Questions: Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 2 m / s, how fast is the area of the spill increasing when the radius is 40 m?

Solution Steps

To solve this problem, we need to use related rates, which involves finding the rate at which one quantity changes with respect to another. Here, we are given the rate of change of the radius of the oil spill and need to find the rate of change of the area. The area \( A \) of a circle is given by the formula \( A = \pi r^2 \). We will differentiate this equation with respect to time \( t \) to find the rate of change of the area \( \frac{dA}{dt} \).

1. Use the formula for the area of a circle \( A = \pi r^2 \).
2. Differentiate both sides with respect to time \( t \) to get \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \).
3. Substitute the given values \( r = 40 \) m and \( \frac{dr}{dt} = 2 \) m/s into the differentiated equation to find \( \frac{dA}{dt} \).

The area \( A \) of a circle is given by the formula: \[ A = \pi r^2 \]

To find how fast the area is changing with respect to time, we differentiate both sides with respect to \( t \): \[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \]

We are given:

• \( r = 40 \, \text{m} \)
• \( \frac{dr}{dt} = 2 \, \text{m/s} \)

Substituting these values into the differentiated equation: \[ \frac{dA}{dt} = 2\pi (40) (2) \]

Calculating the above expression: \[ \frac{dA}{dt} = 2\pi (40) (2) = 160\pi \] Evaluating \( 160\pi \) gives approximately \( 502.6548 \).

Final Answer