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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

1 Answer

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Final answer:

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.

User Aleksander Aleksic
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