Calculus Help - Newton's Law Of Warming?

Question

My teacher threw out this question on a quiz... needless to say I bombed it -_-
Can someone shed a light on it?
*********
A yam is put in a hot oven, maintained at a constant 200°C. Suppose that at the moment t = 30 minutes,the temperature T of the yam is 120°C, and it is increasing at 2°C/min. Newton's Law of Warming tells us that the yam's temperature at time t is given by the formula T(t) = 200-ae^(-bt), where a and b are constants. Find a and b.

Answer 1

You need two equations to solve for the 2 unknowns a and b.

First, T(30) = 120, so 80 = ae^(-30b).

Also, T'(30) = 2. T'(t) = abe^(-bt), so 2 = abe^(-30b).

Substitute ae^(-30b) = 80 into the second equation, and you get 2=80b, so b=1/40.

Then substitute b=1/40 into the first equation to get

80 = ae^(-30/40), so a=169.36.

Answer 2

With two variables, you need two equations. First, you know that
T(30)=200-ae^(-30b)=120. And, you know the rate of increase of the temperature: dT/dt=ab*e^(-bt). At T=30, ab*e^-30b)=2.
From the first equation, ae^(-30b)=80. Divide the equation for dT/dt by that equation-- you get b=1/40. Plugging b=1/40 into the first expression, you get a*e^(-3/4)=80 So, a=80*e^(3/4)

Answer 3

T(t) = 200 - ae^(-bt),
dT/dt = abe^(-bt)
120 = 200 - ae^(-30b)
2 = abe^(-30b)
2/b = ae^(-30b)
120 = 200 - 2/b
80 = 2/b
b = 1/40
120 - 200 = - ae^(-30/40)
80 = ae^(-0.75)
a = 80e^(0.75)
a = 169.36

T(t) = 200-169.39e^(-t/40)

T(30) = 200 - 169.36e^(-.75)
T(30) = 120

Answer 4

T(t) = 200-ae^(-bt)

when t = 30

120 = 200 - ae^(-30b) .... (1)

Also dT/dt = abe^(-bt)
= b(ae^(-bt))
= b(200 - T(t))
So 2 = b(200 - 120)
b = 2/80 = 0.025

Since from (1) ... 120 = 200 - ae^(-30b)
120 = 200 - ae^(-0.75)
So ae^(-0.75) = 80
a = 80 e ^(0.75)
≈169.36

So T(t) = 200 - 169.36e^(-0.025t)

Check When t = 30
T(30) = 200 - 169.36e^(-0.025 * 30)
= 200 - 80.00
= 120

dT/dt = 169.36*-0..025e^(-0.025t)
= 0.025*-169.36e^(-0.025t)
= 0.025 (200 - 120)
= +2