Philip P. answered • 04/13/14

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*An oil slick on a lake is surrounded by a floating circular containment boom. As the boom is pulled in, the circular containment area shrinks. If the radius of the area decreases at a constant rate of 5 m/min, at what rate is the containment area shrinking when the containment area has a diameter of 60 m? please give exact answer*

The area of a circle is:

A = (pi)r

^{2}The rate at which the area decreases as the radius decreases is the first derivative of A wrt time (t):

A' = dA/dt = (pi)*(dr

^{2}/dt)Use the chain rule to evaluate d(r

^{2})/dt = (2r)*(dr/dt) = (2r)(-5) = -5d where d= the diameter of the containment area and dr/dt = -5, the rate that the radius is decreasingA' = (-5)(pi)(d)

When the containment area has a diameter of d = 60m

**A' =**(-5)(pi)(60) =

**-300(pi)**m

^{2}/min ≅ -942.48 m

^{2}/min