Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 5

At what rate is the height of the water increasing?

Given: $\displaystyle r = 5m, \frac{dv}{dt} = 3m^3/min$

Required: $\displaystyle \frac{dh}{dt}$

Solution: Let $V = \pi r^2h$ be the volume of the cylinder

where $r$ = radius

$\qquad h$ = height

It is also stated in the problem that $r$ is a constant so,


$\begin{equation} \begin{aligned} \frac{dv}{dt} =& \frac{dv}{dh} \left( \frac{dh}{dt} \right) = \pi r^2 \left( \frac{dh}{dt} \right) \\ \\ \frac{dv}{dt} =& \pi r^2 \left( \frac{dh}{dt} \right) \\ \\ \frac{dh}{dt} =& \frac{1}{\pi r^2} \left( \frac{dv}{dt} \right) \\ \\ \frac{dh}{dt} =& \frac{1}{\pi (5)^2} (3) \\ \\ \end{aligned} \end{equation}$


$\fbox{$\large \frac{dh}{dt} = \frac{3}{25 \pi} m/min $}$