Answer:
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
