## Calculus: Early Transcendentals 8th Edition

$\frac{dh}{dt} = \frac{3}{25\pi} \frac{m}{min}$
Substitute the $r$ value and differentiate the equation to the volume of a cylinder. $V = \pi r^{2}h$ $V = \pi 5^{2}h$ $V = 25\pi h$ $\frac{dV}{dt} = 25\pi \frac{dh}{dt}$ Now substitute $\frac{dV}{dt}$ and solve it for $\frac{dh}{dt}$. $3 = 25\pi \frac{dh}{dt}$ Now solve for $\frac{dh}{dt}$: $\frac{dh}{dt} = \frac{3}{25\pi} \frac{m}{min}$