Sand 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 radius of the base. How fast is the height of the sand cone increasing when the height is 4cm.

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Filo · Accepted Answer

The volume of a cone (V) with radius (r) and height (h) is given by, V=31​πr2hIt is given that,⇒h=61​r⇒r=6h∴V=31​π(6h)2h=12πh3 The rate of change of volume with respect to time (t) is given by, dtdV​=12xdhd​(h3)⋅dtdh​=12π(3h2)dtdh​=36πh2dtdh​It is also given that dtdV​=12cm3/sTherefore, when h=4 cm, we have: 12=36π(4)2dtdh​⇒dtdh​=36π(16)12​=48π1​Hence, when the height of the sand cone is 4cm, its height is increasing at the rate of 48π1​cm/s

Question Text	Sand is pouring from a pipe at the rate of $12 c m^{3} / s$ . 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 $4 c m .$
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Sand is pouring from a pipe at the rate of $12 c m^{3} / s$ . 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 $4 c m .$

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