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marigirl

  • one year ago

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?

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  1. anonymous
    • one year ago
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    Well, perhaps you can start with your model, the exponential decay function\[H(t) = T_0e^{-Rt}\]

  2. anonymous
    • one year ago
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    $$T_0$$ 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.

  3. marigirl
    • one year ago
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    t=0 cow's body temperature is 22 t=1 cow's body temperature is 19

  4. anonymous
    • one year ago
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    Yes. That allows you to compute R

  5. marigirl
    • one year ago
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    what is with the information about the creek being 5 degrees

  6. anonymous
    • one year ago
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    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

  7. anonymous
    • one year ago
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    This is based on your model though.

  8. marigirl
    • one year ago
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    @freckles I would really appreciate your input :)

  9. marigirl
    • one year ago
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    im a bit lost with finding R @RBauer4

  10. anonymous
    • one year ago
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    Put t = 1, then we get\[22 e^{-R} = 19\]

  11. ganeshie8
    • one year ago
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    morbid use of cows lol

  12. marigirl
    • one year ago
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    actually im very lost

  13. marigirl
    • one year ago
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    oh please help @ganeshie8 .. its bad enough that i have to read about deceased cows :(

  14. marigirl
    • one year ago
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    there is even a picture of a dead cow in the question!!!!! :( :( :( :( :( this is emotionally disturbing me

  15. anonymous
    • one year ago
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    The thing is, is that the farmer left it in the creek for another hour after he found it!

  16. marigirl
    • one year ago
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    and took its temperature

  17. anonymous
    • one year ago
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    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 \[H(1) = 22 e^{-R} = 19\]

  18. anonymous
    • one year ago
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    We can solve for R explicitly for \[e^{-R} = \frac{19}{22} \implies e^{R} = \frac{22}{19} \implies \ln(e^{R}) = R = \ln(22/19)\]

  19. anonymous
    • one year ago
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    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.

  20. freckles
    • one year ago
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    @marigirl are you ok with the equation @RBauer4 has left you with?

  21. marigirl
    • one year ago
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    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

  22. ganeshie8
    • one year ago
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    im getting 4:30 am using newton's law of cooling

  23. marigirl
    • one year ago
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    answer in the book states t=-3.51

  24. ganeshie8
    • one year ago
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    right, they are using newton's law of cooling do you want to know how to setup the differential equation and solve it

  25. marigirl
    • one year ago
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    yes please i will also upload the model answer shown

  26. marigirl
    • one year ago
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    ANSWER

  27. ganeshie8
    • one year ago
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    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: \[y' = k(5-y)\tag{1}\]

  28. ganeshie8
    • one year ago
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    does that look familiar to you if not, you may replace \(y\) by \(T\)

  29. marigirl
    • one year ago
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    thanks, i am not seeing it now, but i will think about it.

  30. ganeshie8
    • one year ago
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    familiar with separation of variables to solve differential equation ?

  31. anonymous
    • one year ago
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    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

  32. ganeshie8
    • one year ago
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    question should explicitly specify the model to use i guess because there are several models...

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