If a circle's area gets increased at the rate of 2 $cm^2. s^{-1}$ .Then what will be the increase rate of it's radius after $\frac{28}{11}$ seconds?
My Try : $\frac{ds}{dt}= 2\pi r \frac{dr}{dt}$ so we can write $2\pi r \frac{dr}{dt} = 2$. Can anyone help me out with this problem?
Yes your way is correct indeed we have
$$A(t)=\pi r(t)^2\implies A'(t)=2 \pi r(t)r'(t)=2 \implies r'(t)=\frac1{\pi r(t)}$$
and therefore
$$\frac{dr}{dt}=\frac1{\pi r} \implies rdr=\frac 1 \pi dt \implies \frac12r^2=\frac t \pi+C \implies r(t)=\sqrt{\frac{2t}\pi+C}$$
and assuming $r(0)=r_0 \implies C=r_0^2$ that is