Chapter 13.3, Problem 33E

(a)

To determine

To Evaluate: The average rates of the water flows from the tank over the time intervals $\left[10,15\right]$ and $\left[15,20\right]$.

Answer to Problem 33E

The average rates of the water flows from the tank over the time intervals $\left[10,15\right]$ and $\left[15,20\right]$ is $38.8$ and $27.8$.

Explanation of Solution

Given:

The volume of the water remaining in the tank after $t$ minutes shows in the table.

A tank holds $1000$ gallons of water.

$t$(min)	$V$(gal)
5	694
10	444
15	250
20	111
25	28
30	0

Formula used:

Definition of the slope of the line:- The average rates volume of the tank is the change in the function between two numbers $t$ and $a$.

$m=\underset{t\to a}{\lim }\frac{f\left(t\right)-f\left(a\right)}{t-a}$ at $t=a$.

Calculation:

Section1:

Use the definition of the slope of the line to find the average rate of water from the tank.

The interval $\left[10,15\right]$ is the value of the time. So, it can be written as:

$m=\underset{t\to 15}{\lim }\frac{f\left(10\right)-f\left(15\right)}{10-15}$

Substitute $444$ for $f\left(10\right)$ and $250$ for $f\left(15\right)$ in the above equation,

$\begin{array}{c}m=\frac{444-250}{10-5}\\ m=-\frac{194}{5}\\ m=-38.8\end{array}$

Thus, The average rates of the water flows from the tank over the time intervals $\left[10,15\right]$ is $-38.8$.

Section2:

To find the average rates of the water flows from the tank over the time intervals $\left[15,20\right]$.

Use the definition of the slope of the line to find the average rate of water from the tank.

The interval $\left[15,20\right]$ is the value of the time. So, it can be written as:

$m=\underset{t\to 20}{\lim }\frac{f\left(15\right)-f\left(20\right)}{15-20}$

Substitute $250$ for $f\left(15\right)$ and $111$ for $f\left(20\right)$ in the above equation,

$\begin{array}{c}m=\frac{250-111}{15-20}\\ m=-\frac{139}{5}\\ m=-27.8\end{array}$

Thus, the average rates of the water flows from the tank over the time intervals $\left[15,20\right]$ is $-27.8$.

(b)

To determine

To find:  The average of the slopes of the secant lines in part (a).

Answer to Problem 33E

The average of the slope of the secant lines in part(a) is $33.3$.

Explanation of Solution

Given:

The volume of the water remaining in the tank after $t$ minutes shows in the table.

A tank holds $1000$ gallons of water.

$t$(min)	$V$(gal)
5	694
10	444
15	250
20	111
25	28
30	0

Formula used:

Definition of the slope of the line:- The average rates volume of the tank is the change in the function between two numbers $t$ and $a$.

$m=\underset{t\to a}{\lim }\frac{f\left(t\right)-f\left(a\right)}{t-a}$ at $t=a$.

Calculation:

Use the definition of the slope of the line to estimate the rate of water from the tank.

$m=\underset{t\to a}{\lim }\frac{f\left(t\right)-f\left(a\right)}{t-a}$

Now, try the values of the $f\left(a\right)$ from the given table at $t=15$.

Substitute $250$ for $f\left(t\right)$ and $694$ for $f\left(a\right)$ and $15$ for $t$ and $5$ for $a$ in the above equation,

$\begin{array}{c}m=\frac{250-694}{15-5}\\ m=-\frac{444}{10}\\ m=-44.4\end{array}$

Substitute $250$ for $f\left(t\right)$ and $444$ for $f\left(a\right)$ and $15$ for $t$ and $10$ for $a$ in the above equation,

$\begin{array}{c}m=\frac{250-444}{15-10}\\ m=\frac{194}{5}\\ m=-38.8\end{array}$

Substitute $250$ for $f\left(t\right)$ and $111$ for $f\left(a\right)$ and $15$ for $t$ and $20$ for $a$ in the above equation,

$\begin{array}{c}m=\frac{250-111}{15-20}\\ m=-\frac{139}{5}\\ m=-27.8\end{array}$

Substitute $250$ for $f\left(t\right)$ and $28$ for $f\left(a\right)$ and $15$ for $t$ and $25$ for $a$ in the above equation,

$\begin{array}{c}m=\frac{250-28}{15-25}\\ m=-\frac{222}{10}\\ m=-22.2\end{array}$

Substitute $250$ for $f\left(t\right)$ and $0$ for $f\left(a\right)$ and $15$ for $t$ and $30$ for $a$ in the above equation,

$\begin{array}{c}m=\frac{250-0}{15-30}\\ m=-\frac{250}{15}\\ m=-16.66\end{array}$

Now take the average of the slopes of two secants between $t=10$ and $t=15$, and between $t=15$ and $t=20$,

$\begin{array}{c}\text{A}=\frac{-38.8-27.8}{2}\\ =-33.3\end{array}$

Thus, the average slopes of the secant line in part(a) is $-33.3$.