# 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?

Given =  $\frac{dV}{dt} = 12 \ cm^{3}/s$      and    $h = \frac{1}{6}r$
To find =  $\frac{dh}{dt}$   at h = 4 cm
Solution:-

Volume of cone(V) = $\frac{1}{3}\pi r^{2}h$
$\frac{dV}{dt} = \frac{dV}{dh}.\frac{dh}{dt} = \frac{d(\frac{1}{3}\pi (6h)^{2}h)}{dh}.\frac{dh}{dt} = \frac{1}{3}\pi\times36\times3h^{2}.\frac{dh}{dt} = 36\pi \times(4)^{2}.\frac{dh}{dt}$
$\frac{dV}{dt} = 576\pi.\frac{dh}{dt}$
$\frac{dh}{dt} = \frac{\frac{dV}{dh}}{576\pi} = \frac{12}{576\pi} = \frac{1}{48\pi} \ cm/s$

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