1. **State the problem:** We have a circular wetted area with area $A$ expanding at a rate of $\frac{dA}{dt} = 4$ mm$^2$/s. We want to find how fast the radius $r$ is expanding, i.e., find $\frac{dr}{dt}$ when $r = 1.49$ mm. 2. **Recall the formula for the area of a circle:** $$ A = \pi r^2 $$ 3. **Differentiate both sides with respect to time $t$:** $$ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} $$ 4. **Solve for $\frac{dr}{dt}$:** $$ \frac{dr}{dt} = \frac{1}{2\pi r} \frac{dA}{dt} $$ 5. **Substitute the known values:** $$ r = 1.49, \quad \frac{dA}{dt} = 4 $$ 6. **Calculate:** $$ \frac{dr}{dt} = \frac{4}{2 \pi \times 1.49} = \frac{4}{2 \times 3.1416 \times 1.49} = \frac{4}{9.363} \approx 0.4274 $$ 7. **Final answer:** The radius is expanding at approximately $0.4274$ mm/s when $r = 1.49$ mm.