The radius $r$ of a circle is increasing at a rate of $2$ meters per minute. Find the rate of change, in $\text{meters}^2/\text{minute}$, of the area when $r$ is $6$ meters.
1. Given that the radius \( r \) of a circle is increasing at a rate of \( \frac{dr}{dt} = 2 \) meters per minute.
2. Let \( A \) be the area of the circle. The area of a circle is given by the formula:
   \[
   A = \pi r^2
   \]
3. To find the rate of change of the area with respect to time, we need to differentiate \( A \) with respect to \( t \). Using the chain rule, we have:
   \[
   \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}
   \]
4. First, compute \( \frac{dA}{dr} \):
   \[
   \frac{dA}{dr} = \frac{d}{dr} (\pi r^2) = 2 \pi r
   \]
5. Now, substitute \( \frac{dA}{dr} \) and \( \frac{dr}{dt} \) into the chain rule equation:
   \[
   \frac{dA}{dt} = 2 \pi r \cdot \frac{dr}{dt}
   \]
6. Given \( \frac{dr}{dt} = 2 \), we have:
   \[
   \frac{dA}{dt} = 2 \pi r \cdot 2 = 4 \pi r
   \]
7. Substitute \( r = 6 \) meters into the equation:
   \[
   \frac{dA}{dt} = 4 \pi \cdot 6 = 24 \pi
   \]

The final answer is \( \boxed{24\pi} \) meters\(^2\)/minute.