Thomas Calculus 13th [Solutions]

37. 0 kt A A e and
Section 7.2 Exponential Change and Separable Differential Equations 519
kt k(24360) ln (0.5)
0.000028454t
A0 10 A 10 e , 5 10e k 0.000028454 A 10 e
,
24360
0.000028454t
then 0.2(10) 10e t ln 0.2 56563 years
0.000028454
38. A A0 e kt
139 139 ln(0.5)
and 1 k 1 k
A
2 0 A0
e e k 0.00499; then
2 139
0.00499t
0.05A ln 0.05
0 A0 e t 600 days
0.00499
kt ( k)(3 k) 3 y0 y0
39. y y0e y0e y0e (0.05)( y
3 20
0 ) after three mean lifetimes less than 5% remains
e
40. (a)
kt 1 2.645k
A A ln 2
0e e k
2 2.645
0.262
(b) 1 3.816 years
k
(c) (0.05) A Aexp ln 2 t ln 20 ln 2 t t 2.645ln 20 11.431 years
2.645 2.645 ln 2
41.
42.
43.
kt 10k 4 10k
T Ts T0 Ts e , T0 90 C, Ts
20 C, T 60 C 60 20 70e e
7
ln
7
4
k 0.05596
10
0.05596t
(a) 35 20 70e t 27.5 min is the total time it will take 27.5 10 17.5 minutes longer to
reach 35 C
kt
0.05596t
(b) T Ts T0 Ts e , T0
90 C, Ts
15 C 35 15 105e t 13.26 min
kt
10k
20k
T 65 T0 65 e 35 65 T0
65 e and 50 65 T0
65 e . Solving
10k
20k
10k
20k
30 T0
65 e and 15 T0
65 e simultaneously T0 65 e 2 T0
65 e
10k
e 2 k ln 2 and
10
65 60 5
ln 2
0 10
10k
T 65 10 ln 2
30 30 e T0 65 T0
65 30 e
e
kt
10k
20k
39 Ts
10k
T Ts To Ts e 39 Ts 46 Ts
e and 33 Ts 46 Ts e e and
46 Ts
33 T 2 33 39
2
s 20k
10k
Ts Ts
2
e e 33 T 46 39
46 Ts 46 Ts 46 T
s Ts Ts
s
2 2
1518 79Ts Ts 1521 78Ts Ts Ts 3 Ts
3 C
44. Let x represent how far above room temperature the silver will be 15 min from now, y how far above room
temperature the silver will be 120 min from now, and t 0 the time the silver will be 10°C above room
temperature. We then have the following time-temperature table:
time in min. 0 20 (Now) 35 140 t 0
temperature T s 70 Ts
60 Ts
x Ts
y T s 10
kt 20k 20k
T T 1 6
s T0 Ts e 60 Ts Ts 70 Ts Ts
e 60 70e k ln 0.00771
20 7
0.00771 t
(0.00771)(35) 0.26985
(a) T Ts T0 Ts e Ts x Ts 70 Ts Ts
e x 70e 53.44 C
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