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Question
The radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec. Find the rate of increase of its surface area, when the radius is 7 cm.
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Solution
\[\text { Let r be the radius and S be the surface area of the spherical ball at any time t.Then },\]
\[S = 4\pi r^2 \]
\[ \Rightarrow \frac{dS}{dt} = 8\pi r\frac{dr}{dt}\]
\[ \Rightarrow \frac{dS}{dt} = 8\pi \times 7 \times 0 . 2 \left[ \because r = 7 \text { cm and } \frac{dr}{dt} = 0 . 2 cm/\sec \right]\]
\[ \Rightarrow \frac{dS}{dt} = 11 . 2\pi {cm}^2 /\sec\]
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