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
Re: Models for Cooling, 9.4 in the Stewart Calculus book, version 6.
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
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!