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Thread: Differential equations and a dead body.

  1. #1
    Super Member Showcase_22's Avatar
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    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:

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

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

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

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

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

    However, my book says k should be $\displaystyle ln \ 2$. What's going wrong?
    Last edited by mr fantastic; Mar 16th 2009 at 12:19 PM.
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  2. #2
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    The problem is that you don't have
    $\displaystyle \frac{d\theta}{dt_1}=-k(28-24)=-4k$
    $\displaystyle \frac{d\theta}{dt_2}=-k(26-24)=-2k$
    (these are notationally and conceptually incorrect), but
    $\displaystyle \left.\frac{d\theta}{dt}\right|_{t=t_1}=-k(28-24)=-4k$
    $\displaystyle \left.\frac{d\theta}{dt}\right|_{t=t_2}=-k(26-24)=-2k$.
    And you cannot turn these into values for $\displaystyle \theta$ like you did.
    Instead, you need to solve the general differential equation
    $\displaystyle \frac{d\theta}{dt}=-k(\theta-\theta_0)$ to get $\displaystyle \theta(t)$ in terms of k and $\displaystyle \theta_0$, and then plug in your two pairs $\displaystyle (t,\theta)$ to obtain two equations that can be solved for the unknowns k and $\displaystyle \theta_0$.

    --Kevin C.
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  3. #3
    Super Member Showcase_22's Avatar
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    $\displaystyle \frac{d \theta}{dt}=-k(\theta-\theta_0)$

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

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

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

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

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

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

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

    Ah, this looks promising!

    So we have:

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

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

    Therefore:

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

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

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

    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?
    Last edited by Showcase_22; Mar 17th 2009 at 04:27 AM.
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  4. #4
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    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.
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  5. #5
    Super Member Showcase_22's Avatar
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    hmm, okay then. Let me see if i'm doing this correctly:

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

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

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

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

    $\displaystyle 13=2^9e^t$

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

    --Kevin C.
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  7. #7
    Super Member Showcase_22's Avatar
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    Thanks Kevin!

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