Question

Sand 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 radius of the base? How fast is the height of the sand cone increasing when the height is $4cm$?

Answer 1

$h=\frac{1}{6}r,h=4cm$
$\frac{dv}{dt}=12c{m}^3/sec$
$v=\frac{1}{3}\pi r^{2}h$
$v=\frac{1}{3}\pi (6h{)}^{2}h$
$v=12\pi h^3$
$\frac{dv}{dt}=12\pi (3h{)}^2\frac{dh}{dt}$
$\frac{12}{12\pi (3\times 16)}=\frac{dh}{dt}\Rightarrow \frac{dh}{dt}=\frac{1}{48\pi }cm/sec$