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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
    $\displaystyle 10 = 30 - Ae^{-k \cdot 0}$
    and you know that the temperature is 15 C at t = 20 min. Thus
    $\displaystyle 15 = 30 - Ae^{-k \cdot 20}$

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

    $\displaystyle 10 = 30 - A \cdot 1$

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

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

    $\displaystyle -15 = -20e^{-20k}$

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

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

    $\displaystyle -20k = ln(3) - ln(5)$

    $\displaystyle 20k = ln(5) - ln(3)$

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

    If you want decimals, k is about 0.02554128.

    So at t = 40 min...

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

    (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: .$\displaystyle f(t) \:= \:30 - A\!\cdot\!e^{-kt}$

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

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

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

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

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


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

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