Chapter 3.4, Problem 8E

To determine

To calculate: The rate at which water running out at the end of 10 min and the average rate of water flows out during the first 10 min.

Answer to Problem 8E

The rate at which the water is running out at the end of 10 min is 8000gallons/min.

The average rate at which the water flows out during the first 10 min is 10000gallon/min.

Explanation of Solution

Given Information: The number of gallons of water in a tank t minutes after the tank has started to drain is $Q\left(t\right)=200{\left(30-t\right)}^2$

.

Calculation:

To find the rate at which the water is running out at the end of 10 min, find derivative of $Q\left(t\right)$ with respect to t at $t=10$.

The derivative of $Q\left(t\right)=200{\left(30-t\right)}^2$

is,

  $\begin{array}{c}\frac{d}{dt}Q\left(t\right)=\frac{d}{dt}\left(200{\left(30-t\right)}^2\right)\\ =\frac{d}{dt}\left(200\left(900-60t+t^2\right)\right)\\ =\frac{d}{dt}\left(180000-12000t+200t^2\right)\\ =-12000+400t\\ {\left[\frac{d}{dt}Q\left(t\right)\right]}_{t=10}=-12000+400\left(10\right)\\ =-12000+4000\\ =-8000\end{array}$

There is a negative sign in the calculation which shows that the water is running out.

Hence, the rate at which the water is running out at the end of 10 min is 8000gallons/min.

Average rate at which the water flows out during the first 10 min is the rate of change between the starting time 0 and the ending point 10.

  $\begin{array}{c}A=\frac{200{\left(30-10\right)}^2-200{\left(30\right)}^2}{10-0}\\ =\frac{200\times {20}^2-200\times 900}{10}\\ =\frac{80000-180000}{10}\\ =10000\text{gal/min}\end{array}$

• Here , A represents the average rate at which the water flows out during the first 10 min.

Conclusion:

The rate at which the water is running out at the end of 10 min is 8000gallons/min.

The average rate at which the water flows out during the first 10 min is 10000gallon/min.