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Find the Surface Area of a Sphere When Its Volume is Changing at the Same Rate as Its Radius

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Question

Find the surface area of a sphere when its volume is changing at the same rate as its radius ?

Sum
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Solution

\[\text{Let r be the radius and V be the volume of the sphere at any time t . Then }, \]

\[V = \frac{4}{3}\pi r^3 \]

\[ \Rightarrow \frac{dV}{dt} = 4\pi r^2 \left( \frac{dr}{dt} \right)\]

\[ \Rightarrow \frac{dV}{dt} = 4\pi r^2 \left( \frac{dV}{dt} \right) \left[ \because \frac{dV}{dt} = \frac{dr}{dt} \right]\]

\[ \Rightarrow 4\pi r^2 = 1 \]

\[ \Rightarrow \text { Surface area of sphere =1 square unit }\]

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Chapter 12: Derivative as a Rate Measurer - Exercise 13.3 [Page 24]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 12 Derivative as a Rate Measurer
Exercise 13.3 | Q 7 | Page 24

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