Simple Differentiation Problem Involving Area Radius and Circumference A stone is dropped into a pool of water, and the area covered by the spreading ripple increases at a rate of $4 m^2 s^{-1} $.
Calculate the rate at which the circumference of the circle formed is increasing 3 seconds after the stone is dropped.
My method: 
$$ \begin{align} 
& \cfrac{dA}{dt}=4\\
& A = \pi \times r \times r \ \text{so} \ \cfrac{dA}{dr} = 2 \times \pi \times r \\
& C = 2 \times \pi \times r \ \text{so} \ \cfrac{dC}{dr} =2 \times \pi \\
& \cfrac{dA}{dt} = \cfrac{dA}{dr} \times \cfrac{dr}{dt} \\
& 4     =  2 \times \pi \times (3) \cfrac{ dr}{dt} \\
& \text{so} \ \cfrac{dr}{dt} = \cfrac{2}{(3 \times \pi)} \\ 
& \text{now} \ \cfrac{dC}{dt} = \cfrac{dC}{dr} \times \cfrac{dr}{dt} \\ 
& \text{so} \  \cfrac{dC}{dt} = 2 \times \pi  \times  \cfrac{2}{(3 \times \pi)}
\end{align} $$
I'm getting $1.3333...$
But the answer is $2.05m/s$ , I don't understand how to get it...

$$ \frac{dC}{dt}=\frac{dA}{dt}\frac{dC}{dA} $$
$$C=2\pi r= 2\frac{\pi r^2}{r}=2\frac{A}{r} $$ (since $A=\pi r^2 $) and
$$A=\pi r^2 \implies r=\sqrt{\frac{A}{\pi}}$$
Therefore using the previous 2 results $$C=2 \frac{A}{\sqrt{\frac{A}{\pi}}}=2\frac{A}{\sqrt{\frac{A}{\pi}}}\times \frac{\sqrt{\pi}}{\sqrt{\pi}}=2\frac{A\sqrt{\pi}}{\sqrt{A}}=2\sqrt{\pi A}$$
$$\therefore \frac{dC}{dA}=\sqrt{\frac{\pi}{A}}$$
$$\implies \frac{dC}{dt}=\frac{dA}{dt}\sqrt{\frac{\pi}{A}} $$
and since A is changing at a constant rate of $4m^2/s$, $A=4t$.
$$\frac{dC}{dt}=4 \sqrt{\frac{\pi}{4t}}$$
So at 3 seconds 
$$\frac{dC}{dt}=4\sqrt{\frac{\pi}{12}}\approx2.0466 \ m/s$$
