Chapter 1.7, Problem 37E

Finding the Model and Solving Water is stored in a conical tank with a faucet at the bottom. The tank has depth 24 in. and radius 9 in., and it is filled to the brim. If the faucet is opened to allow the water to flow at a rate of 5 cubic inches per second, what will the depth of the water be after 2 min?

Solution

To calculate the depth of the water in the tan after $2\text{ min}\text{.}$

The depth of water in the tank after $2\text{ minutes}$ is $h_1=21.364\text{ in}\text{.}$

Given:

Depth of tank $=24\text{ in}$

Radius of tank $=9\text{ in}$

Flow rate of water $=5\text{ cubic inches per second}$

Formula:

  $\text{Volume of cone }=\frac{1}{3}\pi r^{2}h$

Calculation :

The volume of conical tank is:

  $\begin{array}{l}V=\frac{1}{3}\pi r^{2}h\\ V=\frac{1}{3}\times \frac{22}{7}\times 9\times 9\times 24\\ V=\frac{42768}{21}\\ V=2036.57{\text{ in}}^3\end{array}$

The conical tank that will get empty after one second $=5{\text{ in}}^3$

Hence, the amount of water flown after two minutes $=5\times 120=600{\text{ in}}^3$

Hence, the volume of the water left in tank after $2\text{ minutes:}$

  $V_1=2036.57-600=1436.57{\text{ in}}^3$

Now, the radius and height of the cone are related as:

  $\begin{array}{l}\frac{r}{h}=\frac{9}{24}\\ \Rightarrow \frac{r}{h}=\frac{3}{8}\\ \Rightarrow r=\frac{3}{8}h\end{array}$

Hence, the depth of water in the tank is:

  $\begin{array}{l}{V}_1=\frac{1}{3}\pi r_1{}^2{h}_1\\ \Rightarrow V_1=\frac{1}{3}\pi {\left(\frac{3}{8}{h}_1\right)}^2{h}_1\\ \Rightarrow 1436.57=\frac{1}{3}\times \frac{22}{7}\times \frac{9}{64}\times h_1{}^3\\ \Rightarrow h_1{}^3=9751.263\\ \Rightarrow h_1=21.364\text{ in}\end{array}$