Hi! Any help with this ODE word problem will be appreciated!
I'll simplify it to the important data:
We are keeping track of the temperature of coffee. Well, we have the temperature at two times and want a third. I would like to be able to conceptualize what's going on, if anyone can help!
7:30am \(200^\circ\text F\)
7:45am \(120^\circ\text F\)
8:00am \(?^\circ\text F\)
The rate of change of heat is a simplified version of Newton's Law of Cooling, that \(\dfrac{dT}{dt}=k(T-T_r)\) where \(k\) is a proportionality constant and \(T_r\) is the room temperature (\(70^\circ\text F\)).

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- ganeshie8

*

- theEric

hM?

- theEric

Hm?*

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## More answers

- theEric

I figure that I probably need to find out what \(k\) is...

- ganeshie8

- theEric

Can I use \(\Large{\int\limits_{7.5}^{7.75}}\normalsize\dfrac{dT}{dt}dt=\Large{\int\limits_{7.5}^{7.75}}\normalsize k(T-T_r)\ dt=200-120\)?
I'll think on that...

- theEric

I haven't done anything like it in class, that I recall.

- theEric

But I don't have \(T(t)\) to do integration...

- anonymous

You can say nothing about 8:00 AM here. Because We don't know how the temperature function is changing over time.

- theEric

That's what I was thinking as well... But we know the rate-of-change function, two points on the \(T\) vs \(t\) curve, which is one solution to the rate-of-change function...

- theEric

I know that initial value problems require one point...

- phi

I think I would solve the equation then use the given data to find the constants
\[ \frac{dT}{dt}= k(T -T_r) \\ \frac{dT}{(T -T_r)}= k \ dt \\\ln(T -T_r)= k \ t + C
\]
where t is time in minutes from 7:30
make each side the exponent of e
\[ T- T_r= A e^{k\ t} \]
now use 7:30 t=0 T= 200 to find A
and 7:45 t=15 to find k

- theEric

Thank you very much! I'm working on digesting it now.

- theEric

\(A\) is an arbitrary constant like \(C\), right?

- phi

yes A = e^C
we don't care about C per se.

- theEric

This is somewhat related, @phi , but do we always need to look at things like \(t=0\) for initial value problems? Or can we use non-zero values for the independent variable?

- theEric

Thank you!

- phi

you can use non zero values of t. But obviously t=0 makes it especially easy to find A

- theEric

Right, thank you very much! :)

- theEric

In case it is of any viewer's concern:
\(A=130^\circ\text F\)
\(k=-0.0637\ [s^{-1}]\)
and the coffee will be about \(89.2^\circ\text F\) at 8:00am.
:)
The problem was about a professor with an 8am class, who never drinks coffee unless it is at \(90^\circ\text F\) or cooler. Good news for him - he can drink the coffee in class!
I might show off and solve for the time \(t\) at which he can take his first sip. :P But not until I get the real problems done.
Thank you!

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