An accident at an oil drilling platform is causing a circular oil slick. The slick is 0.07 foot thick, and when the radius of the slick is 110 feet, the radius is increasing at the rate of 0.9 foot per minute. At what rate (in cubic feet per minute) is oil flowing from the site of the accident? (Round your answer to two decimal places.)

Answer:[tex]43.54 ft^3/min[/tex]Step-by-step explanation:GivenThickness of oil slick=0.07 footradius of slick=110 ft[tex]\frac{\mathrm{d} r}{\mathrm{d} t}=0.9 ft/s[/tex]Let V be the volume of oil slickso, [tex]V=0.07\pi \cdot r^2[/tex]rate of oil flowing is [tex]\frac{\mathrm{d} V}{\mathrm{d} t}=0.07\times 2\pi \cdot r\frac{\mathrm{d} r}{\mathrm{d} t}[/tex][tex]\frac{\mathrm{d} V}{\mathrm{d} t}=0.07\times 2\pi \times 110\times 0.9=43.54 ft^3/min[/tex]