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AP Calculus AB

An interactive log for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.

Wednesday, April 05, 2006

Hot Coffee, give it a few minutes

Today in class we wrapped the last part of the course, except for the coroner problem which we will be doing next class. We did a problem that is similar to the coroner problem, the problem we did a problem about coffee cooling. The coffee's temperature is 190 degrees, and room temperature is 70 degrees, and coffee is "drinkable" at 120 degrees.

First thing you have to do is identify the differential equation that describes the cooling of the coffee.

T' = k(T-70)

dT/dt = k (T-70)

(1/(T-70))dT = k dt

Now antidifferentiate both sides of the equation.

ln(T-70) = kt+C

T-70 = e^(kt+C)
Now that the variables are separated, use the point (0,190) (T,t) to solve for one of the unknown variables.
T-70 = e^(kt)*e^C
190-70 = e^(k(0))*e^C
120 = 1*e^C
120 = e^C
T-70 = 120e^(kt)
Now that there is only one one unknown "k", use a second point (5,180) to solve the last unknown.
180-70 = 120e^5k
110 = 120e^(5k)
110/120 = e^(5k)
11/12 = e^(5k)
ln(11/12) = 5k
(1/5)ln(11/12) = k
k = -0.0174
T-70 = 120e^(-0.0174t)
Now that the equation has no unknowns besides T, and t, the problem can be answered. Use T=120

120-70 = 120e^(-0.0174t)
50/120 = e^(-0.0174t)
5/12 = e^(-0.0174t)
ln(5/12) = -0.0174t
(1/-0.0174)ln(5/12) = t
t = 50.3143
The coffee is drinkable after 50 minutes.

The next scribe is Sarah

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