g A circular oil slick of uniform thickness is caused by a spill of one cubic meter of oil. The thickness of the oil slick is decreasing at a rate of 0.1 centimeters per hour. At what rate is the radius of the slick increasing when the radius is 8 meters?

the rate of increase of radius is dR/dt = 0.804 m/hour = 80.4 cm/hour

Explanation:

the slick of oil can be modelled as a cylinder of radius R and thickness h, therefore the volume V is

V = πR² * h

thus

h = V / (πR²)

Considering that the volume of the slick remains constant, the rate of change of radius will be

dh/dt = V d[1 / (πR²) ]/dt

dh/dt = (V/π) (-2) / R³ * dR/dt

therefore

dR/dt = (-dh/dt) * (R³/2) * (π/V)

where dR/dt = rate of increase of the radius, (-dh/dt) = rate of decrease of thickness

when the radius is R=8 m, dR/dt is

dR/dt = (-dh/dt) * (R³/2) * (π/V) = 0.1 cm/hour * (8m) ³/2 * π/1m³ * (1m/100 cm) = 0.804 m/hour = 80.4 cm/hour