Question

Sand is falling on the ground forming a cone at the rate of $15{\text{cm}}^3/\text{s}$. The height of cone is always one-fourth the radius of the base. The rate at which the height of sand-cone is increasing when it's height is $5\text{cm}$ is _________

Answer 1

Let $V$ be the volume, $r$ be radius, and $h$ be height of sand-cone at any time $t$.
Given, $r=4h$ and $\frac{dV}{dt}=15{\text{cm}}^3/\text{s}$
$V=\frac{1}{3}\pi r^{2}h$
$V=\frac{1}{3}\pi (4h{)}^{2}h=V=\frac{16}{3}\pi h^3$
$\Rightarrow \frac{dV}{dt}=16\pi h^2\frac{dh}{dt}$
$\Rightarrow \frac{dh}{dt}=\frac{15}{16\times \pi \times 5^2}=\frac{3}{80\pi }$

$\therefore$ Height of sand-cone is increasing at rate of $\frac{3}{80\pi }\text{cm/s}$.