Chapter 2.4, Problem 8E

To determine

To calculate: The rate of draining of water and average rate after $10$ minutes.

Answer to Problem 8E

The rate is $-8000$ gallons and average rate is $-10000$ gallons per minute and the negative sign shows the draining of water.

Explanation of Solution

Given information:

The rate function is $Q\left(t\right)=200{\left(30-t\right)}^2$ , the time is $10$ minutes.

Concept used:

The formula used is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ , $\text{Average rate = }\frac{Q\left(10\right)-Q\left(0\right)}{10-0}$ .

Calculation:

The function is $Q\left(t\right)=200{\left(30-t\right)}^2$ .

Differentiate the function with respect to $t$ .

  $\begin{array}{c}\frac{d}{dt}Q\left(t\right)=\frac{d}{dt}\left[200\left(900+t^2-60t\right)\right]\\ =200\frac{d}{dt}\left(900+t^2-60t\right)+\left(900+t^2-60t\right)\frac{d}{dt}\left(200\right)\\ =200\left(0+2t-60\right)+\left(900+t^2-60t\right)\times 0\\ =200\left(2t-60\right)\end{array}$

Substitute $10$ for $t$ in the derivative $200-6t$ .

  $\begin{array}{c}{Q}^{\prime }\left(10\right)=200\left(2\times 10-60\right)\\ =200\left(20-60\right)\\ =200\left(-40\right)\\ =-8000\end{array}$

The rate is $-8000$ gallons, negative sign shows that the water is draining.

Substitute the values in the formula $\text{Average rate = }\frac{Q\left(10\right)-Q\left(0\right)}{10-0}$ ,

  $\begin{array}{c}\text{Average rate = }\frac{200{\left(30-10\right)}^2-200{\left(30-0\right)}^2}{10-0}\\ =\frac{200\times 400-200\times 900}{10}\\ =\frac{80000-180000}{10}\\ =-10000\end{array}$

The average rate is $-10000$ gallons/minute.

Conclusion: The rate is $-8000$ gallons and average rate is $-10000$ gallons per minute.