Question

Here's one we haven't worked with before: A circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil is decreasing at the rate of 0.1 cm/hr as the slick spreads. (Note: 1 cm=0.01 m.) At what rate is the radius of the slick increasing when it is 8 meters? (You can think of this oil slick as a very very flat cylinder; its volume is given by V= πr2h, where r is the radius and h is the height of this cylinder.) A. 325.201 centimetres per hour B. 0.804 centimetres per hour C. 1.234 meters per hour D. .556 meters per hour E. 0.804 meters per hour

Answer 1

The rate at which the radius of the oil slick is increasing when it is 8 meters is approximately -0.063 meters per hour.

Step-by-step explanation:

In this problem, we are given that a circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil decreases at the rate of 0.1 cm/hr as the slick spreads.

To find the rate at which the radius of the slick is increasing when it is 8 meters, we can use the formula for the volume of a cylinder: V = πr²h.

Since the thickness of the oil slick is decreasing at a constant rate, we can use the derivative to find the rate at which the volume is changing concerning the radius: dV/dt = 2πrh * dr/dt.

Substituting the given values, we have dV/dt = 2π(8)(0.001) * dr/dt.

We know that DV/dt = -1 (since the volume is decreasing at a rate of 1 cubic meter per hour), so we can solve for dr/dt:

-1 = 2π(8)(0.001) * dr/dt.

Simplifying the equation, we have -1 = 0.016π * dr/dt.

Solving for dr/dt, we get dr/dt = -1 / (0.016π).

Converting the value to meters per hour, we have dr/dt = -0.063 meters per hour.