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Question

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?

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Solution

Let r be the radius and h be the height of the cone

Given:

h=r6

Volume of cone,

V=13πr2h

V=13π(6h)2h [h=r6]

V=12πh3

Given,

dVdt=12 cm3/s

ddt(12πh3)=12

12πddt(h3)=12

ddt(h3)=1π

3h2dhdt=1π

dhdt=13πh2

When

h=4 cm

dhdt=13π(4)2

dhdt=148π cm/s

Therefore, the height of the sand cone is increasing at the rate of

148π cm/sec

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