Chapter 5.6, Problem 17E

a.

To determine

To find the speed of the water level falling when the radius is $5m$ deep.

Answer to Problem 17E

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

Explanation of Solution

Given information:

Water is flowing at the rate of $50m^3/\min$ from a concrete conical reservoir at a radius $45m$ and height $6m$ .

Formula:

  $1m=100cm$

Graph:

  

Let the height be $h$ and $r$ be the radius. The vertex is at the origin, the line is connecting it to $\left(45,6\right)$ is,

  $\begin{array}{l}h=\frac{6}{45}r\\ \Rightarrow r=\frac{45}{6}h\end{array}$ ,

Rate of volume change,

  $\frac{dV}{dt}=-50$

Volume of cone is $\pi r^2\frac{h}{3}$

Substitute for $r$ in the cone volume equation,

  $\begin{array}{l}V=\pi r^2\frac{h}{3}\\ V=\frac{1}{3}\pi {\left(\frac{45}{6}h\right)}^{2}h\\ V=\frac{75}{4}\pi h^3\end{array}$

Differentiate $V$ with respect to $t$ ,

  $\frac{dV}{dt}=\frac{225}{4}\pi h^2\frac{dh}{dt}$

At the moment $h=4m$ , Substitute the given values and solve for the rate of change of the height $\frac{dh}{dt}$ for the same moment ,

  $\begin{array}{l}\frac{dV}{dt}=\frac{225}{4}\pi h^2\frac{dh}{dt}\\ -50=\frac{225}{4}\pi {\left(5\right)}^2\frac{dh}{dt}\\ -50=\frac{5625}{4}\pi \frac{dh}{dt}\\ -\frac{200}{5625\pi }=\frac{dh}{dt}\\ \frac{dh}{dt}=-\frac{8}{225\pi }m/\min \cdot \frac{100cm}{m}\\ \frac{dh}{dt}=-\frac{800}{225\pi }cm/\min \approx -1.132cm/\min \end{array}$

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

b.

To determine

To Find:

To find the change in radius of water’s surface.

Answer to Problem 17E

The change in radius is decreased by $8.488cm/\min$ .

Explanation of Solution

Given information:

Water is flowing at the rate of $50m^3/\min$ from a concrete conical reservoir at a radius $45m$ and height $6m$ .

Formula:

  $1m=100cm$

From part (a), we have $\frac{dh}{dt}=-\frac{800}{225\pi }cm/\min =-\frac{32}{9\pi }cm/\min$

The relationship between the height and radius is based on the cones dimension,

  $h=\frac{6}{45}r\Rightarrow r=\frac{15}{2}h$

On differentiating we get,

  $\begin{array}{l}\frac{dr}{dt}=\frac{15}{2}\frac{dh}{dt}\\ \frac{dr}{dt}=\frac{15}{2}\cdot \left(-\frac{32}{9\pi }\right)=-\frac{80}{3\pi }\approx -8.488cm/\min \end{array}$

Therefore, the change in radius is decreased by $8.488cm/\min$ .