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...

$$\begin{align} & \cfrac{dA}{dr} = 2 \pi r  \\ & \cfrac{dC}{dt} = 2\pi r \\ & \cfrac{dA}{dt} = \cfrac{dA}{dr} \times \cfrac{dr}{dt} \\ & 4 = 2 \pi r \times \cfrac{dr}{dt} \\ & \cfrac{dr}{dt} = \cfrac{4}{2\pi r} \\ & \cfrac{dr}{dt} = \cfrac{2} {\pi r} \\ & r dr = \cfrac{2}{\pi } dt \\ & \int r dr = \int \cfrac{2}{\pi } dt \\ & \cfrac{r^2}{2} = \cfrac{2}{\pi} \times t \\ & r^2 = \cfrac{4}{\pi} \times t \\  & \text{Since t = 3 seconds} \ r^2 = \cfrac{4}{\pi} \times 3 = \cfrac{12}{\pi} \\ & r = \sqrt{\cfrac{12}{\pi} } \\ & \text{Since :} \ \cfrac{dC}{dt} = \cfrac{dC}{dr} \times \cfrac{dr}{dt} = 2 \pi \times \cfrac{2}{\pi r} \\ & \cfrac{dC}{dt} = \require{cancel}{2 \cancel{\pi} \times \cfrac{2}{\cancel{\pi}r}} \\  & \text{Put the value of r } \ \cfrac{dC}{dt} = \cfrac{4}{r} = \cfrac{4}{\sqrt{\cfrac{12}{\pi}}} \\ & \cfrac{dC}{dt} = \cfrac{4 \times \sqrt{\pi}}{2\sqrt{3}} \end{align} $$
This is approximately equal to : $2.046 m/s $
