From a cylindrical drum containing oil and kept vertical, the oil is leaking so that the level of the oil is decreasing at the rate of $2$cm/min. If the radius and the height of the drum are $10.5$cm and $40$cm respectively, find the rate at which the volume of the oil is decreasing.
My Attempt:
Given $\textrm {radius} (r)=10.5$cm
$\textrm {height} (h)=40$cm
$\dfrac {dh}{dt}=-2$ cm/min
Also, $V=\pi r^2h$ $\dfrac {dv}{dt}=\pi(2r.\dfrac {dr}{dt} h+r^2\dfrac {dh}{dt})$
HINT: Is the radius of the oil column changing?
If not, then $\frac{dr}{dt}=0$ gives from your equations $$\dfrac {dV}{dt}=\pi r^2\dfrac {dh}{dt}$$
Hope, this helps you.