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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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Question

Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2 cm ?

Sum
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Solution

Let V be the volume of the sphere. Then,

= \[\frac{4}{3}\pi r^3\] 

\[\Rightarrow \frac{dV}{dr} = 4\pi r^2\]

Let S be the total surface area of sphere. Then,

= \[4\pi r^2\]

\[\Rightarrow \frac{dS}{dr} = 8\pi r\]

\[\therefore \frac{dV}{dS} = \frac{\frac{dV} {dr}}{\frac{dS}{dr}}\]

\[ \Rightarrow \frac{dV}{dS} = \frac{4\pi r^2}{8\pi r} = \frac{r}{2}\]

\[ \Rightarrow \left( \frac{dV}{dS} \right)_{r = 2} = \frac{2}{2}\]

\[ = 1 cm\]

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Chapter 12: Derivative as a Rate Measurer - Exercise 13.1 [Page 4]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 12 Derivative as a Rate Measurer
Exercise 13.1 | Q 3 | Page 4

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