Problem Statement
Sand is pouring from a pipe at the rate of $12\text{ cm}^3/\text{s}$. 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 $4\text{ cm}$?
Verified Solution & Marking Scheme
Establish relation between variables
Given rate of change of volume $\frac{dV}{dt} = 12\text{ cm}^3/\text{s}$.
Height of the cone $h = \frac{1}{6}r \implies r = 6h$.
Express volume solely in terms of height h
Volume of cone $V = \frac{1}{3}\pi r^2 h$.
Substitute $r = 6h$:
$V = \frac{1}{3}\pi (6h)^2 h = \frac{1}{3}\pi (36h^2) h = 12\pi h^3$
Differentiate with respect to time t
$\frac{dV}{dt} = \frac{d}{dt}(12\pi h^3) = 36\pi h^2 \frac{dh}{dt}$
Given $\frac{dV}{dt} = 12$ and $h = 4\text{ cm}$:
$12 = 36\pi (4)^2 \frac{dh}{dt} = 36\pi (16) \frac{dh}{dt} = 576\pi \frac{dh}{dt}$
$\frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi}\text{ cm/s}$