# Math Help - calculus - newton's law of cooling?

1. ## calculus - newton's law of cooling?

Hi, i have a calculus hw problem which is the following -
Use Newton's Law of Cooling. The coroner arrives at teh scene of a murder at 2AM. he takes the temperature of the body and finds it to be 61.1 degrees . He waits one hour, takes the temperature again, and finds it to be 57.2 degrees . The body is in a meat freezer, where the temperature is 10 degrees . When was the murder comitted?

I was absent when the teacher went over this and i have no clue what to do. can someone please give me a step by step workthrough of the problem, or if thats too hard to type out, tell me what to do, and what the answer should be so i can check it? thanks for any help!

btw, there is no unit given for degrees, but i dont think that should matter.......

2. Using Newton's Cooling Law, we have

$\frac{dT}{dt}=k(T-10)$

Separate variables:

$\frac{dT}{T-10}=kdt$

Integrate:

$ln(T-10)=kt+c$

$T-10=Ce^{kt}$

Now, use the two given pieces of info to find k and C.

Once you have those you have your formula.

Then, set it equal to, say, 98.6 and solve for t.

Subtract that value from 2AM to find the time the murder was committed.

3. wow u guys are incredible, thanks, i think i got it

so i got an equation that ended up like this:

T=10 + 51.1e^ln0.9237*t

does this look like something that i should come up with?

if this looks good, then now i solve for t in this eqation?

once again thanks so much guys

4. i have another question, if were to solve that equation that i came up with, would e^ln0.9237 * t = 0.9237t because e^lnx = x right?

5. I think you have it.

I get $T=10+\frac{511}{10}e^{ln(\frac{472}{511})t}$

Yes, $e^{ln(\frac{472}{511})t}=(\frac{472}{511})^{t}$

6. wow great

when you write this:

Yes,
is that (472/511) multiplied by t, or (472/511) raised to the t? raised to t right?