To find how fast the area of the circle is increasing with respect to time, we can use the formula for the area of a circle and apply the concept of related rates.
The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]
where \( r \) is the radius of the circle.
Given that the radius \( r \) is increasing at a rate of \( \frac{dr}{dt} = 2 \) cm/s, we want to find \( \frac{dA}{dt} \) when \( r = \frac{1}{2} \) cm.
To find \( \frac{dA}{dt} \), we will differentiate the area function with respect to time \( t \):
\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \pi \cdot \frac{d}{dt}(r^2) \]
Using the chain rule, we differentiate \( r^2 \):
\[ \frac{d}{dt}(r^2) = 2r \cdot \frac{dr}{dt} \]
Substituting this back into our equation for \( \frac{dA}{dt} \):
\[ \frac{dA}{dt} = \pi \cdot 2r \cdot \frac{dr}{dt} = 2\pi r \cdot \frac{dr}{dt} \]
Now we can substitute \( r = \frac{1}{2} \) cm and \( \frac{dr}{dt} = 2 \) cm/s into the equation:
\[ \frac{dA}{dt} = 2\pi \left(\frac{1}{2}\right) \cdot 2 \]
Calculating this gives:
\[ \frac{dA}{dt} = 2\pi \cdot \frac{1}{2} \cdot 2 = 2\pi \text{ cm}^2/\text{s} \]
Thus, the area of the circle is increasing at a rate of \( \boxed{2\pi} \) cm²/s when the radius is \( \frac{1}{2} \) cm.