The area A of a circle is increasing at a constant rate of 1.5 cm\(^2s^{-1}\)....

Explanation

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)