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December 17, 2014

December 17, 2014

Posted by **Anonymous** on Monday, February 25, 2008 at 9:06pm.

We assume that an oil spill is being cleaned up by deploying bacteria that consume the oil at 3 cubic feet per hour. The oil spill itself is modeled in the form of a very thin cylinder as depicted below. Its height is the thickness of the oil slick

Suppose at some moment in time the height is decreasing at .0003 cubic foot per hour, the thickness of the slick is .004 foot, and the cylinder is 600 feet in diameter. At what rate is the area covered by the slick changing at that moment? (That is, the area of the base disc of the cylinder).

- Calc -
**Reiny**, Tuesday, February 26, 2008 at 12:16amV = Pi(r^2)(h)

dV/dt = pi(r^2)dh/dt + pi(h)(2r)dr/dt

sub in the known values

-3 = pi(300^2)(-.0003) + pi(.004)(600)dr/dt

solve for dr/dt

now in A = pi(r^2)

dA/dt = 2pi(r)dr/dt

= 2pi(300)(above value for dr/dt)

=

....

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