Question

Assume that oil spilled from a ruptured tank spreads in a circular pattern whose radius increases at a constant rate of 2 ft/sec. how fast is the area of the spill increasing when the radius of the spill is 60 feet?

Answer 1

The area of the oil spill is increasing at a rate of 240π ft^2/sec when the radius is 60 feet. This is a related rates problem in calculus, involving differentiation of the area formula for a circle with respect to time.

Step-by-step explanation:

The question involves calculating how fast the area of an oil spill is increasing given that the radius of the spill grows at a constant rate. This is a problem in related rates, a concept in calculus.

Finding the rate of change of the area involves using the formula for the area of a circle, which is A = πr^2, where A represents the area and r represents the radius. Since the radius increases at a rate of 2 ft/sec, we can denote this rate as dr/dt = 2 ft/sec. To find the rate of change of the area (dA/dt), we differentiate both sides of the area formula with respect to time t to get dA/dt = 2πr × dr/dt.

When the radius is 60 feet, we substitute r = 60 ft and dr/dt = 2 ft/sec into the differentiated formula:

dA/dt = 2π × 60 ft/sec × 2 ft/sec = 240π ft^2/sec.

This means that the area of the spill is increasing at a rate of 240π square feet per second when the radius is 60 feet.