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Question
The diameter of a circle is increasing at the rate of 1 cm/sec. When its radius is π, the rate of increase of its area is
Options
π cm2/sec
2π cm2/sec
π2 cm2/sec
2π2 cm2/sec2
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Solution
π2 cm2/sec
\[\text { Let D be the diameter and A be the area of the circle at any time t. Then },\]
\[A = \pi r^2 \left( \text { where r is the radius of the cicle } \right)\]
\[ \Rightarrow A=\pi\frac{D^2}{4}\left[ \because r = \frac{D}{2} \right]\]
\[ \Rightarrow \frac{dA}{dt} = 2\pi\frac{D}{4}\frac{dD}{dt}\]
\[ \Rightarrow \frac{dA}{dt} = \frac{\pi}{2} \times 2\pi \times 1 \left[ \because \frac{dD}{dt} = 1 cm/\sec \right]\]
\[ \Rightarrow \frac{dA}{dt} = \pi^2 {cm}^2 /\sec\]
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