Question

(8) 8. the radius of a circular oil spill is increasing at a rate of 3 feet per minute. at what rate is the area increasing when the radius of the circle is 20 feet?

Explanation:

Step1: Recall the formula for the area of a circle

The area \( A \) of a circle is given by the formula \( A = \pi r^2 \), where \( r \) is the radius of the circle.

Step2: Differentiate the area with respect to time \( t \)

We use implicit differentiation with respect to \( t \) (time) on both sides of the equation \( A = \pi r^2 \). Using the chain rule, the derivative of \( A \) with respect to \( t \) is \( \frac{dA}{dt} \), and the derivative of \( \pi r^2 \) with respect to \( t \) is \( 2\pi r \frac{dr}{dt} \) (since \( \pi \) is a constant and we differentiate \( r^2 \) with respect to \( t \) using the chain rule: \( \frac{d}{dt}(r^2) = 2r \frac{dr}{dt} \)). So we get:

$$ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} $$

Step3: Identify the given values

We are given that the radius is increasing at a rate of \( 3 \) feet per minute, so \( \frac{dr}{dt} = 3 \) ft/min. We need to find the rate at which the area is increasing when the radius \( r = 20 \) feet.

Step4: Substitute the given values into the derivative formula

Substitute \( r = 20 \) and \( \frac{dr}{dt} = 3 \) into the equation \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \):

$$ \frac{dA}{dt} = 2\pi (20)(3) $$

Step5: Calculate the result

First, multiply \( 2 \times 20 \times 3 \): \( 2 \times 20 = 40 \), then \( 40 \times 3 = 120 \). So we have:

$$ \frac{dA}{dt} = 120\pi $$

If we use \( \pi \approx 3.14 \), then \( 120\pi \approx 120 \times 3.14 = 376.8 \) square feet per minute. But the exact value is \( 120\pi \) square feet per minute.

Answer:

The rate at which the area is increasing is \( 120\pi \) square feet per minute (or approximately \( 376.8 \) square feet per minute).