Question

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?

Answer 1

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

Step-by-step 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