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Extreme Newton's Law of Cooling D.E.

  1. Apr 27, 2012 #1
    1. The problem statement, all variables and given/known data
    after a very unpleasant valentine's Day, a dead body was found in a downtown warehouse that had no heating or air conditioning. it was February in Florida and we know that the daily temperature in the warehouse fluctuates according to the function T(t)= 63-12sin(∏t/12), where t=0 corresponds to midnight on any given day. The body was discovered at 1:30 am on Feb. 15 and its temperature was 73 degrees F. Two hours later the temperature was 68 degrees F. What was the time of death of this body?


    2. Relevant equations

    dT/dt = k (T - T(m))

    [exp(at)(-BcosBt + asinBt)]/a^2+b^2


    where k is the proportionality constant T is temperature and T(m) is the medium of the environment surrounding the object.

    3. The attempt at a solution

    i set up my differential as dT/dt - kT - 63k = 12ksin(∏t/12)

    after doing the integrating factor method and using the equation to solve the integration i got some huge formula for T

    then i rescaled the time so i could solve for C and k ... for C i got

    C= 73+ (144∏k)/(144k^2+∏^2)

    after that i tried to solve for k but it honestly looks impossible and i'm not sure how to do it
    if someone could help me out or at least check my work so far i would greatly appreciate it....this problem seems close to impossible
     
  2. jcsd
  3. Apr 27, 2012 #2
    Not a 100% sure as I've never solved one with a separate function for ambient air temperature but maybe you need to sub T(t)=63-12sin(∏t/12) into newtons formula rather than equal to it.

    I think the formula should be dT/dt = k [T - {63-12sin(∏t/12)}] where T is the initial temperature given by T(0) = 63.

    So dT/dt = k(12sin(∏t/12))

    and solve that ?
     
    Last edited: Apr 27, 2012
  4. Apr 27, 2012 #3

    HallsofIvy

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    You should not separate the "63k" and "[itex]12ksin(\pi t/12)[/itex]":
    [tex]dT/dt- kT= 63k+ 12ksin(\pi t/12)[/tex]

    The associated homogeneous equation, dT/dt-kT= 0, has solution [itex]T= Ae^{kt}[/itex]. Now, look for a solution of the form [itex]B+ Csin(\pi t/12)+ Dcos(\pi t/12)[/itex]. Put that into the equation and solve for B, C, and D. Then put the entire solution into y(1.5)= 73 and y(3.5)= 68 to solve for A and k.
     
  5. Apr 27, 2012 #4
    ok i see, i'm going to redo it that way, no wonder i was getting nowhere with the way i did it, thanks!
     
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