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...
 A: $$ \frac{dC}{dt}=\frac{dA}{dt}\frac{dC}{dA} $$
$$C=2\pi r= 2\frac{\pi r^2}{r}=2\frac{A}{r} $$ (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 
$$\frac{dC}{dt}=4\sqrt{\frac{\pi}{12}}\approx2.0466 \ m/s$$
A: Start from the equation $$\frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt}$$ using $A(r) = \pi r^2$. We get $dA/dr = 2 \pi r$ and $$4 = 2 \pi r \frac{dr}{dt} \text{ or } r \frac{dr}{dt} = \frac{2}{\pi}.$$ Write $$r \, dr = \frac{2}{\pi} \, dt$$ and integrate, obtaining $$\int_0^r \rho \, d \rho = \frac{r^2}{2} = \int_0^t \frac{2}{\pi} \tau \, d\tau = \frac{2}{\pi} t.$$ Writing $r$ in terms of $t$ we have $$r = 2 \sqrt{\frac{t}{\pi}}.$$ Now differentiating we obtain $$\frac{dr}{dt} = \frac{1}{\sqrt{\pi t}}$$ and substituting into $$\frac{dC}{dt} = \frac{dC}{dr} \frac{dr}{dt}$$ we get $$\frac{dC}{dt} = 2 \pi \frac{1}{\sqrt{\pi t}} = 2 \sqrt{\frac{\pi}{t}}.$$ Taking $t=3$ it results $$\frac{dC}{dt} \approx 2,046 \, \frac{\text{m}}{\text{s}}.$$ EDIT: Perhaps a (non-explicit integrating) way to see it is notice the following: $$\frac{d}{dt} (r(t))^2 = 2 r \frac{dr}{dt} \text{ and } \frac{d}{dt} \left( \frac{2}{\pi} t \right) = \frac{2}{\pi}.$$ Therefore this implies that $$\frac{1}{2} \frac{d}{dt} (r(t))^2 = \frac{d}{dt} \left( \frac{2}{\pi} t \right).$$ Since $r(0) = 0$ you can assert that $$\frac{1}{2} r^2 = \frac{2}{\pi} t$$ and the solution goes as above.
A: $$\begin{align} & \cfrac{dA}{dr} = 2 \pi r  \\ & \cfrac{dC}{dt} = 2\pi r \\ & \cfrac{dA}{dt} = \cfrac{dA}{dr} \times \cfrac{dr}{dt} \\ & 4 = 2 \pi r \times \cfrac{dr}{dt} \\ & \cfrac{dr}{dt} = \cfrac{4}{2\pi r} \\ & \cfrac{dr}{dt} = \cfrac{2} {\pi r} \\ & r dr = \cfrac{2}{\pi } dt \\ & \int r dr = \int \cfrac{2}{\pi } dt \\ & \cfrac{r^2}{2} = \cfrac{2}{\pi} \times t \\ & r^2 = \cfrac{4}{\pi} \times t \\  & \text{Since t = 3 seconds} \ r^2 = \cfrac{4}{\pi} \times 3 = \cfrac{12}{\pi} \\ & r = \sqrt{\cfrac{12}{\pi} } \\ & \text{Since :} \ \cfrac{dC}{dt} = \cfrac{dC}{dr} \times \cfrac{dr}{dt} = 2 \pi \times \cfrac{2}{\pi r} \\ & \cfrac{dC}{dt} = \require{cancel}{2 \cancel{\pi} \times \cfrac{2}{\cancel{\pi}r}} \\  & \text{Put the value of r } \ \cfrac{dC}{dt} = \cfrac{4}{r} = \cfrac{4}{\sqrt{\cfrac{12}{\pi}}} \\ & \cfrac{dC}{dt} = \cfrac{4 \times \sqrt{\pi}}{2\sqrt{3}} \end{align} $$
This is approximately equal to : $2.046 m/s $
