3.A ruptured oil tanker causes a circular oil slick on the surface of the ocean. When its radius is 150 meters, the radius of the slick is expanding at 0.1 m/min, and its thickness is .02 m. At that moment:
a. How fast is the area of the slick expanding?
b. If the circular slick has the same thickness as everywhere, and the volume of oil spilled remains fixed, how fast is the thickness of the slick decreasing?
To answer these questions, we first need to know the formulas for the area of a circle and the volume of a cylinder.
a. How fast is the area of the slick expanding?
The formula for the area of a circle is A = πr², where A represents the area and r represents the radius.
Given that the radius of the slick is expanding at 0.1 m/min, we can differentiate the formula for area with respect to time (t) to find the rate of change of the area with respect to time:
dA/dt = 2πr(dr/dt)
Since we're given that dr/dt = 0.1 m/min and r = 150 meters, we can substitute these values into the equation to calculate the rate of change of the area:
dA/dt = 2π(150)(0.1)
= 2π(15)
= 30π
≈ 94.25 m²/min
Therefore, the area of the slick is expanding at a rate of approximately 94.25 square meters per minute.
b. How fast is the thickness of the slick decreasing?
The volume of a cylinder is given by the formula V = πr²h, where V represents the volume, r represents the radius, and h represents the height (thickness in this case).
Since we are assuming the thickness of the slick is the same everywhere, the volume of oil spilled remains fixed. Therefore, the rate of change of volume with respect to time is zero.
dv/dt = 0
Now let's differentiate the volume formula with respect to time:
dv/dt = d(πr²h)/dt
= 2πrh(dr/dt) + πr²(dh/dt)
Given that dv/dt = 0, we can rearrange the equation as follows:
0 = 2πrh(dr/dt) + πr²(dh/dt)
We are interested in finding the rate at which the thickness of the slick (dh/dt) is decreasing. Rearranging the equation, we get:
πr²(dh/dt) = -2πrh(dr/dt)
(dh/dt) = (-2rh/dr) * (dr/dt)
We know that dr/dt = 0.1 m/min, r = 150 meters, and h = 0.02 meters. Let's substitute these values into the equation to calculate the rate of change of the thickness of the slick:
(dh/dt) = (-2(150)(0.02)/150) * 0.1
= (-0.04) * 0.1
= -0.004 m/min
Therefore, the thickness of the slick is decreasing at a rate of 0.004 meters per minute.