Question

when a bowl of hot soup is left in a room, the soup eventually cools down to room temperature. the temperature t of the soup is a function of time t. the table shows the temperature t (°f) of the soup t minutes after it was left on the table.
find the average rate of change of the temperature of the soup over the first 20 minutes and over the next 20 minutes.

Explanation:

Step1: Recall average rate of change formula

The average rate of change of a function $y = f(x)$ over the interval $[x_1,x_2]$ is $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Here, $x$ is time $t$ (in minutes) and $y$ is temperature $T$ (in $^{\circ}F$).

Step2: Calculate average rate of change for the first 20 minutes

For the first 20 minutes, $t_1 = 0$, $T_1=237$, $t_2 = 20$, $T_2 = 112$. The average rate of change is $\frac{T_2 - T_1}{t_2 - t_1}=\frac{112 - 237}{20-0}=\frac{- 125}{20}=-6.25^{\circ}F/\text{min}$.

Step3: Calculate average rate of change for the next 20 minutes

For the next 20 minutes, $t_1 = 20$, $T_1 = 112$, $t_2=40$, $T_2 = 70$. The average rate of change is $\frac{T_2 - T_1}{t_2 - t_1}=\frac{70 - 112}{40 - 20}=\frac{-42}{20}=-2.1^{\circ}F/\text{min}$.

Answer:

The average rate of change of the temperature of the soup over the first 20 minutes is $-6.25^{\circ}F/\text{min}$ and over the next 20 minutes is $-2.1^{\circ}F/\text{min}$.