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 (a) π cm2/sec (b) 2π cm2/sec (c) π2 cm2/sec (d) 2π2 cm2/sec2

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Solution

## (c) π2 cm2/sec $\text{Let}\mathit{\text{D}}\text{be the diameter and}\mathit{\text{A}}\text{be the area of the circle at any time}\mathit{\text{t}}\text{. Then,}\phantom{\rule{0ex}{0ex}}A=\mathrm{\pi }{r}^{2}\left(\mathrm{where}\mathit{}r\mathrm{is}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{cicle}\right)\phantom{\rule{0ex}{0ex}}⇒\mathit{\text{A}}\text{=π}\frac{{D}^{2}}{4}\text{}\left[\because r=\frac{D}{2}\right]\phantom{\rule{0ex}{0ex}}⇒\frac{dA}{dt}=2\mathrm{\pi }\frac{D}{4}\frac{dD}{dt}\phantom{\rule{0ex}{0ex}}⇒\frac{dA}{dt}=\frac{\mathrm{\pi }}{2}×2\mathrm{\pi }×1\left[\because \frac{dD}{dt}=1\mathrm{cm}/\mathrm{sec}\right]\phantom{\rule{0ex}{0ex}}⇒\frac{dA}{dt}={\mathrm{\pi }}^{2}{\mathrm{cm}}^{2}/\mathrm{sec}$

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