If (A) is the area of a circle of radius \( \underline{\underline{(\mathrm{r}}) \text {, the radius changes at constant rate, then }} \) \( \begin{array}{llll}\text { (a) } \mathrm{A} \text { is constant. } & \text { (b) } \frac{d A}{d t} \text { is constant. } & \text { (C) } \frac{d A}{d t} \propto r & \text { (d) } \frac{d A}{d t} \propto r^{2}\end{array} \)

The area \( A \) of a circle is given by the formula \( A = \pi r^2 \). When the radius \( r \) changes at a constant rate, we can think about how that affects the area. The derivative \( \frac{dA}{dt} \) represents the rate at which the area is changing over time as the radius changes. To derive \( \frac{dA}{dt} \), we can utilize the chain rule: \[ \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 2\pi r \cdot \frac{dr}{dt}. \] Since \( \frac{dr}{dt} \) is constant (as the radius changes at a constant rate), it's clear that \( \frac{dA}{dt} \) is directly proportional to \( r \), not \( r^2 \) or constant. So, the correct choice is (c). Additionally, the area increases as the radius increases, and because of the square relationship, the growth rates diverge more as the radius becomes larger. Be sure to visualize this with a circle growing larger over time; it's like blowing up a balloon where the surface area expands more rapidly as it inflates!