James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

240 Chapter 3 Differentiation Rules
We solve this equation for t by taking the natural logarithm of both sides:
150
_(ln 2)t/1590
m=100e
m=30
0 4000
FIGURE 2
2 ln 2 t − ln 0.3
1590
Thus t − 21590
ln 0.3
ln 2
< 2762 years ■
As a check on our work in Example 2, we use a graphing device to draw the graph of
mstd in Figure 2 together with the horizontal line m − 30. These curves intersect when
t < 2800, and this agrees with the answer to part (c).
Newton’s Law of Cooling
Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the
temperature difference between the object and its surroundings, provided that this difference
is not too large. (This law also applies to warming.) If we let Tstd be the temperature
of the object at time t and T s be the temperature of the surroundings, then we can
formulate Newton’s Law of Cooling as a differential equation:
dT
dt
− ksT 2 T s d
where k is a constant. This equation is not quite the same as Equation 1, so we make the
change of variable ystd − Tstd 2 T s . Because T s is constant, we have y9std − T9std and
so the equation becomes
dy
dt − ky
We can then use (2) to find an expression for y, from which we can find T.
ExamplE 3 A bottle of soda pop at room temperature (72°F) is placed in a refrigerator
where the temperature is 44°F. After half an hour the soda pop has cooled to 61°F.
(a) What is the temperature of the soda pop after another half hour?
(b) How long does it take for the soda pop to cool to 50°F?
SOLUTION
(a) Let Tstd be the temperature of the soda after t minutes. The surrounding temperature
is T s − 448 F, so Newton’s Law of Cooling states that
dT
dt
− ksT 2 44d
If we let y − T 2 44, then ys0d − Ts0d 2 44 − 72 2 44 − 28, so y satisfies
and by (2) we have
dy
dt
− ky ys0d − 28
ystd − ys0de kt − 28e kt
We are given that Ts30d − 61, so ys30d − 61 2 44 − 17 and
28e 30k − 17 e 30k − 17
28
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