Question 14

# Sand is pouring from a pipe at the rate of 12 cm3/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?

The volume of a cone (V) with radius (r) and height (h) is given by,

\begin{align}V=\frac{1}{3}\pi r^2h\end{align}

It is given that,

\begin{align}h=\frac{1}{6} r\Rightarrow r =6h\end{align}

\begin{align}\therefore V=\frac{1}{3}\pi (6h)^2.h = 12\pi h^3\end{align}

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

\begin{align} \frac{dV}{dt}=12 \pi \frac{d}{dh}(h^3).\frac{dh}{dt}[By\; Chain\; Rule]\end{align}

\begin{align}=12 \pi (3h^2).\frac{dh}{dt}\end{align}

\begin{align}=36 \pi h^2.\frac{dh}{dt}\end{align}

It is also given that

\begin{align}\frac{dV}{dt}=12\;cm^3/s \end{align}

Therefore, when h = 4 cm, we have:

\begin{align}12=36\pi (4)^2.\frac{dh}{dt}\end{align}

\begin{align}\Rightarrow \frac{dh}{dt}=\frac{12}{36\pi (16)}=\frac{1}{48\pi}\end{align}

Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of

\begin{align}\frac{1}{48\pi}.\end{align}