8. A right circular cone with a height of 12 inches and a base radius of 3 inches is filled with water and held with its vertex pointing downward. Water flows out through a hole at the vertex at a rate in cubic inches per second numerically equal to the height of the water in the cone. (For example, when the height of the water in the cone is 4 inches, water flows out at a rate of 4 cubic inches per second.) Determine how many seconds it will take for all of the water to flow out of the cone.
Solution:
$$
9 \pi / 2
$$

When the water in the cone is $h$ inches high, it forms a cone similar to the original, so that its base has radius $h / 4$ and its volume is hence $\pi h^{3} / 48$. The given condition then states that
$$
\frac{d}{d t}\left(\frac{\pi h^{3}}{48}\right)=-h \Rightarrow \frac{\pi h^{2}}{16} \cdot \frac{d h}{d t}=-h \Rightarrow 2 h \cdot \frac{d h}{d t}=-\frac{32}{\pi} .
$$

Integrating with respect to $t$, we get that $h^{2}=-32 t / \pi+C$; setting $t=0, h=12$, we get $C=144$. The cone empties when $h=0$, so $0=-32 t / \pi+144 \Rightarrow t=9 \pi / 2$.