Time of death problem using differential equation

The problem states: Give an interval that estimates for the person's time of death. Ts= 21.1 degrees C room temperature, Normal temperature of the body Ta= is an interval of 36.6 to 37.2 degrees C. Temperature of the body when found at midnight is Ti= 34.8 degrees C. The final temperature half hour later of the body Tf= 34.3 degrees C.

So using Newton's cooling equation:

K=(T1-T2)=-ln((T1'-s)/(T2'-s))

where the temperature of the corpse is:

K=-(1/2)ln((Tf - Ts)/(Ti - Ts)

Time of death is:

D=-(1/k)ln((Ta - Ts)/(Ti - Ts))

So i'm confused on where to intergate? I wanted to intergate for the time of death for To but in respects to time but there is not a time variable in that equation so I'm stumped. Thanks to all who can help.

Re: Time of death problem using differential equation

Quote:

Originally Posted by

**nivek0078** The problem states: Give an interval that estimates for the person's time of death. Ts= 21.1 degrees C room temperature, Normal temperature of the body Ta= is an interval of 36.6 to 37.2 degrees C. Temperature of the body when found at midnight is Ti= 34.8 degrees C. The final temperature half hour later of the body Tf= 34.3 degrees C.

So using Newton's cooling equation:

K=(T1-T2)=-ln((T1'-s)/(T2'-s))

where the temperature of the corpse is:

K=-(1/2)ln((Tf - Ts)/(Ti - Ts)

Time of death is:

D=-(1/k)ln((Ta - Ts)/(Ti - Ts))

So i'm confused on where to intergate? I wanted to intergate for the time of death for To but in respects to time but there is not a time variable in that equation so I'm stumped. Thanks to all who can help.

So the first order ode is given by the model

$\displaystyle \frac{dT}{dt} \propto (T_s-T)$ where $\displaystyle T_s$ is the room temperature.

This gives that

$\displaystyle \frac{dT}{dt}=k(21.1-T)$

If we seperate this equation and integrate we getwe get that

$\displaystyle \frac{dT}{T-21.1}=-kdt \implies \ln|T-21.1|=-kt+C \iff T(t)=Ae^{-kt}+21.1$

We are told the temperature at midnight $\displaystyle t_m$ and 30 minutes later $\displaystyle t_m+30$

Using these two data point we get

$\displaystyle 34.8=Ae^{-kt_m}+21.1 \iff Ae^{-kt_m}=13.7$

and

$\displaystyle 34.3=Ae^{-kt_m-30k}+21.1 \iff e^{-30k}=\frac{13.2}{Ae^{-kt_m}}$

puting the first equation into the 2nd gives

$\displaystyle e^{-30k}=\frac{132}{137} \iff k=-\frac{1}{30}\ln \left( \frac{132}{137}\right)$

Puting this back into the orginial equation gives

$\displaystyle T(t)=Ae^{\frac{t}{30}\ln \left( \frac{132}{137}\right)}+21.1 =A \left( \frac{132}{137}\right)^{\frac{t}{30}}+21.1$

Solving this for time gives

$\displaystyle t=\frac{30\ln\left( \frac{T-21.1}{A}\right)}{\ln \left( \frac{132}{137}\right)}$

Since we know the temperature of the body at midnight we can plug that into this equation to get

$\displaystyle t=\frac{30\ln\left( \frac{13.7}{A}\right)}{\ln \left( \frac{132}{137}\right)}$

Now if you plug in the two initial temperatures $\displaystyle A$ it will give you the number of minutes before midnight. Just use the two different values of $\displaystyle A$ given.

Re: Time of death problem using differential equation

Thank you TheEmptySet for your help! You made understanding this problem a lot easier! I have one question though. What do you mean by the initial temperatures A? Is it the intial temperature (Ti) the body was found and final temperature (Tf) or is it the average body temperature (Ta) and the final (Tf)? Sorry I'm sure I'm way over thinking this! If anyone else has the answer please respond.

Re: Time of death problem using differential equation

Quote:

Originally Posted by

**nivek0078** Thank you TheEmptySet for your help! You made understanding this problem a lot easier! I have one question though. What do you mean by the initial temperatures A? Is it the intial temperature (Ti) the body was found and final temperature (Tf) or is it the average body temperature (Ta) and the final (Tf)? Sorry I'm sure I'm way over thinking this! If anyone else has the answer please respond.

Notice that

$\displaystyle T(0)=A+21.1$

This should be the temperature at the begining. You are given a range of values for $\displaystyle T(0)$.

$\displaystyle 36.6 \le T(0) \le 37.2 \implies 36.6 \le A+21.1 \le 37.2 \iff 15.5 \le A \le 16.1$

If you plug these values into the equation above it will give you the set of $\displaystyle t$ values.

Re: Time of death problem using differential equation

Ok that makes sense. I confused myself there for a minute. Thank you again for your help!