Answer
(a) $S'(T)$ is the rate at which the swimming speed changes at the given temperature T. The units are $\frac{(cm/s)}{^{\circ}C}$
(b) $S'(15) \approx 0.5~\frac{(cm/s)}{^{\circ}C}$
When the temperature is $15^{\circ}C$, the swimming speed is changing at a rate of approximately $0.5~\frac{(cm/s)}{^{\circ}C}$
$S'(25) \approx -1.0~\frac{(cm/s)}{^{\circ}C}$
When the temperature is $25^{\circ}C$, the swimming speed is changing at a rate of approximately $-1.0~\frac{(cm/s)}{^{\circ}C}$
(Note: other estimates are possible.)
Work Step by Step
(a) $S'(T)$ is the rate at which the swimming speed changes at the given temperature T. The units are $\frac{(cm/s)}{^{\circ}C}$
(b) On the graph, we can see that the slope at $T = 15^{\circ}F$ is approximately equal to $\frac{\Delta S}{\Delta T} = \frac{10~cm/s}{20^{\circ}C} = 0.5~\frac{(cm/s)}{^{\circ}C}$
$S'(15) \approx 0.5~\frac{(cm/s)}{^{\circ}C}$
When the temperature is $15^{\circ}C$, the swimming speed is changing at a rate of approximately $0.5~\frac{(cm/s)}{^{\circ}C}$
On the graph, we can see that the slope at $T = 25^{\circ}F$ is approximately equal to $\frac{\Delta S}{\Delta T} = \frac{-20~cm/s}{20^{\circ}C} = -1.0~\frac{(cm/s)}{^{\circ}C}$
$S'(25) \approx -1.0~\frac{(cm/s)}{^{\circ}C}$
When the temperature is $25^{\circ}C$, the swimming speed is changing at a rate of approximately $-1.0~\frac{(cm/s)}{^{\circ}C}$
