This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 316 Section 7.5
(d) 0 if we assume a continuous distribution. Between 23. The work to raise a thin disk is
r 2(56)h ( 102 56 (12 y 2)2(56)(10 2 59.5 in. and 60.5 in., the proportion is y) y 60.5 y 2) y. The total work is y)(100 f (x) dx 56 (12 y 2) dy, which evaluates using NINT y)(100 0 to 0.071 (7.1%) 59.5 10 1 28. Use f (x) (x 498)2/(2 1002) e 100 2 967,611 ftlb. This will come to
500 (967,611)($0.005) (a) $4838, so yes, there’s enough money f (x) dx 0.34 (34%) 400 to hire the firm. (b) Take 1000 as a conveniently high upper limit:
1000 24. Pipe radius 1
ft;
6 700
360 12
(62.4)y dy
6 Work to fill pipe
0
385 0.217(300) 58,110,000 ftlb. 30. The proportion of lightbulbs that last between 100 and 800
hours. 58,222,320 ftlb, which will take 35,780,000 31.
58,222,320 1650 110,855 sec 6.5 people 29. Integration is a good approximation to the area (which
represents the probability), since the area is a kind of
Riemann sum. 360 Total work 0.217, which means about 112,320 ftlb. (10)2(62.4)y dy Work to fill tank f (x) dx 6,370,000 31 hr. M
25. (a) The pressure at depth y is 62.4y, and the area of a thin 1000MG
dr
r2 1024, G 5.975 1
r 1000 MG
6.6726 10 35,780,000 , which for
6,370,000 11 evaluates to 1010 J. 5.1446 11
ft, so
horizontal strip is 2 y. The depth of water is
6 32. (a) The distance goes from 2 m to 1 m. The work by an the total force on an end is
11/6 (62.4y)(2 dy) external force equals the work done by repulsion in 209.73 lb. 0 moving the electrons from a 1m distance to a 2m
(b) On the sides, which are twice as long as the ends, the
initial total force is doubled to 419.47 lb. When the tank is upended, the depth is doubled to
11/3 force on a side becomes 11
ft, and the
3 (62.4y)(2 dy) distance:
2 Work 23 dr 2 1 r 23 838.93 lb, 29 10 10 0 1.15 which means that the fluid force doubles.
26. 3.75 in. 5
ft, and 7.75 in.
16 31
ft.
48 31/48 Force on a side p dA (64.5y)
0 5
dy
16 4.2 lb. 27. (a) 0.5 (50%), since half of a normal distribution lies
below the mean.
65 (b) Use NINT to find f (x) dx, where
63 f (x) 1 e 2 2 (x 63.4) /(2 3.2 ) . The result is 3.2 2 0.24 (24%).
(c) 6 ft 72 in. Pick 82 in. as a conveniently high upper
82 limit and with NINT, find f (x) dx. The result is
72 0.0036 (0.36%). 28 10 2 1
r 29 1 J (b) Again, find the work done by the fixed electrons in
pushing the third one away. The total work is the sum
of the work by each fixed electron. The changes in
distance are 4 m to 6 m and 2 m to 4 m, respectively.
6 Work 23 10
r2 4 23 10 7.6667 4 29 dr 29 10 1
r
29 6
4 J. 2 23 10
r2
4
1
r2 29 dr 317 Chapter 7 Review 33. F dv
dt m x
2 W x
1
x
2
x
1
v
2 mv s Chapter 7 Review Exercises dv
, so
dx (pp. 413–415) F(x) dx 5 1. 0.2t 3) dt 0 13
t
3 mv dv 7 v
1 2. 1
mv 2
22 (t 2 0 dv
dx
dx mv 5 v (t) dt 0.001t 4) dt (4 0 1
mv 2
21 10.417 ft
0 7 c(t) dt 5 0.05t 4 0 0.0002t 5 4t 7 31.361 gal
0 34. Work Change in kinetic energy 12
mv .
2 100 3. 0.3125 lb
0.009765625 slug, and
32 ft/sec2
5280 ft
1 hr
90 mph 90
132 ft/sec, so
1 mi
3600 sec
1
Work change in kinetic energy
(0.009765625)(132)2
2 0 1.6 oz 1
(0.003125)(280)2
2 Work
37. 2 oz 2 2 4. (x) dx (11 0 so Work 24 Work
39. 6.5 oz
Work cos t
12 sin t
12 0 12 dt
24 14,400
0 6. 122.5 ftlb.
[ 1, 3] by [ 1, 2] The curves intersect at x
2 x dx 0.02832 slug, so 0 1 12
x
2 1
dx
x2 1
2 109.7 ftlb. 1 lb
/(32 ft/sec2)
16 oz 1
(0.01270)(132)2
2 14 g 0.003125 slug, so 1 lb
14.5 oz
/(32 ft/sec2)
16 oz 6.5 oz 2
0 300 2 0 1 lb
1
/(32 ft/sec2)
slug, and
16 oz
256
5280 ft
1 hr
124 mph
181.867 ft/sec,
1 mi
3600 sec
11
(181.867)2 64.6 ftlb.
2 256 1
(0.02832)(88)2
2 2x 2 11x 24 E(t) dt 5. 1 38. 14.5 oz 4x) dx 0 2 oz 124 mph 1464
0 300 2t
1 lb
/(32 ft/sec2)
16 oz 100 33.333e0.03x 21x 85.1 ftlb.
36. 1.6 oz e0.03x) dx (21 0 2 oz
1/8 lb
1
m
slug, so
32 ft/sec2
32 ft/sec2
256
11
Work
(160)2 50 ftlb.
2 256 35. 0.3125 lb 100 B(x) dx 0.01270 slug, so 1. The area is
1 1
x 0 1
2 2
1 1 1. 7. 110.6 ftlb. 1
1
lb
slug. Compression energy of spring
8
256
12
1
12
ks
(18)
0.5625 ftlb, and final height is
2
2
4
0.5625
given by mgh 0.5625 ftlb, so h
4.5 ft.
(1/256)(32) 40. 2 oz [ 4, 4] by [ 4, 4] The curves intersect at x
1 [3 x2 2 and x
1 (x 1)] dx 2 ( x2 1. The area is
x 2) dx 12
x
2 2x 2 13
x
3
1
3
9
.
2 1
2 2 1
2 8
3 2 4 318 Chapter 7 Review 8. x y 1 implies y x)2 (1 1 2 x x. 12. 2 [ 0.5, 2] by [ 0.5, 1]
1 The area is (1 2 x x) dx 4 3/2
x
3 x 0 12
x
2 1
.
6 by [ 3, 3] The area is 1 (2 sin x 0 sin 2x) dx 2 cos x 0 1
cos 2x
2 4.
0 13. x
.
2 2y 2 implies y 9. x 3
2 , [ 5, 5] by [ 5, 5] The curves intersect at x [ 1, 19] by [ 1, 4] 2.1281 (4 x 2.1281 The curves intersect at x
18 x
dx
2 3
0
3 or 0 y2 10. 4x 1
y
4 x 18
4 x 3/2 3x
23
y
3 2y 2 dy 18. The area is
32 2 2.1281. The area is cos x) dx, which using NINT evaluates to 18, 8.9023. 14. 0 3 18.
0 12
y
4 4 implies x 1, and 4x y 16 implies
[ 4, 4] by [ 4, 4] 4. The curves intersect at x
0.8256 (3 0.8256. The area is sec2 x) dx, x 0.8256 which using NINT evaluates to 2.1043. [ 6, 6] by [ 6, 6] 15. Solve 1
The curves intersect at (3, cos x 2 cos x for the xvalues at the two 4) and (5.25, 5). The area is
ends of the region: x 5 1
y
4 4
5 12
y
4 4
12
y
4 4 12
y
8 425
24 38
3 3 , i.e., symmetry of the area: dy 7 /3 2 1
y
4 13
y
12 1 2 [(1
2 5 dy cos x) 2 cos x)] dx (2 cos x 1) dx 2 5 5y 2 2 sin x x 4 243
8 (2 7 /3 2 30.375. 2
3 3 7 /3
2 1.370. 5 /3 11. 16. [(2
/3 cos x) (1 5 /3 (1 2 cos x) dx /3 x
[ 0.1, 1] by [ 0.1, 1]
/4 The area is (x
0 sin x) dx 2
/4 12
x
2 cos x 2 2 32 0 2 0.0155. 1 2 sin x
3 4
3 5 /3
/3 7.653 cos x)] dx 5
7
or . Use the
3
3 Chapter 7 Review
17. Solve x 3
x x x x2 to find the intersection points at 1 (c) Use cylindrical shells. A shell has radius 4
height 2 x 21/4. Then use the area’s symmetry: 0 and x 2 (4
0 2 0 (x3 1 1
ln (x 2
2 2 x)(2 x x) dx (8 4x 2x 3/2 x 2) dx 2x 2 x 4 5/2
x
5 13
x
3 0 x
x2 x. The total volume is 4 1/4 2 ln ( 2 x) dx
14
x
4 1)
1) 2 1 16 3/2
x
3 2 12
x
2 x and 4 the area is
2 21/4 4 64
.
5 0 (d) Use cylindrical shells. A shell has radius 4 0 319 1.2956. y and y2
. The total volume is
4
4
y2
2 (4 y) y
dy
4
0
4
y3
2
4y 2y 2
dy
4
0
4
23
32
14
2 2y 2
y
y
.
3
3
16
0 height y
18. Use the intersect function on a graphing calculator to
determine that the curves intersect at x 1.8933. The area is
1.8933 31 x2 3
dx,
10 x2 1.8933 which using NINT evaluates to 5.7312. 22. (a) Use disks. The volume is
2 19. Use the x and yaxis symmetries of the area:
4 x sin x dx 4 sin x x cos x 0 4. 0 k (b) 0 0 A(x) r
1 V 1 k, 9 x. 9 x dx x 9 When k 1 y2 2y dy 2 4.
0 k2 0 8 8 k 2 dV (c) Since V
2 0 y2 2y dy 3x 4 and area 20. A cross section has radius r 2 2y)2 dy ( 1, 2. dt dk
dt dk
dt
1 dV
2 k dt 2k .
1
(2)
2 1
1 so the depth is increasing at the rate of 1 21. , unit per second. 23. The football is a solid of revolution about the xaxis. A
[ 5, 5] by [ 5, 5] cross section has radius The graphs intersect at (0, 0) and (4, 4).
r2
(a) Use cylindrical shells. A shell has radius y and height
2 y
. The total volume is
4
4
4
y2
y3
2 ( y) y
dy 2
y2
dy
4
4
0
0
4
13
14
2
y
y
3
16
0
32
.
3 y 12 1 11/2 2 12
0 1 4x 2
and area
121 12 1 4x 2
. The volume is, given the symmetry,
121
11/2
4x 2
4x 2
dx 24
1
dx
121
121
0
11/2
21
2
24 x
x3
3
11
0
11
11
24
2
6 88 276 in3. 24. The width of a cross section is 2 sin x, and the area is
(b) Use cylindrical shells. A shell has radius x and height
2x x. The total volume is 4 4 2 (x)(2 x x) dx 2 0 (2x 3/2 2 x ) dx 0 2 4 5/2
x
5 128
.
15 13
x
3 4
0 12
1
r
sin2 x. The volume is
2
2
1
1
1
sin2 x dx
x
sin 2x
22
4
02 2 0 4 . 320 Chapter 7 Review
29. 25. [ 1, 2] by [ 1, 2] [ 1, 3] by [ 3, 3] Use washer cross sections. A washer has inner radius r
e x/2, and area (R2 outer radius R r 2) (e x 1, 1). 3x 2 y is at x 6x equals zero when x 0, the minimum at x = 2. The distance between
2 them along the curve is The volume is
ln 3 (e x ex 1) dx using NINT evaluates to x
0 (3 ln 3 (2 4.5920
2 ln 3). 1) 26. Use cylindrical shells. Taking the hole to be vertical, a shell
has radius x and height 2 22 3 2x
0 4 2
(4
3 2 x 2 dx x4 27. The curve crosses the xaxis at x
3 ( 2x)2 dx 1
3 1 2 x 4 dx 2
5 x 2 dx 2 2x, so the 13
x
3 5 39.
2 32. (a) (100 N)(40 m) = 4000 J
(b) When the end has traveled a distance y, the weight of
the remaining portion is (40 4x 2 dx, which 3 using NINT evaluates to 5 (F (x))2 dx 1 0 3. y
3 1, so 3 x2)3/2 4
(1 8)
3
28
29.3215 ft3.
3 length is 30. If (b) were true, then the curve y k sin x would have to
get vanishingly short as k approached zero. Since in fact the
curve’s length approaches 2 instead, (b) is false and (a) is
true. 5 3 2 0 4.5920. The time taken is about 2.296 sec. 31. F (x)
x 2) dx y)(0.8) 40 28. 0.8y) dy 0.4y 2 32y 0 The curves intersect at x 0 and x 1. Use the graphs’ 1 4 1
0 (3x 2 640 is (800)(8)
4750 x and yaxis symmetry:
3x 40 640 J. 4640 J 33. The weight of the water at elevation x (starting from x [ 2, 2] by [ 2, 2] x) 0.8y. 0 (c) 4000 2 32 The total work to lift the rope is 19.4942. (32 d3
(x
dx 6x)2 dx, which x 2. The volume of the piece cut out is
2 (x)(2 22 (3x 2 1
0 ln 3 0 0 or 2. The maximum is
0 4750 x/2
4750 128
4750
95 128
4750
95 1
x dx
2 1, and the total perimeter is 0) 1
x . The total work
2 128
4750x
95 12
x
4 4750
0 22,800,000 ftlb. 1)2 dx, which using NINT evaluates to
34. F ks, so k 5.2454. 0.3 Work
0 F
800
80
N/m. Then
s
3
0.3
0.3
800
800 2
x dx
x
12 J.
3
6
0 To stretch the additional meter,
1.3 Work
0.3 800
x dx
3 800 2
x
6 1.3 213.3 J.
0.3 35. The work is positive going uphill, since the force pushes in
the direction of travel. The work is negative going downhill. ...
View Full
Document
 Spring '08
 ALL

Click to edit the document details