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?

- marigirl

- jamiebookeater

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- anonymous

Well, perhaps you can start with your model, the exponential decay function\[H(t) = T_0e^{-Rt}\]

- 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

t=0 cow's body temperature is 22
t=1 cow's body temperature is 19

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## More answers

- anonymous

Yes. That allows you to compute R

- marigirl

what is with the information about the creek being 5 degrees

- 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

This is based on your model though.

- marigirl

@freckles I would really appreciate your input :)

- marigirl

im a bit lost with finding R @RBauer4

- anonymous

Put t = 1, then we get\[22 e^{-R} = 19\]

- ganeshie8

morbid use of cows lol

- marigirl

actually im very lost

- marigirl

oh please help @ganeshie8 .. its bad enough that i have to read about deceased cows :(

- marigirl

there is even a picture of a dead cow in the question!!!!! :( :( :( :( :( this is emotionally disturbing me

- anonymous

The thing is, is that the farmer left it in the creek for another hour after he found it!

- marigirl

and took its temperature

- 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

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

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.

- 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

im getting 4:30 am using newton's law of cooling

- marigirl

answer in the book states t=-3.51

- 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

yes please i will also upload the model answer shown

- marigirl

ANSWER

##### 1 Attachment

- 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

does that look familiar to you
if not, you may replace \(y\) by \(T\)

- marigirl

thanks, i am not seeing it now, but i will think about it.

- ganeshie8

familiar with separation of variables to solve differential equation ?

- 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

question should explicitly specify the model to use i guess because there are several models...

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