#### 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

π cm

^{2}/sec2π cm

^{2}/secπ

^{2}cm^{2}/sec2π

^{2}cm^{2}/sec^{2}

#### Solution

π^{2} cm^{2}/sec

\[\text { Let D be the diameter and A be the area of the circle at any time t. Then },\]

\[A = \pi r^2 \left( \text { where r is the radius of the cicle } \right)\]

\[ \Rightarrow A=\pi\frac{D^2}{4}\left[ \because r = \frac{D}{2} \right]\]

\[ \Rightarrow \frac{dA}{dt} = 2\pi\frac{D}{4}\frac{dD}{dt}\]

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

\[ \Rightarrow \frac{dA}{dt} = \pi^2 {cm}^2 /\sec\]

Is there an error in this question or solution?

Solution 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 (A) π Cm2/Sec Concept: Rate of Change of Bodies Or Quantities.