The area A of a circle is increasing at a constant rate of 1.5 cm $^2s^{-1}$ . Find, to 3 significant figures, the rate at which the radius r of the circle is increasing when the area of the circle is 2 cm $^2$ .

0.200 cms $^{-1}$

0.798 cms $^{-1}$

0.300 cms $^{-1}$

0.299 cms $^{-1}$

Correct answer is D

Area of a circle (A) = $\pi r^2$

Given

$\frac{dA}{dt} = 1.5cm^2s^{-1}$

$\frac{dr}{dt}$ = ?

A = 2cm $^2$

Now

2 = $\pi r^2$

= r $^2 = \frac {2}{\pi}$

r = $\sqrt \frac {2}{\pi}$ cm = 0.798cm

$\frac {dr}{dt} = \frac {dA}{dt} \times \frac {dr}{dt}$

$\frac {dA}{dr} = 2\pi r$ (differentiating A = $\pi r^2)$

$\frac {dr}{dA} = \frac {1}{2\pi r}$

$\frac {dr}{dt} = 1.5 \times \frac {1}{2\times \pi \times 0.798} = 1.5 \times 0.199$

$\frac {dr}{dt} = 0.299cms^{-1}$ (to 3 s.f)

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