Thomas Calculus 13th [Solutions]

340 Chapter 4 Applications of Derivatives
25. ds
kt
ks ds k dt ln s kt C s s
dt s
0e
the 14th century model of free fall was
exponential; note that the motion starts too slowly at
first and then becomes too fast after about 7 seconds.
26. Two views of the graph of y 1000 1 (.99) x 1 are shown below.
x
At about x = 11 there is a minimum. There is no maximum; however, the curve is asymptotic to y = 1000. The
curve is near 1000 when x 643.
27. (a) a( t) s ( t) k ( k 0) s ( t) kt C 1,
where s (0) 88 C 1 88 s ( t) kt 88. So
2
2
2
s( t) kt 88t C
2
2 where s(0) 0 C 2 0 so s( t) kt 88 t . Now s( t ) 100 when kt 88t
100.
2
2
2
88 88
Solving for t we obtain
200 2
k
88 88 200k
t . At such t we want s ( t ) 0, thus k 88 0 or
k
k
2
88 88 200k
2
k 88 0. In either case we obtain 88 200k 0 so that k
k
88 2 2
38.72 ft/sec .
200
(b) The initial condition that s (0) 44 ft/sec implies that s ( t) kt 44 and s( t) kt 44t where k is as
2
above. The car is stopped at a time t such that s ( t) kt 44 0 t 44 . At this time the car has
k
2
2
traveled a distance s 44 k 44 44 44 44 968 968 200 25 feet. Thus halving the initial
k 2 k k 2k
k
2
88
velocity quarters stopping distance.
2
28.
2 2
h( x) f ( x) g ( x) h ( x) 2 f ( x) f ( x) 2 g( x) g ( x) 2[ f ( x) f ( x) g( x) g ( x)]
2[ f ( x) g( x) g( x)( f ( x ))] 2 0 0. Thus h( x) c,
a constant. Since h(0) 5, h( x ) 5 for all x in the
domain of h. Thus h(10) 5.
29. Yes. The curve y
dy
x satisfies all three conditions since
dx
1 everywhere, when x 0, y 0, and
everywhere.
2
0
2
d y
dx
30.
2
y 3x 2 for all
3
x y x 2x C where
3 3
1 1 2 1 C C 4 y x 2x
4.
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