Thomas Calculus 13th [Solutions]

218 Chapter 3 Derivatives
k k k
k dx dx
k 1
k du d v u d v .
dx dx
k dx
k 1
Therefore the formula (c) holds for n ( k 1) whenever it holds for n k.
k k 1 k 1 k
1
k 1 2
du d v d v d u
1 ( 1) d u dv d u d v
k u v k
1 1 2
...
k k k k k 1 2
dx dx dk dx dx dx
k k 1
27. (a)
(b)
2 2 2
2
2
4
T g (1sec )(32.2 ft/sec )
T L L L L
g
2 2
4 4
2
0.8156ft
2
T 4 L T 2 L; dT 2 1 dL dL; dT
(0.01 ft) 0.00613 sec.
g g g L Lg
2 2
(0.8156ft)(32.2ft/sec )
(c) Since there are 86,400 sec in a day, we have we have (0.00613 sec)(86, 400 sec/day)
8.83 min/day; the clock will lose about 8.83 min/day.
529.6 sec/day, or
28.
3 2 2
v s dv 3 s ds k (6 s ) ds 2 k . If s
dt dt dt
0 the initial length of the cubes side, then s 1 s 0 2k
2k s 0 s 1 . Let t the time it will take the ice cube to melt. Now, 1/3
s0 s0 ( v0
)
t
2k s 1/3
1/3
0 s1
3
( v0 ) v
4 0
1 11 hr.
1/3
1
3
4
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