Chapter 4.1, Problem 28E

To determine

To find: The fast is the water level rising when the depth at the deepest point is $5ft$ .

Answer to Problem 28E

The water level is rising at rate is $0.0013$ .

Explanation of Solution

Given information:

The given expression is $20ft$ wide $40ft$ long at rate of $0.8\raisebox{1ex}{$f{t}^3$}\!\left/ \!\raisebox{-1ex}{$\min $}\right.$ .

Calculation:

Calculate the expression.

From the swimming pool as shown in the figure (1),

  

Figure (1)

The total value is

  $\begin{array}{c}v=20\left[\frac{1}{2}\left(b_1+b_2\right)h\right]\\ v=20\left[\frac{1}{2}\left(12+x_1+12+x_2\right)h\right]\\ =10\left(24+x_1+x_2\right)h\mathrm{....}\left(1\right)\end{array}$

Find $x_{1}and{x}_2$

From figure the first -1 construct the following triangle as shown.

  

Figure -2 figure -3

Use the similar triangle properties to the figure-2

  $\begin{array}{c}\frac{x_1}{h}=\frac{6}{6}\\ =1\\ x_1=h\end{array}$

Again Use the similar triangle properties to the figure-3

  $\begin{array}{c}\frac{x_2}{h}=\frac{16}{6}\\ =\frac{8}{3}\\ x_2=\frac{8}{3}h\end{array}$

Substitute the value of $x_{1}and{x}_2$ in (1)

  $\begin{array}{c}v=10\left(24+x_1+x_2\right)h\\ =10\left(24+h+\frac{8}{3}h\right)h\\ =10\left(24+\frac{11}{3}h\right)h\\ =240h+\frac{110}{3}{h}^2\end{array}$

Differential both sides of $v=240h+\frac{110}{3}{h}^2$ with respect to t.

  $\begin{array}{c}\frac{dv}{dt}=\frac{d}{dt}\left(240h+\frac{110}{3}{h}^2\right)\\ =240\frac{d}{dt}\left(h\right)+\frac{110}{3}\frac{d}{dt}\left(h^2\right)\\ =240\frac{dh}{dt}+\frac{110}{3}\left(2h\frac{dh}{dt}\right)\\ =\left(240+\frac{220}{3}h\right)\frac{dh}{dt}\\ 0.8=\left(240+\frac{220}{3}5\right)\frac{dh}{dt}\\ 0.8=\left(240+\frac{110}{3}\right)\frac{dh}{dt}\\ 0.8\left(\frac{3}{1820}\right)=\frac{dh}{dt}\\ 0.0013=\frac{dh}{dt}\end{array}$

Therefore, the water level is rising at rate is $0.0013$ .