<s>[INST] Growth of a circular ink blot grows at the rate of $2\text{ cm}^2$ per second A circular ink blot grows at the rate of $2\text{ cm}^2$ per second. Find the rate at which the radius is increasing after $2\frac{6}{{11}}$  seconds. Use $\pi  = \frac{{22}}{7}$.
My solution is as follow let $A = \pi {r^2}$ where $A$ is the volume
$\frac{{dA}}{{dt}} = 2\pi r\frac{{dr}}{{dt}}$, given $\frac{{dA}}{{dt}} = 2\text{ cm}^2/\sec $ and $r = 2\frac{6}{{11}}\text{ cm} = \frac{{28}}{{11}}\text{ cm}$. Solving we get $2\text{ cm}^2/\sec  = 2 \times \frac{{22}}{7} \times \frac{{28}}{{11}}\frac{{dr}}{{dt}} \Rightarrow \frac{{dr}}{{dt}} = \frac{1}{8}\text{ cm}/\sec $.
But the correct answer is $\frac{1}{4}\text{ cm}/\sec$.
Where am I commiting mistake?
 [/INST] Area after ${28 \over 11} \mbox{sec} = 2 \times {28 \over 11} = \pi r^2$
Simplifying,
$$
r^2 = {56 \over 11 \pi}
$$
Thus,
$$
r = \sqrt{ {56 \over 11 \pi}}
$$
Since $A = \pi r^2$, we have
$$
{dA \over dt} = 2 \pi r {dr \over dt}
$$
Thus,
$$
2 = 2 \pi \sqrt{ {56 \over 11 \pi}} {dr \over dt}
$$
Thus,
$$
1 = \sqrt{56 \pi \over 11} \ {dr \over dt}
$$
or
$$
{dr \over dt} = \sqrt{11 \over  56 \pi}
$$
Using the approximation $\pi \approx {22 \over 7}$, we get
$$
{11 \over 56 \pi} = {11 \over 56 \times {22 \over 7}} = {11 \over 56} \times {7 \over 22} = {1 \over 16}
$$
Hence,
$$
{dr \over dt} = \sqrt{1 \over 16} = {1 \over 4}
$$
</s>