Thomas Calculus 13th [Solutions]

Section 3.11 Linearization and Differentials 195
43.
44.
45.
47.
1
2
f ( x) x , x0
0.5, dx 0.1 f ( x)
x
(a) f f ( x0 dx) f ( x0
) f (.6) f (.5) 1
3
(b) df f ( x0
) dx ( 4) 1 2
10 5
(c) | f df | | 1 2 | 1
3 5 15
3
2
f ( x) x 2x 3, x0
2, dx 0.1 f ( x) 3x
2
(a) f f ( x0 dx) f ( x0
) f (2.1) f (2) 1.061
(b) df f ( x0
) dx (10)(0.10) 1
(c) | f df | |1.061 1| .061
4 3 2
V r dV 4 r
3
0 dr 46.
2
S 6x dS 12x0dx
3 2
V x dV 3x 0 dx
48.
2 2
S r r h
dS
2r
2 2
0 h
2 2
0 h
r
2 2 1/2
r( r h ) , h constant dS
dr
dr, h constant
2 2 1/2 2 2 1/2
( r h ) r r ( r h ) dS
dr
2 2 2
r h r
r
2 2
h
49.
2 ,
V r h height constant dV 2 r0
h dr 50. S 2 rh dS 2 r dh
51. Given r 2m, dr .02m
2
(a) A r dA 2 r dr
(100%) 2%
(b)
.08
4
2
2 (2)(.02) .08 m
52. C 2 r and dC 2 in. dC 2 dr dr
1
the diameter grew about
2 in.; A r 2
dA
2 r dr 2 (5) 10 in.
1
2
2 .
53. The volume of a cylinder is V r h When h is held fixed, we have dV 2 rh
dr , and so dV 2 rh dr .
For h 30 in., r 6 in., and dr 0.5 in., the volume of the material in the shell is approximately
3
dV 2 rh dr 2 (6)(30)(0.5) 180 565.5 in .
54. Let angle of elevation and h height of building. Then h 30 tan , so dh 30 sec d . We want
| dh| 0.04 h , which gives: |30sec | 0.04 |30 tan |
5 5
12 12
2
d
1
2
cos
0.04sin
cos
| d | | d | 0.04sin cos
| d | 0.04sin cos 0.01 radian. The angle should be measured with an error of less than
0.01 radian (or approximately 0.57 degrees), which is a percentage error of approximately 0.76%.
dr
55. The percentage error in the radius is
dt
r
100 2%.
dC
dt
dr
dt
(a) Since C 2 r 2 . The percentage error in calculating the circles circumference is dt
100
2
2
dr
dt
r
100
dr
dt
r
100 2%.
2
dC
C
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