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theEric
Group Title
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 \(\dfrac{dT}{dt}=k(TT_r)\) where \(k\) is a proportionality constant and \(T_r\) is the room temperature (\(70^\circ\text F\)).
 one year ago
 one year ago
theEric Group Title
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 \(\dfrac{dT}{dt}=k(TT_r)\) where \(k\) is a proportionality constant and \(T_r\) is the room temperature (\(70^\circ\text F\)).
 one year ago
 one year ago

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theEric Group TitleBest ResponseYou've already chosen the best response.1
I figure that I probably need to find out what \(k\) is...
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
@sauravshakya
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Can I use \(\Large{\int\limits_{7.5}^{7.75}}\normalsize\dfrac{dT}{dt}dt=\Large{\int\limits_{7.5}^{7.75}}\normalsize k(TT_r)\ dt=200120\)? I'll think on that...
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
I haven't done anything like it in class, that I recall.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
But I don't have \(T(t)\) to do integration...
 one year ago

Saeeddiscover Group TitleBest ResponseYou've already chosen the best response.0
You can say nothing about 8:00 AM here. Because We don't know how the temperature function is changing over time.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
That's what I was thinking as well... But we know the rateofchange function, two points on the \(T\) vs \(t\) curve, which is one solution to the rateofchange function...
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
I know that initial value problems require one point...
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.4
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
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Thank you very much! I'm working on digesting it now.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
\(A\) is an arbitrary constant like \(C\), right?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.4
yes A = e^C we don't care about C per se.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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 nonzero values for the independent variable?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Thank you!
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.4
you can use non zero values of t. But obviously t=0 makes it especially easy to find A
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Right, thank you very much! :)
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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!
 one year ago
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