The area of the circle is increasing at a uniform rate of $2 \text{ cm}^2/\text{s}$. How fast is the circumference of the circle increasing when the radius $r = 5 \text{ cm}$?

$0.4 \text{ cm/s}$

The correct answer is Option (2) → $0.4 \text{ cm/s}$ ##

Let radius of the circle be $r \text{ cm}$.

Given, $\frac{dA}{dt} = 2 \text{ cm}^2/\text{s}$

Since $A = \pi r^2$

$∴\frac{dA}{dt} = 2\pi r \frac{dr}{dt} \dots (i)$

Also, circumference, $C = 2\pi r$

$∴\frac{dC}{dt} = 2\pi \frac{dr}{dt} \dots (ii)$

From (i), $2 = 2\pi r \frac{dr}{dt}$

$⇒\frac{dr}{dt} = \frac{1}{\pi r}$

Now, substituting the value of $\frac{dr}{dt}$ in eq. (ii) we get

$\frac{dC}{dt} = 2\pi \frac{1}{\pi r} = \frac{2}{r}$

Now, $\left. \frac{dC}{dt} \right|_{r=5} = \frac{2}{5} = 0.4$

Thus, circumference of circle increases at the rate of $0.4 \text{ cm/s}$.