Question

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?

Answer 1

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

Answer 2

$$T_0$$ 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.

Answer 3

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

Answer 4

Yes. That allows you to compute R

Answer 5

what is with the information about the creek being 5 degrees

Answer 6

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

Answer 7

This is based on your model though.

Answer 8

@freckles I would really appreciate your input :)

Answer 9

im a bit lost with finding R @RBauer4

Answer 10

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

Answer 11

morbid use of cows lol

Answer 12

actually im very lost

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

and took its temperature

Answer 17

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 \[H(1) = 22 e^{-R} = 19\]

Answer 18

We can solve for R explicitly for \[e^{-R} = \frac{19}{22} \implies e^{R} = \frac{22}{19} \implies \ln(e^{R}) = R = \ln(22/19)\]

Answer 19

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.

Answer 20

@marigirl are you ok with the equation @RBauer4 has left you with?

Answer 21

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

Answer 22

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

Answer 23

answer in the book states t=-3.51

Answer 24

right, they are using newton's law of cooling
do you want to know how to setup the differential equation and solve it

Answer 25

yes please i will also upload the model answer shown

Answer 26

ANSWER

Answer 27

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:
\[y' = k(5-y)\tag{1}\]

Answer 28

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

Answer 29

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

Answer 30

familiar with separation of variables to solve differential equation ?

Answer 31

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

Answer 32

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