一块厚度可忽略的圆形水膜正在蒸发。该水膜的半径 $r$ cm 正以 0.2 cm/s 的恒定速率减小。当水膜面积为 $225\pi$ cm$^2$ 时，求其面积 $A$ cm$^2$ 的变化率。

1. Find the radius of the patch when the area is $225\pi$ cm$^2$.

$A = \pi r^2$
$225\pi = \pi r^2$
$r^2 = 225$
$r = 15$ (since $r > 0$)

2. Differentiate the area formula with respect to $r$.

$\frac{dA}{dr} = 2\pi r$

3. Apply the chain rule to find the rate of change of the area.

$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$
$\frac{dA}{dt} = 2\pi r \times (-0.2)$

4. Substitute $r = 15$ to find the final rate of change.

$\frac{dA}{dt} = 2\pi (15) \times (-0.2)$
$\frac{dA}{dt} = 30\pi \times (-0.2)$
$\frac{dA}{dt} = -6\pi$