Chapter 4.6, Problem 18E

$\text{(a)}$

To determine

To find: The rate is the water level changing when the water is $8m$ deep.

Answer to Problem 18E

The answer: $-0.01327 \text{m}/\text{min}$

Explanation of Solution

Given information:

The figure is:

  

Calculation:

Radius $R=13 \text{m}$ .

Hemisphere volume when water is $y$ units deep:

  $\begin{array}{c}V=\frac{\pi }{3}{y}^2(3R-y)\\ =\frac{\pi }{3}{y}^2(3(13)-y)\\ =\frac{\pi }{3}\left(39y^2-y^3\right)\end{array}$

Rate of water flow, negative because it is flowing out

  $\frac{dV}{dt}=-6 {\text{m}}^3/\text{min}$

Differentiate $V$ with respect to time $t$ :

  $\begin{array}{c}\frac{dV}{dt}=\frac{\pi }{3}\left(2\cdot 39y\frac{dy}{dt}-3y^2\frac{dy}{dt}\right)\\ =\frac{\pi }{3}\left(78y-3y^2\right)\frac{dy}{dt}\end{array}$

At the moment the water is $y=8 \text{m}$ deep. Plug in all the known values to find the rate of water level change:

  $\begin{array}{c}-6=\frac{\pi }{3}\left(78(8)-3{(8)}^2\right)\frac{dy}{dt}\\ \Rightarrow -6=144\pi \frac{dy}{dt}\\ \Rightarrow \frac{dy}{dt}=-\frac{6}{144\pi }\\ =-\frac{1}{24\pi }\\ \approx -0.01327 \text{m}/\text{min}\end{array}$

Therefore, the required rate is the water level changing when the water is $8m$ deep is $-0.01327 \text{m}/\text{min}$ .

$\text{(b)}$

To determine

To find: The radius $r$ of the water's surface when the water is $ym$ deep.

Answer to Problem 18E

The answer: $r=\sqrt{26y-y^2}$

Explanation of Solution

Given information:

The figure is:

  

Calculation:

Radius $R=13 \text{m}$ .

The goal of the exercise is to compute the radius of the water's surface in a hemispherical bowl $r$ when the water is $ym$ deep in that bowl.

Also, it is given that the radius of the hemispherical bowl is $s=13 \text{m}$ .

It is given that,

  $\text{AB=y}$

And

  $OB=13 \text{m}$ being the radius of the hemispherical bowl. This implies that

  $\begin{array}{c}\text{OA=OB-AB}\\ \text{=13-y}\end{array}$

Now apply the Pythagoras theorem to find the value of $r$ :

  $\begin{array}{c}O{C}^2=O{A}^2+A{C}^2\\ $ \Rightarrow {(13)}^2={(13-y)}^2+r^2$\Rightarrow 169={13}^2+y^2-2(13)(y)+r^2\\ =169+y^2-26y+r^2\\ \Rightarrow r^2=169-169-y^2+26y\\ =-y^2+26y\\ \Rightarrow r=\sqrt{26y-y^2} \text{m}\end{array}$

Therefore, the required radius $r$ of the water's surface when the water is $ym$ deep is $r=\sqrt{26y-y^2}$ .

$(c)$

To determine

To find: The rate is the radius $r$ changing when the water is $8m$ deep.

Answer to Problem 18E

The answer: $\frac{-5}{288\pi }\text{ meters }/\text{ minute }$

Explanation of Solution

Given information:

The figure is:

  

Calculation:

  $\begin{array}{c}r=\sqrt{26y-y^2}\\ \Rightarrow r={\left(26y-y^2\right)}^{-1/2}\\ \Rightarrow \frac{dr}{dt}=\frac{1}{2\sqrt{26y-y^2}}\cdot \left(26\frac{dy}{dt}-2y\frac{dy}{dt}\right)\\ \Rightarrow \frac{dr}{dt}=\frac{1}{2\sqrt{26(8)-{(8)}^2}}\cdot \left[\left(26\cdot \frac{-1}{24\pi }\right)-\left(2(8)\cdot \frac{-1}{24\pi }\right)\right]\\ \Rightarrow \frac{dr}{dt}=\frac{-5}{288\pi }\text{ meters }/\text{ minute }\end{array}$

Therefore, the required rate is the radius $r$ changing when the water is $8m$ deep is $\frac{-5}{288\pi }\text{ meters }/\text{ minute }$ .