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#1
09-13-2007, 03:10 PM
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stuck on a diff eq problem
Ok so I've thought about this one for a while but I seem to be stuck. It seems like there isn't enough information, but I'm probably just dumb.
"Sherlock Holmes and Dr. Watson discover the body of a recently deceased homicide victim. At 10:00 PM sharp, Watson records the victim's body temperature to be 31 degrees Celsius and makes a similar recording of 29 degrees Celsius exactly one hour later... Determine to the nearest second the exact time of the murder." It says to assume the guy's body temperature was 37 Celsius and that the room is always maintained at 15 Celsius. So you set it up like dT/dt=k(T-15) where T is temp of body at time t through separation of variables and integrating I get T=Ae^(kt) + 15. then plugging in T(0)=37 I get A to be 22 so T=22e^(kt) + 15. I think i need to find k but I don't see how the given temperature recordings help me since I don't know when 10 PM was in relation to when the guy died. Any help would be appreciated. 
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#2
09-13-2007, 03:26 PM
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Re: stuck on a diff eq problem
Can you not just work it out given that you've got two different times and two different temperatures? It's like seven years since I last did ode's so I forget how to do this kind of thing, but you definitely have enough information to shove in.
No help I guess
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#3
09-13-2007, 04:39 PM
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Re: stuck on a diff eq problem
10pm is t=0, increment t by the hour.
at t=0, T=31 at t=1, T=29 You already used the first equation to find the constant A. You should be able to use the second to solve for k, no?
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#5
09-13-2007, 06:36 PM
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Re: stuck on a diff eq problem
Let 10pm be time X. Then 11pm is time X+1...
31 = 22e^(kx) + 15 ... 16/22 = e^(kx) ... ln (8/11) = kx k = ln (8/11) / x Now, plug in x+1... 29 = 22e^(kx+k) + 15 ... 14/22 = e^(kx+k) ln (7/11) = k (x+1) ln (7/11) = ln (8/11) * (1 + 1/x) I don't feel like doing the rest, but solve for x obv.
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#6
09-13-2007, 07:41 PM
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Re: stuck on a diff eq problem
Thank you that was easy enough. I actually ended up doing it a different way. Someone else suggested doing T=31 as t(0) and t(1)=29. I got -2.38486 instead of 2.38486 from your method, but the negative doesnt matter i guess. so i guess the answer is 7:37:05 PM.
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