To solve this problem, let's use related rates, given that the radius $r$ is increasing at a rate of $\frac{dr}{dt} = 4$ ft/min. We need to find the rate at which the area $A$ is changing, denoted $\frac{dA}{dt}$, when $r = 2$ ft.

1. Formula for the area of a circle: $A = \pi r^2$

2. Differentiate both sides with respect to time $t$: Using the chain rule, we get: $\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}$

3. Substitute the given values: We know $r = 2$ ft and $\frac{dr}{dt} = 4$ ft/min. Substitute these values into the differentiated formula: $\frac{dA}{dt} = 2\pi (2) \cdot 4$

4. Calculate: $\frac{dA}{dt} = 2\pi \cdot 2 \cdot 4 = 16\pi$

Thus, the rate at which the area is changing when the radius is 2 ft is: $\frac{dA}{dt} = 16\pi \ \text{ft}^2/\text{min}$

Summary:

The area of the circle is increasing at a rate of $16\pi$ square feet per minute when the radius is 2 feet.

Would you like further clarification on any of these steps?

Related Questions:

1. What would the rate of area change be if the radius were increasing at a different rate?
2. How does changing the rate of $\frac{dr}{dt}$ affect the rate of area change?
3. What is the rate of area change when the radius reaches 5 feet?
4. How does this rate of area change compare when using different radii?
5. What would the rate of area change be if the radius were halved?

Tip:

In related rates problems, always identify the variables changing with time and set up relationships using differentiation before plugging in values.