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Question

Sound is pouring from a pipe at the rate of 12cm3/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 base. How fast height of the sand cone increasing when the height is 4 cm?

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Solution

Let r be the radius, h be the height and V be the volume of sand cone at any time t.
It is given that dVdt=12cm3/sandh=16rr=6h
V=13πr2h=13π(6h)2h=12πh3
On differentiating w.r.t. t, we get
dVdt=(12π)(3h2dhdt)=36πh2dhdt12=36π(4)2dhdt (h=4 cm anddhdt=12cm3/s)dhdt=1236π×16=148πcm/s
Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of 148πcm/s


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