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We are given that the area of the circle is increasing at a rate of $2~cm^{2}/sec$. This can be written as $\frac{dA}{dt} = 2$. We are also given that the radius $r = 5$ cm. We need to find how fast the circumference is increasing, which is $\frac{dC}{dt}$.
The area of a circle is given by $A = \pi r^2$. Differentiating both sides with respect to time $t$, we get: $$ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt} $$
We know that $\frac{dA}{dt} = 2$. Substituting this into the equation from Step 2, we have: $$ 2 = 2\pi r \frac{dr}{dt} $$ $$ \frac{dr}{dt} = \frac{2}{2\pi r} = \frac{1}{\pi r} $$ When $r = 5$, we have: $$ \frac{dr}{dt} = \frac{1}{5\pi} $$
The circumference of a circle is given by $C = 2\pi r$. Differentiating both sides with respect to time $t$, we get: $$ \frac{dC}{dt} = \frac{d}{dt}(2\pi r) = 2\pi \frac{dr}{dt} $$
We found that $\frac{dr}{dt} = \frac{1}{5\pi}$ when $r = 5$. Substituting this into the equation from Step 4, we have: $$ \frac{dC}{dt} = 2\pi \left(\frac{1}{5\pi}\right) = \frac{2}{5} $$
The circumference of the circle is increasing at a rate of $\frac{2}{5}~cm/sec$.
Final Answer: $\frac{2}{5}$ cm/sec
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