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Homework Help: Modeling with exponential and logarithmic functions help?

  1. Apr 30, 2011 #1
    Modeling with exponential and logarithmic functions help???

    1. The problem statement, all variables and given/known data
    Use Newton's Lay of Cooling, T = C + (T0 - C)e-kt, to solve this exercise. At 9:00 A.M., a coroner arrived at the home of a person who had died during the night. The temperature of the room was 70 degrees F, and at the time of death the person had a body temperature of 98.6 degrees F. The coroner took the body's temperature at 9:30 A.M., at which time it was 85.6 degrees F, and again at 10:00 A.M., when it was 82.7 degrees F. At what time did the person die?

    2. Relevant equations
    T = C + (T0 - C)e-kt
    If you do not know what the variable's mean...these are their meanings:
    T = temperature of a heated object
    C = constant temperature of the surrounding medium (the ambient temp)
    T0 = initial temperature of the heated object
    k = negative constant associated with the cooling object
    t = time (in minutes)

    3. The attempt at a solution
    I tried solving for k by doing:
    Steps (I plugged all the values in to their corresponding places):

    85.6 = 70 + (98.6 - 70)e-k(30)
    15.6 = 28.6e-30k
    0.5454545455 = e-30k
    ln(0.5454545455) = ln(e-30k)
    ln(0.5454545455) = -30k
    k = (ln(0.5454545455)/(-30))
    k = 0.0202045268

    After getting this....i do not know what to do next...of even if I did the process of anything correctly as yet.
  2. jcsd
  3. Apr 30, 2011 #2


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    Re: Modeling with exponential and logarithmic functions help???

    Excellent! Writing them all out like this is a very good idea- even on your home work paper. Not every one uses the same symbols. One criticism- you say "t= time (in minutes)" but do not say "time from what?" From what you write later, t appears to be the time (in minutes) after 9:00. You should say that.

    NO. You don't know what the temperature of the body was at 9:00- you are told that " at the time of death the person had a body temperature of 98.6 degrees F" and you don't know when the time of death was. If you had said "t= time (in minutes) after death until the Coroner arrived, at 9:00", you would have
    85.6= 70+ (98.6- 70)e-k(t+ 30)
    82.7= 70+ (98.6- 70)e-k(t+ 60)

    That gives you two equations to solve for k and t. Subtracting t minutes from 9:00 gives you the time of death.

  4. Apr 30, 2011 #3
    Re: Modeling with exponential and logarithmic functions help???

    Which values of k ant t do I use to find the time of death???
    Do I use both; do I add both t values together and then subtract from 9:00 ??
  5. Apr 30, 2011 #4


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    Homework Helper

    Re: Modeling with exponential and logarithmic functions help???

    camp3r101: while what HallsofIvy said is correct, I wouldn't do it this way, because you defined T0 as the initial temperature of the object... that is to say, the temperature at t = 0. I would define t as simply the number of minutes after death. So the equations I would write are
    [tex]85.6 = 70 + (98.6 - 70)e^{-kt}[/tex]
    for 09:30am, and
    [tex]82.7 = 70 + (98.6 - 70)e^{-k(t + 30)}[/tex]
    for 10:00am.

    No. (The hints I'm going to give may be too much, so apologies to the mods in advance.)

    Take my 1st equation and simplify it a bit:
    [tex]15.6 = 28.6e^{-kt}[/tex]

    Take my 2nd equation and simplify:
    [tex]12.7 = 28.6e^{-k(t + 30)}[/tex]
    [tex]12.7 = 28.6e^{-kt - 30k}[/tex]

    Use property of exponents:
    [tex]12.7 = \frac{28.6e^{-kt}}{e^{30k}}[/tex]

    Notice the numerator? That equals 15.6, from the first equation, so we substitute:
    [tex]12.7 = \frac{15.6}{e^{30k}}[/tex]

    Now I'll shut up. Solve for k, and plug this in back into
    [tex]15.6 = 28.6e^{-kt}[/tex]
    and solve for t. Remember that this t value corresponds to 09:30am, so subtract to find the time of death.
    Last edited: Apr 30, 2011
  6. May 6, 2011 #5
    Re: Modeling with exponential and logarithmic functions help???

    DANNGGG! You guys are soo much help!! Much thanks to you!
    I see what steps I didn't take. Thanks again! :)
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