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

The volume of a cone $\left(V\right)$ with radius (r) and height $\left(h\right)$ is given by,
$V=\frac{1}{3}\pi r^{2}h$

It is given that,

$\Rightarrow h=\frac{1}{6}r$

$\Rightarrow r=6h$

$\therefore V=\frac{1}{3}\pi (6h{)}^{2}h=12\pi h^3$

The rate of change of volume with respect to time $\left(t\right)$ is given by,

$\frac{dV}{dt}=12x\frac{d}{dh}\left(h^3\right)\cdot \frac{dh}{dt}=12\pi \left(3h^2\right)\frac{dh}{dt}=36\pi h^2\frac{dh}{dt}$

It is also given that $\frac{dV}{dt}=12c{m}^3/s$

Therefore, when $h=4$ cm, we have:

$12=36\pi (4{)}^2\frac{dh}{dt}$

$\Rightarrow \frac{dh}{dt}=\frac{12}{36\pi \left(16\right)}=\frac{1}{48\pi }$

Hence, when the height of the sand cone is $4cm$, its height is increasing at the rate of $\frac{1}{48\pi }$cm/s