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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?

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

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