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Math Help - Need some help with a hard question

  1. #1
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    Need some help with a hard question

    Ok the question is

    Coroners estimate the time of death from body temperature using the simple rule that a body cools about 1C in the first hour after death and about 1/2C for each addictional hour.

    Assuming an air temperature of 20C and a living body temperature of 37C, the temperature, T(t), in degrees Celcius is given by T(t) = 20 + 17e^-kt, where t=0 is the instant that death occurred..

    a) For what value of k will the body cool by 1C in the first hour?
    b) Using the value of k found in part a) after how many hours will the temperature of the body be decreaing at a rate of 1/2 C per hour?
    c) Using the value of k found in part a) Show that 24h after death, the coroner's rule gives approx the same temp as the forumula.

    I know I need an equation to do this but Im not sure what it is, I understood my other log study questions but this one stumps me. Any help would be appreciated greatly.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Skor View Post
    Ok the question is

    Coroners estimate the time of death from body temperature using the simple rule that a body cools about 1C in the first hour after death and about 1/2C for each addictional hour.

    Assuming an air temperature of 20C and a living body temperature of 37C, the temperature, T(t), in degrees Celcius is given by T(t) = 20 + 17e^-kt, where t=0 is the instant that death occurred..

    a) For what value of k will the body cool by 1C in the first hour?
    b) Using the value of k found in part a) after how many hours will the temperature of the body be decreaing at a rate of 1/2 C per hour?
    c) Using the value of k found in part a) Show that 24h after death, the coroner's rule gives approx the same temp as the forumula.

    I know I need an equation to do this but Im not sure what it is, I understood my other log study questions but this one stumps me. Any help would be appreciated greatly.
    a) You have:

    T(t) = 20 + 17e^{-kt}

    Now if we measure time in hours we have if 1 is the time to cool by 1 degree C

    <br />
36=20+17 e^{-k}<br />

    or:

    <br />
e^{-k}=16/17<br />

    or:

    <br />
k=-\ln(16/17)\approx 0.060625<br />

    b) The rate of fall of temprature is:

    \frac{dT}{dt}=-k \times 17 e^{-kt}

    If the temprature is falling at a rate of 1/2 a degree per hour we have:

    -k \times 17 e^{-kt}=-1/2

    so:

    e^{-kt}=0.5/(k \times 17)

    or:

    -kt=\ln(0.5/(k \times 17))

    so:

    t=-(1/k)\ln(0.5/(k \times 17))\approx 11.28 \mbox{ hours}

    RonL
    Last edited by CaptainBlack; January 23rd 2007 at 05:16 AM.
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