A circle has an initial radius of 50 ft when the radius begins decreasing at the rate of \( 5 \mathrm{ft} / \mathrm{min} \). What is the rate in the change of area at the instant that the radius is 20 ft ? The rate of change of the area is (Type an exact answer in terms of \( \pi \).)

To find the rate of change of the area of the circle as the radius decreases, we can use the formula for the area of a circle, \( A = \pi r^2 \). First, we'll differentiate the area with respect to time: \[ \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 2\pi r \cdot \frac{dr}{dt} \] Given that \( \frac{dr}{dt} = -5 \, \text{ft/min} \) (the radius is decreasing) and we need to evaluate this when \( r = 20 \, \text{ft} \), we plug in the values: \[ \frac{dA}{dt} = 2\pi (20) \cdot (-5) = -200\pi \, \text{ft}^2/\text{min} \] So, the rate of change of the area when the radius is 20 ft is \( -200\pi \, \text{ft}^2/\text{min} \).