Chapter 5.6, Problem 18E

a.

To determine

To determine rate of the water level changing when the water is $8m$ deep.

Answer to Problem 18E

The water level is decreasing by $-1.132cm/\min$ .

Explanation of Solution

Given information:

Water is flowing at the rate of $6m^3/\min$ from a hemispherical reservoir at a radius $13m$ .

Formula:

  $1m=100cm$

The bowl radius is $R=13m$

Hemisphere volume when water is $y$ units deep.

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

Rate of water flow is negative as it is flowing out,

  $\frac{dV}{dt}=-6m^3/\min$

Differentiate $V$ with respect to $t$ ,

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

At the moment the water is $y=8m$ , $\begin{array}{l}-6=\frac{\pi }{3}\left(78\left(8\right)-3{\left(8\right)}^2\right)\frac{dy}{dt}\\ -6=\frac{432\pi }{3}\frac{dy}{dt}\\ -6=144\pi \frac{dy}{dt}\\ \frac{dy}{dt}=-\frac{6}{144\pi }=-\frac{1}{24\pi }\approx -0.01327m/\min \\ \frac{dy}{dt}=-1.327cm/\min \end{array}$

Therefore, the water level is decreasing by $-1.132cm/\min$ .

b.

To determine

To Find:

To find the radius of water’s surface when the water is $ym$ deep .

Answer to Problem 18E

The radius is found to be $r=\sqrt{26y-y^2}m$ .

Explanation of Solution

Given information:

Water is flowing at the rate of $6m^3/\min$ from a hemispherical reservoir at a radius $13m$ .

Formula:

  $1m=100cm$

Pythagorean theorem $a^2+b^2=c^2$

Binomial theorem ${\left(a-b\right)}^2=a^2-2ab+b^2$

The bowl radius is $R=13m$

Side $=13-y$ ,

Using Pythagorean theorem find $r$ ,

  $\begin{array}{l}{\left(13-y\right)}^2+r^2=169\\ r=\sqrt{169-{\left(13-y\right)}^2}\end{array}$

  ${\left(13-y\right)}^2=169-26y+y^2$

  $\begin{array}{l}r=\sqrt{169-\left(169-26y+y^2\right)}\\ r=\sqrt{169-169+26y-y^2}\\ r=\sqrt{26y-y^2}\end{array}$

Therefore, the radius is found to be $r=\sqrt{26y-y^2}m$ .

c.

To determine

To Find:

To find the rate of radius changing when the water is $8m$ deep .

Answer to Problem 18E

The rate of radius is found to be $\frac{dr}{dt}=-\frac{5}{288\pi }m/\min$ .

Explanation of Solution

Given information:

Water is flowing at the rate of $6m^3/\min$ from a hemispherical reservoir at a radius $13m$ .

Formula:

  $1m=100cm$

From part (b), we have $r=\sqrt{26y-y^2}m$

  $r=\sqrt{26y-y^2}$

Implicitly differentiating to derive $r$ ,

  $\begin{array}{l}\frac{dr}{dt}=\frac{1}{2\sqrt{26y-y^2}}\cdot \left(26\frac{dy}{dt}-2y\frac{dy}{dt}\right)\\ \frac{dr}{dt}=\frac{1}{2\sqrt{26\left(8\right)-{\left(8\right)}^2}}\cdot \left[\left(26\cdot -\frac{1}{24\pi }\right)-\left(2\left(8\right)\cdot -\frac{1}{24\pi }\right)\right]\\ \frac{dr}{dt}=-\frac{5}{288\pi }m/\min \end{array}$

Therefore, the rate of radius is found to be $\frac{dr}{dt}=-\frac{5}{288\pi }m/\min$ .