We assume that an oil spill is being cleaned up by deploying bacteria that consume the oil at 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.

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).

V = 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)
=
....

To find the rate at which the area covered by the slick is changing at that moment, we need to find the rate at which the height of the slick is changing.

Given:
Rate of change of height = -0.0003 cubic feet per hour
Thickness of the slick = 0.004 feet
Diameter of the cylinder = 600 feet

We can calculate the initial volume of the slick using the formula for the volume of a cylinder:

Volume = π * radius^2 * height

Since the slick is very thin, we can assume the radius is equal to half of the diameter. So, the radius (r) in this case is half of 600 feet:

r = 600 feet / 2 = 300 feet

Using the given thickness of the slick, we can calculate the initial height (h) of the slick:

h = 0.004 feet

Therefore, the initial volume (V) of the slick is:

V = π * (300 feet)^2 * 0.004 feet = 1130.973 cubic feet

Next, we need to find the rate at which the volume is changing with respect to time. This can be calculated using the chain rule of calculus:

dV/dt = dV/dh * dh/dt

dV/dh is the rate at which the volume changes with respect to the height of the slick, which can be calculated using the formula for the volume of a cylinder:

dV/dh = π * radius^2

Substituting the given radius of 300 feet:

dV/dh = π * (300 feet)^2

Now, we can find the rate at which the volume changes with respect to time (dV/dt) by multiplying dV/dh by the rate of change of height:

dV/dt = π * (300 feet)^2 * (-0.0003 cubic feet per hour)

Simplifying the equation:

dV/dt = -0.283 cubic feet per hour

Finally, to find the rate at which the area covered by the slick is changing, we can use the formula for the area of the base disc of the cylinder:

A = π * radius^2

Substituting the given radius of 300 feet:

A = π * (300 feet)^2

Therefore, the rate at which the area covered by the slick is changing at that moment is:

A * dV/dt = π * (300 feet)^2 * -0.283 cubic feet per hour

To find the rate at which the area covered by the slick is changing at a given moment, we need to use the formula for the surface area of a cylinder. The formula is:

A = 2πr^2 + 2πrh

Where A is the surface area, r is the radius of the base disc, and h is the height of the cylinder.

Given that the diameter of the cylinder is 600 feet, the radius (r) is half of that, which is 300 feet. The height (h) is given as 0.004 feet.

To find the rate at which the area covered by the slick is changing, we need to take the derivative of the formula with respect to time. This will give us the rate of change of the area (dA/dt).

dA/dt = d/dt(2πr^2 + 2πrh)

To find the rate at the given moment, we need to plug in the values into the equation. We are given that the height is decreasing at a rate of -0.0003 cubic feet per hour. This means that dh/dt = -0.0003 ft^3/hour.

Now, substitute the given values into the equation and solve for dA/dt:

dA/dt = d/dt(2πr^2 + 2πrh)
= 2π(2r(dr/dt) + h(dA/dt))

Substitute the given values:

dA/dt = 2π(2(300)(0)(dr/dt) + 0.004(dA/dt))
= 2π(0 + 0.004(dA/dt))

Now, we can solve for dA/dt:

dA/dt = 2π(0 + 0.004(dA/dt))
dA/dt = 0.008π(dA/dt)

Divide both sides by 0.008π:

dA/dt = (dA/dt) / (0.008π)

So, the rate at which the area covered by the slick is changing at that moment is equal to (dA/dt) / (0.008π).