The radius of a circular oil slick expands at a rate of \( 4 \mathrm{~m} / \mathrm{min} \). (a) How fast is the area of the oil slick increasing when the radius is 22 m ? \( \frac{d A}{d t}=\square \mathrm{m}^{2} / \min \) (b) If the radius is 0 at time \( t=0 \), how fast is the area increasing after 3 mins ? \( \frac{d A}{d t}=\square \mathrm{m}^{2} / \min \)

To determine how fast the area of the oil slick is increasing, we need the formula for the area of a circle, which is \( A = \pi r^2 \). By using related rates, we can find \( \frac{dA}{dt} \). (a) The formula for the area is \( A = \pi r^2 \). To find the rate of change of area with respect to time, we differentiate both sides with respect to \( t \): \[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \] Substituting \( r = 22 \, \text{m} \) and \( \frac{dr}{dt} = 4 \, \text{m/min} \): \[ \frac{dA}{dt} = 2\pi(22)(4) = 176\pi \] Calculating \( 176\pi \approx 553.58 \), so: \[ \frac{dA}{dt} \approx 553.58 \, \text{m}^2/\min \] (b) After 3 minutes, the radius will increase by \( 3 \times 4 = 12 \, \text{m} \). Therefore, the radius at this time is: \[ r = 12 \, \text{m} \] Again, using the area expansion formula: \[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi(12)(4) = 96\pi \] Calculating \( 96\pi \approx 301.59 \), so: \[ \frac{dA}{dt} \approx 301.59 \, \text{m}^2/\min \]