Thomas Calculus 13th [Solutions]

674 Chapter 9 First-Order Differential Equations
9.3 APPLICATIONS
1. Note that the total mass is 66 7 73 kg, therefore, v v0e ( k / m) t
v
3.9 t/73
9e
3.9 t/73 3.9 /73
(a) ( ) 9
2190 t
s t e dt e
13
C
Since s (0) 0 we have C 2190 and
2190 3.9 t/73
lim s( t) lim 1 e
2190
13
t
t
13 13
168.5
The cyclist will coast about 168.5 meters.
3.9/73
(b) 1 9e 3.9t
ln 9 t 73ln 9
73 3.9
41.13 sec
It will take about 41.13 seconds.
2.
( k / m) t (59,000/51,000,000) t 59 t/51,000
v v0e v 9e v 9e
59 t/51,000 459,000 59 t/51,000
(a) s( t) 9e dt e C
59
459,000
459,000 59 t/51,000
459,000
Since s (0) 0 we have C and lim s( t) lim 1 e 7780 m
59
t
t
59 59
The ship sill coast about 7780 m, or 7.78 km.
59 t/51,000 51,000ln 9
(b) 1 9e 59t
ln 9 t 1899.3 sec
51,000 59
It will take about 31.65 minutes.
v0 m (2.75)(39.92)
3. The total distance traveled 4.91 k 22.36. Therefore, the distance traveled is given
k
k
(22.36/39.92) t
by the function s( t) 4.91 1 e . The graph shows s( t ) and the data points.
v0m
(0.80)(49.90)
4. coasting distance
1.32 k 998
k
k
33
v0m
We know that 1.32 and k 998 20
k
m 33(49.9) 33 .
v0 m ( k / m) t 20 t/33 0.606t
Using Equation 2, we have: s( t) 1 e 1.32 1 e 1.32 1 e
k
5. y mx y xy y y
0 .
x
m y 2
x
x
So for
2
dy
2
orthogonals: x
y
y dy x dx x C
dx y
2 2
2 2
x y C 1
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