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United States24709 Posts
Edit: I was using the integrating factor method incorrectly (forgetting to integrate in the last step) so I fixed that. It's too late to type all the steps out but I get:
t1 = t2 * ln((T1-Te)/(T0-Te)) / ln((T2-Te)/(T1-Te))
A big thanks to Gene for catching my errors and discussing it with me (for hours), and also to the others on irc who helped.
********** End edit
I remember doing a problem like this way back in calc2 and finding it pretty easy but I tried to go through it right now and I must be getting rusty already ._.
You find a dead body outside on a cool evening (temperature of environment significantly lower than a human body's normal temperature). You notice that the body is still warm and want to determine how many hours ago the body was at 98.6 degrees F to approximate the time of death. How do you do it?
I believe you can find it by taking the temperature of the body at that moment, then waiting a while and taking it again to see how much the body temperature has gone down. At that point it becomes a calculation that should be fairly trivial to the mathematically inclined, I'd believe. Here's my work I put together right now without looking anything up.
dT/dt = -k(T - Te)
where T is temperature as a function of time
and k is a constant determined by the body/environment that is constant for this problem
and Te is the environmental temperature towards which the body's temperature approaches. This is Newton's Law of Cooling IIRC.
I solved it as follows:
multiply out the RHS and move all terms on the same side as the differential:
dT/dt + kT = k*Te
Solve using the integrating factor method:
u(t) = e^(kt)+C
T*e^(kt) = k * Te * e^(kt)+C
T = k*Te + C*e^(-kt)
This should be my general solution for which I will apply an initial condition. Specifically, at time 0 the temperature is measured to be 98.6 F, denoted T0. I apply the statement T(0)=T0 and get
T0 = k*Te+C
C = T0-k*Te
T = k*Te + (T0-k*Te)e^(-kt)
T = k*Te + T0*e^(-kt) - k*Te*e^(-kt)
I measured the temperature when the body was found to be T1 so I can apply T(t1)=T1 where t1 is the duration that has passed since the person died.
T1 = k*Te + T0*e^(-k*t1) - k*Te*e^(-k*t1)
I now have an equation where everything is known except for k, and t1. With one more equation I should be able to solve this. I can apply that after an additional wait of some time period, the temperature has become T2. Let's call the time period t2, thus with respect to the time of death, the temperature was measured at time t1+t2 with t2 known (and was suggested to be 1 hour earlier in the post). I can apply T(t1+t2)=T2:
T2 = k*Te + T0*e^(-k(t1+t2)) - k*Te*e^(-k(t1+t2))
In this equation again only k and t1 are unknown (with t1 being the final answer I am looking for). I think if I solve each equation for e^(-k*t1) and set the results equal to each other, I should be able to solve the resulting equation for k, but I don't see how D:
edit: someone in a math channel in IRC suggested http://en.wikipedia.org/wiki/Lambertw when I showed him the work I was doing to try to solve the resulting equation for k, but I don't recognize it.
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is awesome32277 Posts
dT/dt = -k(T - Te)
dt = [-k(T - Te)/dT] integer both sides
t between 0 and t (t time you found him death? 0 time he died). T between temperature that you found him, and To lower human body temperature limit.
0 - t = -k [ Ln T - Ln Te]
0 - t = -k Ln[ T/Te]
And there you go. I only skimmed your post, i don't know if this was what you wanted, but remember that bodies not only lose heat by direct heat transfer, but by radiation too. And to consider radiations on the human body you have to do a lot of approximations for it to be easier.
http://en.wikipedia.org/wiki/Heat_transfer
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is awesome32277 Posts
I may be setting the integer limits wrong because I remember conduction and convection having difference in signs but can't remember which one is each. :p
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United States24709 Posts
As per our IRC discussion, k is unknown D:
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you have to make an assumption of the temperatures of the person. or are we doing the 98.6 thing? and the temperature would decline differently according to the outside weather and material the body is touching... or are we doing some sort of deep internal temperature measuring once again?
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calculus is some crazy and amazing stuff. i don't think they teach it well enough so people can use it's applications - just the rules do this do that... what ur doing here is great. i found an interesting application having to do with the rate of healing of a wound based on the surface area of the wound... i'd love to be able to be "proficient" at calculus.
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United States24709 Posts
On May 21 2008 10:51 gwho wrote:
you have to make an assumption of the temperatures of the person. or are we doing the 98.6 thing? and the temperature would decline differently according to the outside weather and material the body is touching... or are we doing some sort of deep internal temperature measuring once again?
The formula should, to a first order approximation, compensate for the environment so long as it doesn't change drastically. We are assuming the temperature started at 98.6 and decayed towards the environment's temperature.
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On May 21 2008 09:25 micronesia wrote:
You find a dead body outside on a cool evening (temperature of environment significantly lower than a human body's normal temperature). You notice that the body is still warm and want to
I stopped reading there.
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You can do that long formula thing or you could just ask a pathologist and they have some easy formula thing for determining time since death using temp :D
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United States24709 Posts
On May 21 2008 12:05 Slaughter)BiO wrote:
You can do that long formula thing or you could just ask a pathologist and they have some easy formula thing for determining time since death using temp :D
It's probably less accurate.
BTW I made a mistake in the OP, my sign was wrong on the q(t) for the integrating factor, but it has only a tiny influence.
Edit: I fixed the error but I'm still having trouble getting the answer :/
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United States24709 Posts
OMG FOUND THE MOTHERQUAKING ERROR UPDATING OP.
Updated.
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