The diameter of a circle is increasing at the rate of 1 cm/sec. When its radius is $\pi$, the rate of increase of its area is

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KollegeApply · Accepted Answer

$\pi$2 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=$\pi$D^$\frac{2}{4}$\left[ \because r = D/2 \right] \Rightarrow dA/dt = 2$\pi$D/4dD/dt \Rightarrow dA/dt = $\pi$/2 $\times$ 2$\pi$ $\times$ 1 \left[ \because dD/dt = 1 cm/\sec \right] \Rightarrow dA/dt = $\pi$^2 {cm}^2 /\sec

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