91Ó°ÊÓ

A thermometer reading 72°F is placed in a refrigerator where the temperature is a constant

38°¹ó.

The temperature u of a heated object at a given time t is given by the function

$u\left(t\right)=T+\left(u_0-T\right)e^{kt}$where $k<0$, T is the constant temperature of the surrounding medium, $u_0$ is the initial temperature of the heated object, and k is a negative constant.

Substitute role="math" localid="1647352366126" $u\left(t\right)=60,u_0=72,T=38,t=2$in the function role="math" localid="1647351730224" $u\left(t\right)=T+\left(u_0-T\right)e^{kt}$

$$\begin{gathered} u\left(t\right)=T+\left(u_0-T\right)e^{kt} \\ 60=38+\left(72-38\right)e^{k×2} \\ 22=34e^{k×2} \\ \frac{22}{34}=e^{k×2} \\ \ln \left(\frac{11}{17}\right)=\ln \left(e^{k×2}\right) \\ \ln \left(\frac{11}{17}\right)=k×2 \\ k=\frac{\ln \left(\frac{11}{17}\right)}{2} \\ k=-0.217 \end{gathered}$$

Reading in the thermometer after 7 minutes can be calculated by

Substituting role="math" localid="1647351965657" $u_0=72,T=38,k=-0.217,t=7$in the function role="math" localid="1647352434797" $u\left(t\right)=T+\left(u_0-T\right)e^{kt}$

role="math" localid="1647352162079" $$\begin{gathered} u\left(t\right)=T+\left(u_0-T\right)e^{kt} \\ u\left(t\right)=38+\left(72-38\right)e^{-0.217×7} \\ u\left(t\right)=38+34e^{-0.217×7} \\ u\left(t\right)=45.{44}^{0}F \end{gathered}$$

The time at which the thermometer reads 39°F can be calculated by substituting

$u\left(t\right)=39,u_0=72,T=38,k=-0.217$in the function$u\left(t\right)=T+\left(u_0-T\right)e^{kt}$

$$\begin{gathered} u\left(t\right)=T+\left(u_0-T\right)e^{kt} \\ 39=38+\left(72-38\right)e^{-0.217t} \\ t=16.25 \end{gathered}$$

The relation between temperature and time is given by $y=38+34e^{-0.217x}$

Where y is the temperature and x is the time.

By graphing $y_1={45}^{0}F$and $y_2=38+34e^{-0.217x}$

where x is time, and using INTERSECT, we find that it takes 7.28 minutes by the thermometer to read 45°F.

The trace of the function for a larger value of time is

We can notice that as t increases, the value of $e^{-0.217t}$approaches ${38}^{0}F$, which is the temperature of the refrigerator.