Sand is pouring from a pipe at the rate of $12 \; cm^3 /s$. 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 cm?
Let volume of sand pouring out $=V$
Then, rate at which the sand is pouring out $= \dfrac{dV}{dt}=12 \; cm^3/s$ $\;\; \cdots$ (1)
Let radius of the cone formed $=r$
Then height of the cone $=h=\dfrac{r}{6}$
$\therefore$ Radius of cone formed $=r=6h$ $\;\; \cdots$ (2)
Now, volume of cone formed = volume of sand pouring out $=V$
Volume of cone $=V = \dfrac{1}{3}\pi r^2 h$
Substituting the value of r from equation (2) gives
$V = \dfrac{1}{3} \times \pi \times 36h^3$
$\implies$ $V = 12 \pi h^3$ $\;\; \cdots$ (3)
Differentiating equation (3) w.r.t time t gives
$\dfrac{dV}{dt} = 12\pi \times 3h^2 \dfrac{dh}{dt}$
i.e. $\dfrac{dh}{dt} = \dfrac{1}{36\pi h^2}\dfrac{dV}{dt}$ $\;\; \cdots$ (4)
When $h = 4 \; cm$, in view of equation (1), equation (4) becomes
$\dfrac{dh}{dt} = \dfrac{12}{36 \times \pi \times 4^2}$
i.e. $\dfrac{dh}{dt} = \dfrac{1}{48\pi} \; cm/s$
$\therefore$ The height of the sand cone is increasing at the rate of $\dfrac{1}{48\pi} \; cm/s$