Chapter 4.6, Problem 17E

(a)

To determine

To find : How fast the water level falling when the water is $5m$ deep.

Answer to Problem 17E

The answer: decreasing at $1.132\text{cm}/\text{min}$

Explanation of Solution

Given information:

Water is falling at the rate of $50{\text{m}}^3/\text{min}$

Base radius $545\text{m }$ and height $6\text{m}$

Calculation:

  

The cone's half from the side.

Using the measurements of the cone, determine the relationship between the height (the depth of the water level) and the radius $r$ . The origin serves as the vertex, and the line going from there to $(45,6)$ is:

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

Change in volume rate (negative since it is declining):

  $\frac{dV}{dt}=-50$

Substitute for $r$ in the cone volume equation

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

Differentiate with respect to time $t$ :

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

  $h=5\text{m}$ ,Substitute all the known values and solve for the rate of change of the height $\frac{dh}{dt}$ for the same moment.

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

Negative rate meaning it is decreasing.

Therefore the required decreasing at $1.132\text{cm}/\text{min}$ fast the water level falling when the water is $5m$ deep.

(b)

To determine

To find: How fast the radius of the water’s surface changing at the moment.

Answer to Problem 17E

The answer: decreasing at $8.488\text{cm}/\text{min}$

Explanation of Solution

Given information:

Water is falling at the rate of $50{\text{m}}^3/\text{min}$

Base radius $545\text{m }$ and height $6\text{m}$

Calculation:

From part $(a)$ :

  $\begin{array}{c}\frac{dh}{dt}=-\frac{800}{225\pi }\\ =-\frac{32}{9\pi }\text{cm}/\text{min}\end{array}$

  $\begin{array}{c}h=\frac{6}{45}r\\ \Rightarrow r=\frac{15}{2}h\end{array}$

Differentiate this to,

  $\frac{dr}{dt}=\frac{15}{2}\frac{dh}{dt}$

Then plug in the result from part $(a)$ :

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

Negative rate meaning it is decreasing.

Therefore the required decreasing at $8.488\text{cm}/\text{min}$ fast the radius of the water’s surface changing at the moment.