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Math Help - Models for Cooling, 9.4 in the Stewart Calculus book, version 6.

  1. #1
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    Exclamation 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
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    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
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    That's the part I'm confused on..could you show me how to start that?
    Thanks for your reply

    And sorry I guess I posted in the wrong subforum, I'm new :/
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    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!
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