1. **State the problem:** We are given a circle whose radius $r$ increases at a rate of $\frac{dr}{dt} = 4$ m/s. We need to find the rate at which the area $A$ of the circle increases when the radius is $r = 2$ m. 2. **Formula and rules:** The area of a circle is given by the formula: $$A = \pi r^2$$ To find how fast the area changes with respect to time, we differentiate both sides with respect to $t$: $$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt}$$ This uses the chain rule since $r$ is a function of time. 3. **Substitute known values:** Given $r = 2$ m and $\frac{dr}{dt} = 4$ m/s, substitute into the derivative: $$\frac{dA}{dt} = 2\pi (2)(4)$$ 4. **Calculate:** $$\frac{dA}{dt} = 16\pi$$ 5. **Interpretation:** The area of the circle is increasing at a rate of $16\pi$ m$^2$/s when the radius is 2 m. **Final answer:** $$\boxed{16\pi \text{ m}^2/\text{s}}$$