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Question
Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2 cm ?
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Solution
Let V be the volume of the sphere. Then,
V = \[\frac{4}{3}\pi r^3\]
\[\Rightarrow \frac{dV}{dr} = 4\pi r^2\]
Let S be the total surface area of sphere. Then,
S = \[4\pi r^2\]
\[\Rightarrow \frac{dS}{dr} = 8\pi r\]
\[\therefore \frac{dV}{dS} = \frac{\frac{dV} {dr}}{\frac{dS}{dr}}\]
\[ \Rightarrow \frac{dV}{dS} = \frac{4\pi r^2}{8\pi r} = \frac{r}{2}\]
\[ \Rightarrow \left( \frac{dV}{dS} \right)_{r = 2} = \frac{2}{2}\]
\[ = 1 cm\]
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