related rate question

Question

a leak at an undersea oil well is producing a cylindrical oil slick, which is about one inch thick. when the radius of the slick is 150 feet, the radius is increasing at the rate of 2 feet per minute. at what rate is oil leaking from the wellhead?

Answer 1

Hi Karen.

 

The question asks "at what rate is the oil leaking from the wellhead?" Translation: what is dV/dt?

 

The problem says the slick forms the shape of a cylinder. The volume of a cylinder is:

 

V = pi*r²*h

 

In order to get dV/dt, implicitly differentiate both sides of the equation with d/dt. Don't forget the product rule is in place because r and h are both variables.

 

dV/dt = pi*(2r*dr/dt*h + r²*dh/dt)

 

dV/dt is the rate of change of the volume over time; it's what we're looking for.

 

r is the radius of the oil slick. You're interested in figuring out dV/dt when r is 150 feet. so r = 150.

 

h is the height of the oil slick. The problem claims it's 1 inch, but since we're dealing in units of feet, we need to say it's 1/12 ft.

 

dr/dt is the rate of change of the radius of the oil slick. The problem says it's increasing at a rate of 2 ft/ min, so dr/dt = 2.

 

dh/dt is the rate of change of the height of the oil slick. Since this never changes (the oil slick stays 1 inch thick throughout the expansion) dh/dt = 0.

 

Let's plug all these values in and see how fast the oil is flooding out.

 

dV/dt = pi*(2*150*2*1/12 + (150)²*0)

 

dV/dt = pi*(600/12 + 0)

 

dV/dt = 50pi cubic feet/ min

 

Hope this helps.