An ocean oil well is leaking oil into the ocean at a rate of 1 m3/hr resulting in an oil slick in the form of a circular
disk (like a quarter or dime). If the slick remains in the shape of a disk with a constant thickness of 1/100 m as the disk expands, determine the rate at which the radius of the slick increases when the radius is 10 meters
To find the rate at which the radius of the slick increases, we can related the rate of change of the slick's volume to the rate of change of its radius. We can do this by using the formula for the volume of a cylinder:
V = πr^2h
where V is the volume, r is the radius, and h is the height (thickness) of the slick.
Since we are given that the slick has a constant thickness of 1/100 m, we can substitute h = 1/100 into the formula:
V = πr^2(1/100)
The rate of change of the volume with respect to time (dv/dt) is equal to the rate at which oil is leaking into the ocean:
dv/dt = 1 m^3/hr
To find the rate at which the radius is increasing, we need to find dr/dt. We can relate dv/dt to dr/dt using the chain rule:
dv/dt = d/dt(πr^2(1/100))
= π * 2r * dr/dt * (1/100)
Since we want to find dr/dt when the radius is 10 meters, we can substitute r = 10 into the equation:
1 m^3/hr = π * 2(10) * dr/dt * (1/100)
Simplifying the equation:
1 = 20π * dr/dt * (1/100)
dr/dt = 1 / (20π * (1/100))
dr/dt = 1 / (2π)
Therefore, the rate at which the radius of the slick increases when the radius is 10 meters is 1/(2π) meters per hour.