin
r cos 3 sin 8
y 1 x
x 3 y 8
y x
y 64. x 2 y 2 arctan x
62. x 2 y 2 4 x 0 r 2 4r cos 0
2
a2
r 2 2 a 2
r 4 cos
66.
π
68. r 3 csc , r sin 3, y 3
2
12
π
2
Horizontal line
Line 0 1
2
0 1
70. r 3 4 cos
Limaçon Symmetric to polar axis π
2
2
0
r
1
2
2 3
3
1
3
5
7
0
2
3
4
467
Rev iew Ex erc ises for Cha pter 9
72. r 2
Spiral Symmetric to 2 π
2
0 2
4
8
0
r
0
4
2
3 4 3 2
5
2
5 4 5 2
3 2 3
74. r cos 5
π
2
Rose curve with five petals Symmetric to polar axis
2 3 4 , 1, , 1, , 1, 5 5 5 5 3 7 9 Tangents at the pole: , , , , 10 10 2 10 10 Relative extrema: 1, 0, 1,
76. r 2 cos2
0
r
±1
Lemniscate Symmetric to the polar axis Relative extrema: ± 1, 0 Tangents at the pole:
4
,
±
6
4
2 2
0 1
π
2
0 0 1 2
3 4
78. r 2 sin cos 2
80. r 4sec cos
0.75
3
Bifolium
Semicubical parabola
Symmetric to 2
Symmetric to the polar axis −1
−1
5
1
r ⇒
as ⇒
− 0.25
r ⇒
as ⇒
2
−3
2
82. r 2 4 sin2
(a) 2r
dr 8 cos2 d
(b)
dr 4 cos2 d r
Tangents at the pole: 0, (c)
2
−3
dy r cos 4 cos 2 sin r dx r sin 4 cos 2 cos r
3
2
cos2 sin sin2 cos cos2 cos sin2 sin
Horizontal tangents: dy 0 when cos2 sin sin 2 cos 0, dx
tan tan 2 , 0,
3
, 0, 0,
2 3,
±
3
Vertical tangents when cos 2 cos sin 2 sin 0: −2
tan 2 tan 1, 0,
6
, 0, 0,
±
2 3,
6
