# Math Help - Differential equations and a dead body.

1. ## Differential equations and a dead body.

Let's assume we find a dead body in a room with a constant temperature of 24 degrees.

At 8a.m, the body has a temperature of 28 degrees. At 9a.m it has a temperature of 26 degrees. Assuming normal body temperature is 37 degrees, when was the person killed?
This is what I have so far:

$\frac{d \theta}{dt}=-k(\theta-\theta_0)$
I need to find k.

$\frac{d \theta}{dt_1}=-k(28-24)=-4k$
$\frac{d \theta}{dt_2}=-k(26-24)=-2k$

Both these equations give:
$\theta=-4kt_1+C$
$\theta=-2kt_2+C$

Hence:
$28=-4k(8)+C \Rightarrow \ 28=-32k+C$
$26=-2k(9)+C \Rightarrow \ 26=-18k+C$

$2=-14k \Rightarrow k=-\frac{1}{7}$

However, my book says k should be $ln \ 2$. What's going wrong?

2. The problem is that you don't have
$\frac{d\theta}{dt_1}=-k(28-24)=-4k$
$\frac{d\theta}{dt_2}=-k(26-24)=-2k$
(these are notationally and conceptually incorrect), but
$\left.\frac{d\theta}{dt}\right|_{t=t_1}=-k(28-24)=-4k$
$\left.\frac{d\theta}{dt}\right|_{t=t_2}=-k(26-24)=-2k$.
And you cannot turn these into values for $\theta$ like you did.
Instead, you need to solve the general differential equation
$\frac{d\theta}{dt}=-k(\theta-\theta_0)$ to get $\theta(t)$ in terms of k and $\theta_0$, and then plug in your two pairs $(t,\theta)$ to obtain two equations that can be solved for the unknowns k and $\theta_0$.

--Kevin C.

3. $\frac{d \theta}{dt}=-k(\theta-\theta_0)$

$\frac{d \theta}{dt}=-k \theta+k\theta_0$

$\frac{d \theta}{dt}+k \theta=k \theta_0$

$R=e^{\int k dt}=e^{kt}$

$e^{kt} \frac{d \theta}{dt}+ke^{kt} \theta=k \theta_0 e^{kt}$

$\frac{d}{dt} (e^{kt} \theta)=k \theta_0e^{kt}$

$e^{kt} \theta= \theta_0 e^{kt}+C$

$\theta=\theta_0+Ce^{-kt}$

Ah, this looks promising!

So we have:

$28=24+Ce^{-8k}$ and $26=24+Ce^{-9k}$.

$28=24+Ce^{-8k}\Rightarrow C=4e^{8k}$

Therefore:

$2=4e^{8k}e^{-9k} \Rightarrow \ 2=4e^{-k}$

$\frac{1}{2}=e^{-k}$

$-ln \ 2=-k \Rightarrow \ k=ln \ 2$

Thanks!

One more thing, when I substitute my value for k back into the equation and work everything out I get t=-2.28 hours. Which time do I add this to?

Do I do 8:00-2.28 or 9-2.28?

4. Neither. You did t=8 for 8:00 AM and t=9 for 9:00 AM, thus t=0 is midnight, and t=-2.28 is 2.28 hours before midnight, or about 9:43 PM the previous night.

--Kevin C.

5. hmm, okay then. Let me see if i'm doing this correctly:

$C=4e^{8ln\ 2}=2^{10}$

$\theta=24+2^{10}e^{-ln 2 \ t}$

$\theta=24+2^9e^t$

$37=24+2^9e^t$

$13=2^9e^t$

$t=ln\frac{37}{2^9}=-2.627$ which is approximately -2 hours 38 minutes (sorry, I mistyped it in the previous post:s)

Hence the time should be 9:22pm the previous day.

Unfortunately my book has done 9:00-2 hours 38 minutes giving 6:22 am.

Does this mean my book is wrong?

6. Well, first I can see that when you went from $\theta=24+2^{10}e^{-(\ln{2})t}$ to the next line, you made an error:
$e^{-(\ln{2})t}$ is not $\frac{1}{2}e^{t}$. Instead, you have
$e^{-(\ln{2})t}=\left(e^{\ln{2}}\right)^{-t}=2^{-t}$,
so you should go from
$\theta=24+2^{10}e^{-(\ln{2})t}$ to $\theta=24+2^{10-t}$ (which, rechecking, still gives $theta=28$ at t=8 and $theta=26$ at t=9).
Letting $37=24+2^{10-t}$
we get
$13=2^{10-t}$
$10-t=\log_2(13)$
$t=10-\log_2(13)$

--Kevin C.

7. Thanks Kevin!

I actually managed to get the right answer now, HUZZAH!!!