Section 4.3– Number 38
The temperature, T, in oC, of a yam put into a 200oC oven is given as a function
of time, t, in minutes, by
T = a(1 – e−kt) + b
a)
If the yam starts at 20oC, find a and b.
Solution:
Since the yam starts at 20oC, we have that T = 20 when t = 0. Substituting into
the formula, we get
20 = a(1 – e−k(0)) + b
20 = a(1 – e0) + b
20 = a(1 – 1) + b
20 = a(0) + b
20 = b
Since e−kt =
1

lim a  1 − kt
e

x →∞
1
e kt
and since the oven is 200oC, we have

 + 20 = a(1 – 0) + 20 = a + 20 = 200

So a = 180.
Thus the formula is T = 180(1 – e−kt) + 20
b)
If the temperature of the yam is initially increasing at 2oC per minute, find
k.
Solution:
What this means is that T '(0) = 2. So we need to find T '(t).
T(t) = 180 – 180e−kt + 20 = 200 – 180e−kt
T '(t) = –180e−kt(−k) = 180ke−kt
T '(0) = 2 = 180ke−k(0) = 180k(1) = 180k
So k =
1
90
Thus the formula is T = 180(1 – e
−
1
t
90
) + 20