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Question
The volume of a sphere is increasing at 3 cubic centimeter per second. Find the rate of increase of the radius, when the radius is 2 cms ?
Solution
\[\text { Let r be the radius and V be the volume of the sphere at any time t.Then },\]
\[V = \frac{4}{3}\pi r^3 \]
\[ \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\]
\[ \Rightarrow \frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV}{dt}\]
\[ \Rightarrow \frac{dr}{dt} = \frac{3}{4\pi \left( 2 \right)^2} \left[ \because r = 2 \text { cm and } \frac{dV}{dt} = 3 {cm}^3 /\sec \right]\]
\[ \Rightarrow \frac{dr}{dt} = \frac{3}{16\pi} \text{cm} /\sec\]
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