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 \(\frac{1}{48π}cm/s.\)