Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 23

How fast is the water being pumped into the tank?
Recall that the volume of the cone is $\displaystyle v = \frac{1}{3} \pi r^2 h$

By using similar triangles we have,


$
\begin{equation}
\begin{aligned}

\frac{r}{h} =& \frac{2}{6}
\\
\\
r =& \frac{h}{3}

\end{aligned}
\end{equation}
$


Substituting the value of $r$ to the volume we get,


$
\begin{equation}
\begin{aligned}

v =& \frac{1}{3} \pi \left( \frac{h}{3} \right)^2 h
\\
\\
v =& \frac{1}{27} \pi h^3

\end{aligned}
\end{equation}
$


Taking the derivative with respect to time,

$\displaystyle \frac{dv}{dt} = \frac{\pi}{27} \cdot (3h^2) \frac{dh}{dt}$

We know that $\displaystyle \frac{dv}{dt} = \frac{dp}{dt} - 10,000$ since the water is being pumped and leaked simultaneously so...


$
\begin{equation}
\begin{aligned}

\frac{dp}{dt} - 10,000 =& \frac{\pi}{27} \cdot 3 \left( 2 \cancel{m} \cdot \frac{100cm}{1 \cancel{m}} \right)^2 (20)
&& \text{We use the measurement in $cm$ to be consistent with the units}
\\
\\
\frac{dp}{dt} =& \frac{\pi}{9} (200)^2 (20) + 10,000
\\
\\
\frac{dp}{dt} =& 289,252.68 cm^3/min

\end{aligned}
\end{equation}
$


This means that the water is being pumped at a rate of $289.252.68 cm^3/min$.