The radius of a circle increases at a rate of 3 cm s^{-1}. Find, in terms of $\pi$, the rate of increase of the area when

1. State the formula for the area of a circle and differentiate it with respect to the radius.

Let $A$ be the area and $r$ be the radius of the circle.
$A = \pi r^2$
$\frac{dA}{dr} = 2\pi r$

2. Apply the chain rule to express the rate of change of area with respect to time.

Using the chain rule for connected rates of change:
$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$
Given that $\frac{dr}{dt} = 3$,
$\frac{dA}{dt} = (2\pi r)(3) = 6\pi r$

3. Substitute the given radius to find the specific rate of increase of the area.

When $r = 5$,
$\frac{dA}{dt} = 6\pi (5) = 30\pi$

1. Find the radius when the area is $4\pi\ \text{cm}^2$

$A = \pi r^2$
$4\pi = \pi r^2$
$r^2 = 4$
$r = 2$ (since $r > 0$)

2. Use the chain rule to find the rate of increase of the area

From part (i), $\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} = 2\pi r \times 3 = 6\pi r$
Substitute $r = 2$:
$\frac{dA}{dt} = 6\pi (2)$

3. Calculate the final rate of increase

$\frac{dA}{dt} = 12\pi$