Question

Sound is pouring from a pipe at the rate of $12c{m}^3/s$. 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 base. How fast height of the sand cone increasing when the height is 4 cm?

Answer 1

Let r be the radius, h be the height and V be the volume of sand cone at any time t.
It is given that $\frac{dV}{dt}=12c{m}^3/sandh=\frac{1}{6}r\Rightarrow r=6h$
$\therefore V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (6h{)}^{2}h=12\pi h^3$
On differentiating w.r.t. t, we get
$\frac{dV}{dt}=\left(12\pi \right)\left(3h^2\frac{dh}{dt}\right)=36\pi h^2\frac{dh}{dt}\Rightarrow 12=36\pi (4{)}^2\frac{dh}{dt}(\because h=4cmand\frac{dh}{dt}=12c{m}^3/s)\Rightarrow \frac{dh}{dt}=\frac{12}{36\pi \times 16}=\frac{1}{48\pi }cm/s$
Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of $\frac{1}{48\pi }$cm/s