Hungerford T.W., Shaw D.J. Contemporary Precalculus: A Graphing Approach
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(e) It has the same shape (a bit like a witch’s hat), but its
peak is at (0, 6).
53. Local minimum at (6, 2)
55. Local maximum at (4, 5)
57. (a) Calculator is in degree mode for this graph.
(b) t 3.23 seconds
(c) y 41.67 feet
59. (a)
(b) The arrow will go over the wall.
61. v 105.29 ft/sec.
63. (a)
(b) 40°
(c) It appears that the ball would go farthest for a 45° angle.
65. (a) OT 3t (see page 737). The figure shows that
cos
t
p
2
C
3
Q
and sin
t
p
2
P
3
Q
. Hence,
CQ 3 cos
t
p
2
and PQ 3 sin
t
p
2
. So x
OT CQ 3t 3 cos
t
p
2
and y CT PQ
3 3 sin
t
p
2
.
(b) cos(t p/2) cos t cos p/2 sin t sin p/2
(cos t)(0) (sin t)(1) sin t. Therefore, 3t
3 cos(t p/2) 3t 3 sin t 3(t sin t).
Sin(t p/2) sin t cos p/2 cos t sin p/2
(sin t)(0) (cos t)(1) cos t. Therefore, 3
3 sin(t p/2) 3 3 cos t 3(1 cos t).
0
200
0 350
0
0 250
120
0
50
0 350
1036 ANSWERS
67. (a) OT 3t (see page 737). The figure shows that
cos
t
3
2
p
C
3
Q
and sin
t
3
2
p
P
3
Q
. Hence,
CQ 3 cos
t
3
2
p
and PQ 3 sin
t
3
2
p
.
So x OT CQ 3t 3 cos
t
3
2
p
and y
CT PQ 3 3 sin
t
3
2
p
.
(b) cos(t 3p/2) cos t cos 3p/2 sin t sin 3p/2
(cos t)(0) (sin t)(1) sin t. Therefore,
3t 3 cos(t 3p/2) 3t 3 sin t 3(t sin t).
Sin(t 3p/2) sin t cos 3p/2 cos t sin 3p/2
(sin t)(0) (cos t)(1) cos t. Therefore, 3
3 sin(t 3p/
2) 3 3 cos t 3(1 cos t).
69. (a) Particles A and B do not collide.
(b) t 1.10.
(c) A and C are closest near t 1.13, but do not collide.
(d) The given function is obtained by applying the distance
formula to the points (8 cos t, 5 sin t) A and
(3t, 4t) C. y
(8 cos
t 3t)
2
(5
sin t
4t)
2
There is a minimum at t 1.1322; however, the mini-
mum is not zero.
71. (a) As shown on next page, the center Q of the small circle
is always at distance a b from the origin O. Suppose
t is the angle that OQ makes with the x-axis. Then
the coordinates of Q are x (a b) cos t,
y (a b) sin t. Examining the smaller circle in detail,
we see that the change in x-coordinate from Q to P is
b cos u, where u is the angle that PQ makes with the
−1
02
8
0
0
12
6
0
0 250
120
positive x-axis. Likewise, the change in y-coordinate from
Q to P is b sin u. Therefore, the coordinates of P are
x (a b) cos t b cos u
y (a b) sin t b sin u.
The angles t and u are related by the fact that the inner
circle must roll without “slipping.” This means the arc
length that P has moved around the inner circle must
equal the arc length that the inner circle has moved
along the circumference of the larger circle. In other
words, the arc length from P to W must equal the arc
length from S to V. Since the length of a circular arc is
the radius times the angle, this means
bu (a b)t, or u (a b)t/b.
Substituting this for u in the above equations will give
the desired parametric equations.
(b)
y
x
24
−2−4
2
4
−2
−4
P
W
Q
u
b
y
x
a − b
a − b
a
S
0
P
t
V
Q
b
ANSWERS 1037
(c)
Section 10.6, page 752
1. P:
2,
p
4
, Q:
3,
2
3
p
, R: (5, p), S:
7,
7
6
p
, T:
4,
3
2
p
,
U:
6,
p
3
, V: (7, 0)
3.
5.
Many answers are possible in Exercises 7–12 in addition to
those given here.
7.
3,
7
3
p
,
3,
5
3
p
,
3,
4
3
p
,
3,
2
3
p
9.
2,
4
3
p
,
2,
8
3
p
,
2,
p
3
,
2,
5
3
p
11.
3
,
11
4
p
,
3
,
5
4
p
,
3
,
7
4
p
,
3
,
p
4
123
(−3, −5π/3)
(2, −3π/4)
123
y
x
24
−2−4
2
4
−2
−4
13.
3
2
,
3
2
3
15.
2
3
,
1
2
17. (.4255, 1.438) 19. (3.604, 1.736)
21.
6,
p
6
23.
2
,
p
4
25.
6,
p
3
27. (25
, 1.107)
29. (5.59, 2.6679) 31.
2,
3
2
p
33. (4.47, 2.0344) 35. r 5
37. r 12 sec u 39. r
sin u
1
2 cos u
41. x
2
y
2
9 43. y
x
3
45. x 1 47. x
3
xy
2
y
49. x
2
y
2
2y 51. x
2
16 8y
53.
55.
57.
−13
1
−2
2
4
−4
2
−2−3
θ= 0
θ= π/2
1 radian
−13
1
−2
2
4
−4
2
−2−3
θ= 0
θ= π/2
−π/3
−24
2
−2
2
4
−4
−4
u 0
u p/2
1038 ANSWERS
59.
61.
63.
65.
−1
1
−1 1
θ= 0
θ= π/2
θ= 3π/2
−1
1
−2 −1
θ= 0
θ= π
θ= π/2
θ= 3π/2
−0.5 1
0.5
−1
−1.5
−2
−0.5
−1
θ= 0
θ= π/2
θ= 3π/2
−510
5
−7.5
−10
−5
2.5
7.5
5
−2.5
−10
θ= 0
θ= −3π/2
θ = −π/2
67.
69.
71.
73.
75.
−.75
−1
.75
1
θ= 0
θ= π/2
θ= 3π/2
1.2
−.2
1.2
1
θ= 0
θ= π/2
θ= 3π/2
1
−3
−2
−1
3
2
1
23456
θ= 0
θ= π/2
θ= 3π/2
1−1−2
−.2
.2
.4
.6
−.4
−.6
2
θ= 0
θ= π/2
θ= 3π/2
−.75
−1
.5
.75
θ= 0
θ= π/2
θ= 3π/2
ANSWERS 1039
77.
79.
81.
83.
(a)
(b)
−3
3
−44
−3
3
−44
−.2
−.1
−.2
.5
.4
.3
.2
.1
.2 .4
θ= 0
θ= π/2
θ= 3π/2
6
4
2
−6 −4 −2
θ= 0
θ= π/2
.25
−.4
−.2
.4
.2
.5
θ= 0
θ= π/2
θ= 3π/2
(c)
85. r a sin u b cos u
r
2
ar sin u br cos u
x
2
y
2
ay bx
x
2
bx
b
4
2
y
2
ay
a
4
2
a
2
4
b
2
x
b
2
2
y
a
2
2
a
2
4
b
2
This is the equation of a circle.
87. The distance from (r, u) to (s, b) is given by the Law of
Cosines.
d
2
r
2
s
2
2rs cos(u b)
d
r
2
s
2
2r
s cos(u
b)
Section 10.7, page 761
1. (d) 3. (c) 5. (a)
7. Hyperbola, e 4/3 9. Parabola, e 1
11. Ellipse, e 2/3 13. .1
15. 5
17. 5/4
19. (a)
(b) 15/4, 10/4, 2
/4
(c) The smaller the eccentricity, the closer the shape is to
circular.
In the graphs for Exercises 21–32, the x- and y-axes with
scales are given for convenience, but coordinates of points are
in polar coordinates.
21.
5
(4, π)
10−5−10
−5
−10
5
10
−4
4
−66
−3
3
−44
1040 ANSWERS
23.
25.
27.
29.
31.
33.
r
1
6
cos u
35. r
5
1
3
6
sin u
37. r
1 2
3
cos u
39. r
1 4
8
cos u
41. r
1
3
sin u
43. r
2
2
cos u
510−5−10
−5
5
10
15
(
−10,
)
3π
2
(
2,
)
π
2
510−5−10
−5
−10
5
10
(
,
)
π
2
3
2
51015
(15, 0)
(3, π)
−5
−5
−10
5
10
510
−5−10
−5
−10
5
10
(
10,
)
π
2
(
,
)
3π
2
10
7
5
(−2, 0)
10−5−10
−5
−10
5
10
(
, π
2
3
)
45. r
1 2
2
cos u
47. Since 0 e 1, 0 1 e
2
1 as well. The formulas for
a
2
and b
2
show that a
2
b
2
/(1 e
2
), so a
2
b
2
. Since a
and b are both positive, a b.
49. r
Chapter 10 Review, page 764
1. Ellipse, vertices 0, 25
, foci (0, 2)
3. Ellipse, vertices (1, 1) and (1, 7), foci (1, 0) and (1, 6)
5. Focus: (0, 5/14), directrix: y 5/14
7.
9.
11.
Asymptotes: y 4
5
2
(x 1)
−10
−20
−30
10
20
x
y
−2−4246
12 34 56
−4
−3
−6
−7
−5
2
−2
−4
4
x
y
−112−2
3
10
7
1 cos u
ANSWERS 1041
13.
15.
17.
19.
21. Center: (4, 5), radius: 22
23. (4, 6) 25.
(x
4
3)
2
( y
2
1)
2
1
27.
y
4
2
(x
16
3)
2
1 29.
y
1
2
2
1
2
x
3
2
31. x 3 cos t 5 and y 3 sin t 2(0 t 2p)
2
2
4
4
2
6
1
3
5
6
3
1
5
46 102
x
y
8135 97
−1
−2
1
2
x
y
23
451
−1
−2
−3
−4
3
2
1
0
x
y
24
6810
5
10
15
20
x
y
12
3456
−1
−2
1
2
x
y
24
68
33. x 2 cos t and y 5 sin t (0 t 2p)
35. x 3 cos t 3 and y 2 sin t 5(0 t 2p)
37. x 2 tan t 1 and y
co
5
s t
4(0 t 2p)
39. x (t 4)
2
2 and y t (any real number t)
2
2
4
4
2
6
10
8
6
10
8
4 6 10 12 14 16 1882
x
y
y
2−2
−20
−10
10
4
x
y
21 43 65
−2
−1
−4
−3
−6
−5
−8
−7
x
2
−2
−4
4
x
y
−112−2
1042 ANSWERS
41. 6 feet
43. The receiver should be placed at the focus,
1
9
6
feet or
6.25 inches from the vertex.
45. Ellipse
47. Hyperbola
49. 6 x 6 and 4 y 4
51. 9 x 9 and 6 y 6
53. 15 x 10 and 10 y 20
55. x
1
2
u
2
3
v, y
2
3
u
1
2
v
57. 45
59. 15 x 15 and 10 y 10
61. 35 x 32 and 2 y 16
63. x 3 2y
65. y 2 2x
2
67. (b) and (c)
69.
(−3, −2π/3)
(2, 3π/4)
123
θ= 0
θ= π/2
0
2
−11
−10
10
−10 10
71.
73.
75.
77.
79.
81.
3
2
,
3
2
3
.5−.5−1
−.1
.1
.2
.3
−.2
−.3
1
θ= 0
θ= π/2
−12
1
−2
−3
−4
−1
−2
θ= 0
θ= π/2
−10 20
10
−15
−20
−10
5
15
10
−5
−20
θ= 0
θ= π/2
−13
1
−2
2
4
−4
2
−2−3
u 0
u p/2
2π/3
2 4−4 −2
−2
−4
2
4
θ= π
θ= π/2
ANSWERS 1043
83. Eccentricity
2
3
0.8165
85. Ellipse
87. Hyperbola
89. r
5
2
c
4
os
91. r
1
2
cos
Chapter 10 Test, page 766
1. (a) Focus (0, 5/4); directrix y 5/4
(b) x
2
36y
2. (a) Hyperbola
(b) Center (0,0); vertices (0, 3), (0, 3);
foci (0, 45), (0, 45)
(c)
2
2
4
4
2
6
10
8
6
10
8
46 10846108 2
x
y
Pole
(−2, π)
(4, 0)
θ= 0θ=π
Pole
θ= 0θ=π
π
2
θ=
3π
2
θ=
3π
2
4,
()
π
2
12,
()
3. Ellipse; center (1, 3)
4. (a) (2, 2)
(b)
5.
(x
4
6)
2
(y
16
1)
2
1
6.
y
4
2
x
9
2
1
7. (a) Parabola
(b) Hyperbola
8. (a) Ellipse
(b) Center (1, 2); vertices (3, 2), (5, 2)
9. 41.42 ft
10. (a) Ellipse
(b) Center (0, 0); vertices (0, 7
), (0, 7
);
foci (0, 3
), (0, 3
)
(c)
11. Hyperbola; 15 x 50 and 200 y 75
12. S
7,
7
6
p
; T
5,
3
2
p
13. x 5 cos t 3 and y 5 sin t 4(0 t 2p)
14. x 3 (t 4)
2
5 and y t (t any real number)
15. (a) (32
, 32
)
(b)
18,
p
6
16. x y
2
3y 2
2
2
4
3
4
2
1
5
3
1
5
431545 2 13
x
y
2
2
4
4
2
6
8
6
10
8
464610 8 2
x
y
1044 ANSWERS
17.
18.
(a) x 402
t and y 402
t 16t
2
5
(b) About 3.62 seconds later
19. x
co
3
s t
and y 4 tan t (0 t 2p)
20. x 2 cos t 4 and y 2 sin t 2(0 t 2p)
21.
22.
r
9
2
4
0
sin
Chapter 11
Section 11.1, page 781
1. The values are a solution.
3. The values are a solution.
5. The values are not a solution.
7. x 6/5, y 8/5 9. x 2/7, y 11/7
11. r 5/2, s 5/2 13. x
3
2
c, y d
1
2
c
15. x 28, y 22 17. x 2, y 1
19. Inconsistent
21. x b, y
4b
5
2
, where b is any real number.
23. x 6, y 2 25. x 13.2, y 3.6
27. x 7/11, y 7 29. x 1, y 1/2
2
2
4
4
2
6
10
8
6
10
8
46 10846108 2
x
y
0
2
2
2
4
4
2
6
6
4646 2
x
y
31. (a) Solve each equation for y, and observe that a positive
slope represents an increasing population; a negative
slope represents a decreasing population.
(b) 1988
33. 2115
35. x
a
rd
d
s
b
b
c
, y
a
a
d
s
r
b
c
c
37. Since the two lines have slopes
1
2
and 2, they are perpen-
dicular and intersect at one point, regardless of c. Therefore,
there is exactly one solution for the system. Alterna-
tively, solve the system to find that the only solution is
x
3c
1
5
8
, y
6c
1
5
4
.
39. c 3, d 1/2
41. Equilibrium quantity is 20,000; equilibrium price is $17.
43. Equilibrium quantity is 4000; equilibrium price is $180.
45. 145 adults; 55 children
47. Boat: 18 mph; current: 2 mph
49. 1
1
2
lb cashews and 4
1
2
lb peanuts
51. 24 grams of 50% alloy; 16 grams of 75% alloy
53. (a) Electric: y 2000 960x, solar: y 14,000 114x
(b) Electric: $6800, solar: $14,570
(c) The costs will be the same when x 14.2. Electric heat-
ing will be cheaper before that, solar afterwards.
55. (a) R 250,500x
(b) C 1,295,000 206,500x
(c) P R C 44,000x 1,295,000
(d) R 109,000x, C 440,000 82,000x,
P R C 27,000x 440,000
(e) 50.3 weeks, $918,000
(f ) It would be better to open off Broadway if a run of 50
weeks or less is expected.
57. (a) R 60x
(b) 400 hedge trimmers
59. Since the break-even point is above the maximum number
that can be sold, the product should not be produced.
61. $14,450 at 9%, $6,450 at 11%
63. (a) y .373x 27.02
(b) y .624x 23.65 (c) 2003
65. 80 bowls and 120 plates
Special Topics 11.1.A, page 790
1. x 3, y 9 or x 1, y 1
3. x
1
4
41
, y
21
8
41
or x
1
4
41
,
y
21
8
41
5. x 7, y 3 or x 3, y 7
7. x 0, y 3 or x 8, y 1
ANSWERS 1045
9. x 2, y 0 or x 4, y 2
11. x 4, y 3 or x 4, y 3 or x 21, y 2 or
x 21, y 2
13. x 1.6237, y 8.1891 or x 1.3163, y 1.0826 or
x 2.8073, y 2.4814
15. x 1.9493, y .4412 or x .3634, y .9578 or
x 1.4184, y .5986
17. x .9519, y .8145
19. No solutions
21. x 2 5
, y 1 5
or x 2 5
, y 1 5
23. x 4.8093, y 19.3201 or x 3.1434, y 7.7374 or
x 2.1407, y 7.7230 or x 2.8120, y 11.7195
25. x 3.8371, y 2.2596 or
x .9324, y 7.7796
27. x 1.4873, y .0480 or x .0480, y 1.4873
or x .0480, y 1.4873 or x 1.4873, y .0480
29. The intersection point nearest the origin is (2.4, 3.2).
31. Two possible boxes: One is 2 by 2 by 4 meters and the other
is approximately 3.1231 by 3.1231 by 1.6404 meters.
33. 13 and 4 35. 30 and 4.25; 4.25 and 30
37. 16 and 14 39. 16.5 feet by 10 feet
41. 32.1 cm and 14.0 cm
Section 11.2, page 803
1.
3.
5. 3x 5y 4
9x 7y 2
7. x z 1
x y 4z 2w 3
4x 2y 5z 2
9. x 3/2, y 5, z 2, w 0
11. x 3 2t, y 5 3t, z 2, w t, where t is any real
number.
13. y 3/2, x 3/2, z 3/2
15. z t, y 1
1
3
t, x 2
4
3
t, where t is any real number.
17. x 14, y 6, z 2
19. x 100, y 50, z 50
21. z t, y 2t 1/2, x t, where t is any real number.
23. x 2, y 3
25. No solution
0
0
0
2
3
4
1
2
5
2
1
4
3
2
1
0
1
0
3
4
6
4
3
2
7
2
1
3