1. **State the problem:** We are given a circle with radius $r$ increasing at a rate of $\frac{dr}{dt} = 5$ cm/s. We need to find the rate at which the area $A$ of the circle is increasing when $r = 3$ cm. 2. **Formula and rules:** The area of a circle is given by the formula: $$ A = \pi r^2 $$ To find the rate of change of the area with respect to time, we differentiate both sides with respect to $t$: $$ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) $$ Using the chain rule: $$ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} $$ 3. **Substitute known values:** Given $r = 3$ cm and $\frac{dr}{dt} = 5$ cm/s, substitute these into the formula: $$ \frac{dA}{dt} = 2 \pi \times 3 \times 5 = 30 \pi $$ 4. **Interpretation:** The area of the circle is increasing at a rate of $30 \pi$ cm$^2$/s when the radius is 3 cm. **Final answer:** $$ \frac{dA}{dt} = 30 \pi \text{ cm}^2/\text{s} $$