Question

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 $4 cm$?

Answer 1

The correct answer is $\frac{1}{48π}cm/s.$
The volume of a cone $(V)$ with radius $(r)$ and height $(h)$ is given by,
$v=\frac{1}{3}πr^2h$
It is given that
$h=\frac{1}{6} r\implies r=6h$
$∴v=\frac{1}{3}π(6h)^2h=12πh^3$
The rate of change of volume with respect to time $(t)$ is given by,
$\frac{dV}{dt}=12π\frac{d}{dh}(h^3).\frac{dh}{dt}$ [By chain rule]
$12π(3h^2) \frac{dh}{dt}$
$=36πh^2 \frac{dh}{dt}$
It is also given that $\frac{dv}{dt}=12 cm^3/s$
Therefore, when $h = 4 cm$, we have:
$12=36π(4)^2 \frac{dh}{dt}$
$=\frac{dh}{dt}=\frac{12}{36π(16)}=\frac{1}{48π}$
Hence, when the height of the sand cone is $4 cm$, its height is increasing at the rate of $\frac{1}{48π}cm/s.$