# 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?

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, $${dr \over dt} = \sqrt{1 \over 16} = {1 \over 4}$$