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Math Help - Exponential functions - assistance needed :)

  1. #1
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    Exponential functions - assistance needed :)

    Forensic scientists use the following model to determine the time of death of accident or murder victims. If T (in degrees centigrade) denotes the temperature of a body t hours after death, then

    T(t) = T0 + (T1 -T0)(0.97)^t

    where T0 is the air temperature and T1 is the body temperature at the time of death. A man was found dead at midnight in his house. Assume that the room temperature remained constant at 21 deg C, and that his body temperature at the time of death was 37 deg C.

    a) Show that the model above may be written as T(t) = 21+16(0.97)^t

    T(t) = 21+(37-21)(0.97)^t

    b) What would the body temperature be 5 hours after death according to this model?

    T(5) = 21+16(0.97)^5 =34.74 (2dp)

    The body temperature would be 34.74 deg C 5 hours after death.

    c) Calculate how many hours have passed if the body temperature was 26 deg C when the man’s body was found (give your answer to 1 d.p.)?Use your answer to estimate the time of death.

    T(t) = 21+16(0.97)^t = 26

    Am I on the right track here? Can you help me find t, and how to use this to esimate the time of death?


    d) Predict what will happen to the body temperature in the long run.
    (Hint: Let t take on large values, say t = 100, 200, 1000 hours etc)
    T(100) = 21+16(0.97)^100 = 21.7608
    T(200) = 21+16(0.97)^200 = 21.03617
    T(1000) = 21+16(0.97)^1000 = 21
    T(2000) = 21+16(0.97)^2000 = 21

    In the long run the body temp will drop to a constant value.


    Thanks to mr fantastic and badgerigar, for your helpful answers
    Last edited by newatthis; January 25th 2009 at 02:25 AM.
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  2. #2
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    Quote Originally Posted by newatthis View Post
    Forensic scientists use the following model to determine the time of death of accident or murder victims. If T (in degrees centigrade) denotes the temperature of a body t hours after death, then

    T(t) = T0 + (T1 -T0)(0.97)^t

    where T0 is the air temperature and T1 is the body temperature at the time of death. A man was found dead at midnight in his house. Assume that the room temperature remained constant at 21 deg C, and that his body temperature at the time of death was 37 deg C.

    a) Show that the model above may be written as T(t) = 21+16(0.97)^t

    T(t) = 21+(37-21)(0.97)^t

    b) What would the body temperature be 5 hours after death according to this model?

    T(5) = 21+16(0.97)^5 =34.74 (2dp)

    The body temperature would be 34.74 deg C 5 hours after death.

    c) Calculate how many hours have passed if the body temperature was 26 deg C when the manís body was found (give your answer to 1 d.p.)?Use your answer to estimate the time of death.

    T(t) = 21+16(0.97)^t = 26

    Am I on the right track here? Can you help me find t, and how to use this to esimate the time of death?

    d) Predict what will happen to the body temperature in the long run.
    (Hint: Let t take on large values, say t = 100, 200, 1000 hours etc)
    T(100) = 21+16(0.97)^100 = 21.7608
    T(200) = 21+16(0.97)^200 = 21.03617
    T(1000) = 21+16(0.97)^1000 = 21
    T(2000) = 21+16(0.97)^2000 = 21

    In the long run the body temp will drop to a constant value.


    Thanks for looking!
    c) 21+16 (0.97)^t= 26 \Rightarrow 16 (0.97)^t= 5 \Rightarrow (0.97)^t = \frac{5}{16}.

    Take \log_{10} of both sides:

    \log_{10} (0.97)^t = \log_{10} \frac{5}{16}

    \Rightarrow t \log_{10} (0.97) = \log_{10} \frac{5}{16}

    \Rightarrow t = \frac{\log_{10} \frac{5}{16}}{\log_{10} (0.97)} \approx \, ....

    where you would use a calculator here.
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  3. #3
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    Edit: sorry Mr. Fantastic, didn't see you there.


    In the long run the body temp will drop to a constant value.
    Yes it will. You should also note that this constant value is the same as the air temperature.

    Edit: past this point I'm just repeating what Mr. Fantastic said

    T(t) = T0 + (T1 -T0)(0.97)^t

    where T0 is the air temperature and T1 is the body temperature at the time of death. A man was found dead at midnight in his house. Assume that the room temperature remained constant at 21 deg C, and that his body temperature at the time of death was 37 deg C.

    a) Show that the model above may be written as T(t) = 21+16(0.97)^t

    T(t) = 21+(37-21)(0.97)^t

    b) What would the body temperature be 5 hours after death according to this model?

    T(5) = 21+16(0.97)^5 =34.74 (2dp)

    The body temperature would be 34.74 deg C 5 hours after death.
    excellent

    c) Calculate how many hours have passed if the body temperature was 26 deg C when the man’s body was found (give your answer to 1 d.p.)?Use your answer to estimate the time of death.

    T(t) = 21+16(0.97)^t = 26
    If you haven't studied logarithms yet you are probably expected to just pull the answer out of your calculator. At my school, asking for an answer to a set number of decimal places meant we were just meant to enter the equation into a computer.

    If you do want to solve it exactly, just proceed in the normal method of solving an equation, starting as far from the t as possible and working in:

    Subtract 21 from both sides of the equation
    16(0.97)^t = 5
    divide by 16
    <br />
0.97^t = 5/16
    then use the definition of log
    t = \log_{0.97}(5/16)
    Last edited by badgerigar; January 24th 2009 at 10:37 PM. Reason: Mr. fantastic got there first.
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