Filling a conical tank

Solution 1:

Think about what is happening. The water (volume) is being poured in at a constant rate. This relates to how the water level ($h$) changes and how the width of the water in the tank at that level ($r$) changes. Further, $r$ and $h$ are related.

The volume of a cone is

$$V = \frac13 \pi \, r^2 \, h$$

How is $h$ related to $r$? You know that, at the top, the radius is $2$ and $h=4$. Because this is a cone, we can say that $r = h/2$ at all levels of the cone. Thus,

$$V(h) = \frac{1}{12} \pi \, h^3$$

We may then differentiate with respect to time; use the chain rule here

$$\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$$

You are given $dV/dt$ and the height $h$ at which to evaluate; solve for $dh/dt$.