Calculus Yam Help.?

Question

A yam is put in a hot oven, maintained at a constant 200Celsius. Suppose that the yam's initial temperature is 20C, and it is initially increasing at 2C/min. After t minutes in the oven, the yam's temperature is T(t) = a[1-e^(-kt)]+b, where a, b, and k are constants.
Find a, b, and k.

Answer 1

Using the initial and final temperatures given, we can write

T(t) = 200[1 - e^(-kt)] + 20

We thus have a and b.

Now, we further know that at t=0, the temperature is increasing at the rate of 2 degrees C/minute. This will allow us to find k. Jason glibly says that k is the initial rate, and he may be right.

To check, let's take the first derivative of the function and set it equal to 2 degrees C/minute at t=0.

Expanding the function gives

T(t) = a - ae^(-kt) + b
T'(t) = ake^(-kt)

Putting 2 for T'(t) and t=0, we get

2 = 200k
k = .01

Which value for k is more reasonable, Jason's or mine? After one second in the oven, the temperature of the yam ought to be around 22 degrees C.

Jason's T(1) = 200[1 - e^(-2)] + 20 = 193 degrees C
My T(1) = 200[1 - e^(-.01)] + 20 = 21.99 degrees C

Jason's value for k is way too high, 200 times larger than mine. k is not the initial rate; it is the initial rate divided by the final value. Jason's value for k implies an initial rate of increase of 400 degrees C/second. The only reason Jason's yam does not explode is that the rate of temperature increase starts to fall off immediately as the yam gets hotter. Within the first second, Jason's yam has gotten so close to the final temperature of 200 degrees that it does not continue to get hot so fast.

Answer 2

You have 3 conditions that the equation governs:
>> the initial temperature - set t=0
>> the initial rate - take the derivative, then set t=0
>> the final temperature set t=infinity

Answer 3

a is the temperature of oven
k is the rate of change
b is initial temperature

i think
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oops sorry, i forgot to tell you the values

a = 200C
k = 2C/min
b = 20C