# Thread: word problem

1. ## 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

2. Originally Posted by startingover
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

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