Answer
$\displaystyle \frac{3}{25\pi}$ m/min
Work Step by Step
Given information: $r = 5$ m, $\displaystyle \frac{dV}{dt} = 3 $ $m^3$/min,
What we're trying to find: Change in height with respect to time ($\displaystyle \frac{dh}{dt}$)
---
Volume of a cylinder: $V = πr^2h$
Implicitlyifferentiate with respect to time using the product rule:
$\displaystyle \frac{dV}{dt} = \pi[h(2r\cdot\frac{dr}{dt}) + r^2\cdot\frac{dh}{dt}]$
$\displaystyle \frac{dV}{dt} = 2πhr\frac{dr}{dt} + πr^2\frac{dh}{dt}$
Now you might be wondering what to do with the $ \displaystyle \frac{dr}{dt}$ since we weren't given a value for it, but we actually don't need one. Because the radius of the cylinder is not physically changing, $ \displaystyle \frac{dr}{dt} = 0$ (this can be hard to conceptualize but just imagine a cylinder being filled up with water–only the height of the water in the cylinder changes as water is being poured in). Thus, that gives us
$\displaystyle \frac{dV}{dt} = πr^2\frac{dh}{dt}$
We can now plug in our givens
$\displaystyle 3 = π(5)^2\frac{dh}{dt}$
$\displaystyle 3 = 25π\frac{dh}{dt}$
$\displaystyle \frac{3}{25π} $m/min $\displaystyle= \frac{dh}{dt}$
