Struggling with coffee cup calculus problem

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Sanshin77
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Struggling with coffee cup calculus problem

Post by Sanshin77 »

Hey! I'm in AP Calculus in HS and can't figure out what I'm doing wrong in this problem.It's about the rate at which a coffe cup filled with coffee cools down.
Here it is:
Coffe Cup Problem
Data given:
D = Coffee temperature in F above room temp.
t = time in minutes
At t=0, D = 130
At t=3, D = 117
dD/dt = kD
D varies exponentially with t

Write the general solution for D. Use the given data to find the two constants in this particular solution.
This is what I've done:
dD/dt = kD
dD*D = k*dt Take integral of both sides
1/2D^2 = k*t
D^2 = 2kt

D^2 = 2*k*t

If I try to plug in t=0, D=130, I'll end up with 130 = 0 so obviously something went wrong... What did I do?
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Sanshin77
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Re: Struggling with coffee cup calculus problem

Post by Sanshin77 »

This is the original worded problem:

Coffee Cup Problem:

Suppose you pour a cup of coffee. When it is poured, it is at D = 130 degrees Fahrenheit above room temperature. Three minutes later it has cooled to 117 degrees F above room temperature. As the coffee cools the instantaneous rate of change of D with respect to time, t, in minutes is directly proportional to D.

If dD/dt is directly proportional to D, then D varies exponentionally with t. Use this fact to write the general solution for D. Use the given information to find the two constants in the particular solution.
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Re: Struggling with coffee cup calculus problem

Post by qpHalcy0n »

It appears to me that you have made a trivial mistake:


1) dD/dt = kD
2) dD = kD dt ( divide the D out )
3) dD/D = k dt ( variables are isolated, and k is constant )
4) 1/D dD = k dt ( alternate form of #3 above )
5) integrate
5) ln|D| = kt ( the constant k comes out, dt = 1dt = t -> kt )
6) turn into exponential form
7) D = e^kt ( fundamental form of newton's law of cooling, where k is some constant of cooling and t is time )


This is newton's law of cooling. Go thenceforth from here, young padiwan :]
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