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  1. #1
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    Post word problem

    On a hot summer day, a cool drink is removed from a refrigerator whose temperature is 10 degrees Celsius and is placed in a room where the temperature is 30 degrees Celsius. The temperature of the drink after t minutes is given by a function of the form f(t) = 30 - Ae^-kt. If the temperature of the drink is 15 degrees Celsius after 20 minutes, what will it be after 40 minutes?

    I understand that the t stands for time, but I am unsure of the significance of the A
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  2. #2
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    Quote Originally Posted by startingover View Post
    On a hot summer day, a cool drink is removed from a refrigerator whose temperature is 10 degrees Celsius and is placed in a room where the temperature is 30 degrees Celsius. The temperature of the drink after t minutes is given by a function of the form f(t) = 30 - Ae^-kt. If the temperature of the drink is 15 degrees Celsius after 20 minutes, what will it be after 40 minutes?

    I understand that the t stands for time, but I am unsure of the significance of the A
    You need to find a value for k and A. You know that the initial temperature is 10 C, so
    10 = 30 - Ae^{-k \cdot 0}
    and you know that the temperature is 15 C at t = 20 min. Thus
    15 = 30 - Ae^{-k \cdot 20}

    The first equation gives a value for A:
    10 = 30 - Ae^{-k \cdot 0}

    10 = 30 - A \cdot 1

    Thus A = 20. (The unit is degrees C.)

    So the second equation gives a value for k:
    15 = 30 - 20e^{-k \cdot 20}

    -15 = -20e^{-20k}

    e^{-20k} = \frac{-15}{-20} = \frac{3}{5}

    -20k = ln \left ( \frac{3}{5} \right )

    -20k = ln(3) - ln(5)

    20k = ln(5) - ln(3)

    k = \frac{1}{20} \cdot (ln(5) - ln(3) )

    If you want decimals, k is about 0.02554128.

    So at t = 40 min...

    f(40) = 30 - 20e^{-0.02554128 \cdot 40} = 22.8

    (It's a good exercise to carry this through with the exact value of k.)

    -Dan
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  3. #3
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    Hello, startingover!

    On a hot summer day, a cool drink is removed from a refrigerator
    whose temperature is 10 C and is placed in a room where the temperature is 30 C.
    The temperature of the drink after t minutes is given by the function: . f(t) \:= \:30 - A\!\cdot\!e^{-kt}

    If the temperature of the drink is 15 C after 20 minutes,
    what will it be after 40 minutes?
    Both A and k are constants to be determined.

    We assume that, when t = 0, the termperature is 10.
    . . That is: . f(0) = 10
    So we have: . 30 - A\!\cdot\!e^0 \:=\:10\quad\Rightarrow\quad A = 20

    The function (so far) is: . f(t) \:=\:30 - 20\!\cdot\!e^{-kt}

    We are told that when t = 20, the termperature is 15.
    . . That is: . f(20) = 15
    So we have: . 30 - 20\!\cdot\!e^{-k(20)} \:=\:15\quad\Rightarrow\quad -20\!\cdot\!e^{-20k} \:=\:-15
    . . e^{-20k} \:=\:0.75\quad\Rightarrow\quad -20k\:=\:\ln(0.75)\quad\Rightarrow\quad k = \frac{\ln(0.75)}{-20} \:\approx\:0.0144

    The function is: . f(t) \:=\:30 - 20\!\cdot\!e^{-0.0144t}


    When t = 40:\;f(40) \:=\:30 - 20\!\cdot\!e^{-0.0144(40)} \;=\:18.7571511 \:\approx\:19^o

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