# Thread: Differential Equation Word Problem

1. ## Differential Equation Word Problem

Hello.

My calculus class just began a new unit on differential equations and my teacher assigned a problem for extra credit. It's a word problem involving Newton's Law of Cooling. We only went through one quick example problem in class so I'm unsure of how to proceed on this. Any help is greatly appreciated.

A cake comes out of a 350-degree oven and is set aside to cool in a 70-degree kitchen. If 10 minutes after removal from the oven, the temperature of the cake is 280 degrees, how long after the cake was removed from the oven will it be safe to frost the cake, given the frosting process requires that the cake be no hotter than 100 degrees?

2. Excellent. Please state Newton's Law of Cooling and then label some parts according to your given information.

3. - "Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the temperature of its surroundings."

Going by the example I did in class:

$\text {Let T = temperature of the cake, and t = time.}$

$\frac{ dT}{dt } = k(T - 70)$

$\int \frac{dT}{T-70} = \int kdt$

$ln(T-70) = kt + C$

$T-70 = Ce^{kt}$

$T = Ce^{kt} + 70$

Am I on the right track?...

4. Originally Posted by Status
- "Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the temperature of its surroundings."

Going by the example I did in class:

$\text {Let T = temperature of the cake, and t = time.}$

$\frac{ dT}{dt } = k(T - 70)$

$\int \frac{dT}{T-70} = \int kdt$

$ln(T-70) = kt + C$

$T-70 = Ce^{kt}$ Mr F says: No! You have already used C in the previous line See below.

$T = Ce^{kt} + 70$

Am I on the right track?...
Almost. You should say $T-70 = Ae^{kt}$ where $A = e^{C}$

I'm closing this thread now because an Extra credit question is meant to be all your own work (rule #6).