theEric
  • theEric
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 (\(70^\circ\text F\)).
Differential Equations
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SOLVED
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schrodinger
  • schrodinger
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ganeshie8
  • ganeshie8
*
theEric
  • theEric
hM?
theEric
  • theEric
Hm?*

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theEric
  • theEric
I figure that I probably need to find out what \(k\) is...
ganeshie8
  • ganeshie8
@sauravshakya
theEric
  • 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
  • theEric
I haven't done anything like it in class, that I recall.
theEric
  • theEric
But I don't have \(T(t)\) to do integration...
anonymous
  • anonymous
You can say nothing about 8:00 AM here. Because We don't know how the temperature function is changing over time.
theEric
  • 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
  • theEric
I know that initial value problems require one point...
phi
  • 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
  • theEric
Thank you very much! I'm working on digesting it now.
theEric
  • theEric
\(A\) is an arbitrary constant like \(C\), right?
phi
  • phi
yes A = e^C we don't care about C per se.
theEric
  • 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
  • theEric
Thank you!
phi
  • phi
you can use non zero values of t. But obviously t=0 makes it especially easy to find A
theEric
  • theEric
Right, thank you very much! :)
theEric
  • 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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