Sand is pouring from a pipe at the rate of $12 \mathrm{~cm}^{3} / \mathrm{s}$. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is $4 \mathrm{~cm}$ ?
The volume of a cone (V) with radius (r) and height (h) is given by,
$V=\frac{1}{3} \pi r^{2} h$
It is given that,
$h=\frac{1}{6} r \Rightarrow r=6 h$
$\therefore V=\frac{1}{3} \pi(6 h)^{2} h=12 \pi h^{3}$
The rate of change of volume with respect to time (t) is given by,
$\frac{d V}{d t}=12 \pi \frac{d}{d h}\left(h^{3}\right) \cdot \frac{d h}{d t}$ [By chain rule]
$=12 \pi\left(3 h^{2}\right) \frac{d h}{d t}$
$=36 \pi h^{2} \frac{d h}{d t}$
It is also given that $\frac{d V}{d t}=12 \mathrm{~cm}^{3} / \mathrm{s}$.
Therefore, when h = 4 cm, we have:
$12=36 \pi(4)^{2} \frac{d h}{d t}$
$\Rightarrow \frac{d h}{d t}=\frac{12}{36 \pi(16)}=\frac{1}{48 \pi}$
Hence, when the height of the sand cone is $4 \mathrm{~cm}$, its height is increasing at the rate of $\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s}$.