# Newtons cooling law DE, mathematical approach

1. Jan 29, 2006

### Pengwuino

I have this one problem dealing with Newton's law of cooling:

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

I'm basically trying to determine at what time someone died! The info I have is that at time of death, the temperature was 98.6 degrees, at 10 AM it was at 92 degrees, and at 2PM it was at 86 degrees. The surrounding temperature, A, was 78 degrees and constant. Unfortunately, the book does not give an example as to how this DE works.

I have a feeling I need to do this:

$$\frac{{dT}}{{(T - 78)}} = - kdt$$

And integrate it… but I'm not sure how I would do that… especially with it just being k*dt. Any suggestions?

2. Jan 29, 2006

### HallsofIvy

What? you have a problem with kdt? What is the integral of a constant?

If dT/(T-78), try a simple substitution: let u= T- 78 so du= dT. Can you integrate du/u??

3. Jan 29, 2006

### benorin

$$\int\frac{{dT}}{{(T - 78)}} = - k\int dt\Rightarrow\ln (T - 78) = - kt+C$$

Last edited: Jan 29, 2006
4. Jan 29, 2006

### arildno

Well, your approach looks perfectly fine to me.
Alternatively, you could introduce the the dependent variable u(t)=T(t)-A,
and recognizing that u obeys the diff. eq $\frac{du}{dt}=-ku$
whereby you see that we have:
$u(t)=Ce^{-kt}\to{T}(t)=A+Ce^{-kt}$
If you let t=0 correspond to 10 AM, then you can determine C and k from the info known at that time, along with the info known at 2P.M.
Finally, use the info about the body temp. at time of death to determine when he died.

5. Jan 29, 2006

### Pengwuino

I got to the second line benorin stated and now im stuck again... I feel I need to find 'k' but I don't have an initial condition to work with, only a time change...

6. Jan 29, 2006

### Pengwuino

Oh so are you trying to say I could use 10:00AM as a t=0 and simply use the negative value (would I get a negative value?) to determine how long before 10:00am he died?

7. Jan 29, 2006

### arildno

Yup!

And yes, you'll get a negative value.

8. Jan 29, 2006

### Pengwuino

Ahhhh there we go! -2.75 hours later... so 7:14am!