A potato is put into an oven that has been heated to 350°F. Its temperature as a function of time is given by T(t)=a(1-e^kt)+b. The potato was 50F when it was first put into the oven.

Question

1.If the potato is 60°F after 2 minutes, what is the value of k?
2.When will the potato reach 150°F?
I got 290 for k and the second one i got 3 min.

Answer 1

T(0) = a (1 - e^0) + b = 50
T(0) = a(1-1) + b = 50
so
b = 50

T(oo) = a(1 - e^k*oo) +50 = 350
so k better be negative
350 = a + 50
a = 300
so
T = 300 (1-e^kt) + 50

T(2) = 300 (1-e^2k) + 50 = 60
1- e^2k = 1/30
e^2k = .96667
2 k = ln .96667
k = -.01695

150 = 300 (1 - e^-.01695 t ) + 50
1/3 = 1 - e^-.01695 t
e^-.01695 t = 2/3 = 0.6667
-.01695 t = -.4055
t = 23.9 minutes