# Models for Cooling, 9.4 in the Stewart Calculus book, version 6.

• June 12th 2011, 02:24 PM
CalculusABStudent
Models for Cooling, 9.4 in the Stewart Calculus book, version 6.
1.#7 Page 618: A peach pie is removed from the oven at 5:00 PM. At that time it is piping hot, 100 C. At 5:10 PM its temperature is 80 C; at 5:20 PM it is 65 C. What is the temperature of the room?

The answer is 20 degrees Celsius but I'm not sure how to get there from this step

(0, 100): 100= x-u
(10, 80): 80 = x-ue^-10k
(20, 65): 65= x- ue^-20k

solving for x "simultaneously" is what yields 20 degrees Celsius but I don't know how to do that with those equations to get 20.

btw the original equation that got me to those 3 equations was dT/dt= -k(T-x) then you make it to dt/dT= 1/(k(x-T)) then separate variables etc

any help is apprecfiated
• June 12th 2011, 02:39 PM
pickslides
Solving the equation using separation of variables gives

$\displaystyle T = Ae^{kt}+T_a$

Use the 3 points you have to solve for $A, k$ and $T_a$
• June 12th 2011, 02:57 PM
CalculusABStudent
That's the part I'm confused on..could you show me how to start that?

And sorry I guess I posted in the wrong subforum, I'm new :/
• June 14th 2011, 10:06 PM
pickslides
Re: Models for Cooling, 9.4 in the Stewart Calculus book, version 6.
100= x-u
80 = x-ue^-10k
65= x- ue^-20k

To solve this system, you can easily make 3 variables and 3 equations into 2 variables and 2 equations

x=u+100 therefore

80 = u+100-ue^(-10k)
65= u+100- ue^(-20k)

Take 100 from both sides, giving

-20 = u-ue^(-10k)
-35= u- ue^(-20k)

Factor out u from both

-20 = u(1-e^(-10k))
-35= u(1- e^(-20k))

So making u the subject for both we end up with

-20 /(1-e^(-10k)) = -35/(1- e^(-20k))

Now you have one variable with one equation, go get em!