marigirl
  • marigirl
Integration application question: Please help me, Where can I start? I know it will be an exponential equation eventually. A farmer goes out at 7.30 am to check his stock and finds one of his cows dead in the creek. The temperature of the cow is 22 degrees Celsius and the temperature of the creek is 5 degrees Celsius. One hour later the temperature of the cow is 19 degrees Celsius. The normal body temperature of a healthy cow is 38.6 degrees Celsius. When did the cow die?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Well, perhaps you can start with your model, the exponential decay function\[H(t) = T_0e^{-Rt}\]
anonymous
  • anonymous
$$T_0$$ will probably be the cows temperature when the farmer found it dead and $$R$$ will be the rate at which heat dissipates from the cows body.
marigirl
  • marigirl
t=0 cow's body temperature is 22 t=1 cow's body temperature is 19

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anonymous
  • anonymous
Yes. That allows you to compute R
marigirl
  • marigirl
what is with the information about the creek being 5 degrees
anonymous
  • anonymous
This is a good question, however, it seems to be erroneous information since the rate is fixed given the conditions at t = 0 and t = 1
anonymous
  • anonymous
This is based on your model though.
marigirl
  • marigirl
@freckles I would really appreciate your input :)
marigirl
  • marigirl
im a bit lost with finding R @RBauer4
anonymous
  • anonymous
Put t = 1, then we get\[22 e^{-R} = 19\]
ganeshie8
  • ganeshie8
morbid use of cows lol
marigirl
  • marigirl
actually im very lost
marigirl
  • marigirl
oh please help @ganeshie8 .. its bad enough that i have to read about deceased cows :(
marigirl
  • marigirl
there is even a picture of a dead cow in the question!!!!! :( :( :( :( :( this is emotionally disturbing me
anonymous
  • anonymous
The thing is, is that the farmer left it in the creek for another hour after he found it!
marigirl
  • marigirl
and took its temperature
anonymous
  • anonymous
Here is what I would do, take\[H(t) = 22 e^{-Rt}\]. Then the problem states that at t = 1, the cows temperature was 19, so equivalently \[H(1) = 22 e^{-R} = 19\]
anonymous
  • anonymous
We can solve for R explicitly for \[e^{-R} = \frac{19}{22} \implies e^{R} = \frac{22}{19} \implies \ln(e^{R}) = R = \ln(22/19)\]
anonymous
  • anonymous
Ok, now that we have R, all we need to do is put H(t) = 38.6 and find the value of t this corresponds to.
freckles
  • freckles
@marigirl are you ok with the equation @RBauer4 has left you with?
marigirl
  • marigirl
it makes sense but i am still considering why the paddock temperature is stated. ill take a photo of the answer and send it to you guys
ganeshie8
  • ganeshie8
im getting 4:30 am using newton's law of cooling
marigirl
  • marigirl
answer in the book states t=-3.51
ganeshie8
  • ganeshie8
right, they are using newton's law of cooling do you want to know how to setup the differential equation and solve it
marigirl
  • marigirl
yes please i will also upload the model answer shown
marigirl
  • marigirl
ANSWER
ganeshie8
  • ganeshie8
Let \(y(t)\) represent the temperature of cow \(t\) hours after \(7:30\) am, (time after it was observed for the first time) then the temperature in cow follows the newton's law of cooling: \[y' = k(5-y)\tag{1}\]
ganeshie8
  • ganeshie8
does that look familiar to you if not, you may replace \(y\) by \(T\)
marigirl
  • marigirl
thanks, i am not seeing it now, but i will think about it.
ganeshie8
  • ganeshie8
familiar with separation of variables to solve differential equation ?
anonymous
  • anonymous
Not that I have never solved such an ODE before, but I was never taught that this particular ODE models cooling. Thanks for the clarification @ganeshie8
ganeshie8
  • ganeshie8
question should explicitly specify the model to use i guess because there are several models...

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