A bicyclist leaves from her home at 9 A.M. and rides to a beach \(40 \mathrm{mi}\) away. Because of a breeze off the ocean, the temperature at the beach remains \(60^{\circ} \mathrm{F}\) throughout the day. At the cyclist's home the temperature increases linearly with time, going from \(60^{\circ} \mathrm{F}\) at 9 A.M. to \(80^{\circ} \mathrm{F}\) by 1 P.M. The temperature is assumed to vary linearly as a function of position between the cyclist's home and the beach. Determine the rate of change of temperature observed by the cyclist for the following conditions: (a) as she pedals 10 mph through a town 10 mi from her home at 10 A.M.; (b) as she eats lunch at a rest stop \(30 \mathrm{mi}\) from her home at noon; (c) as she arrives enthusiastically at the beach at 1 P.M., pedaling 20 mph.

First, express the temperature changes over time and space as linear functions. Since the temperature at home increases linearly, from 60°F at 9 A.M. to 80°F by 1 P.M., the rate of change is \((80 - 60) / (1 - 9) = 20/4 = 5 \)°F/hour. The temperature at the beach remains constant at 60°F. As for the spatial temperature change, since temperature varies linearly between the cyclist's home and the beach, the rate of change is \((60 - 60) / (0 - 40) = 0°F/mile.

Condition (a): The cyclist is 10 miles away from home at 10 A.M.. At 10 A.M., the temperature at home is 60°F + 1*5 = 65°F. Since the temperature varies linearly in space, the temperature at the cyclist's location is 65 - 10*0 = 65°F. Because the cyclist pedals at 10 mph, the rate of change of temperature observed by the cyclist is 10*mile/hour * 0°F/mile = 0°F/hour. Condition (b): The cyclist is 30 miles away from home at noon. At noon, temperature at home is 60°F + 3*5 = 75°F. The temperature at the cyclist's location is 75 - 30*0 = 75°F. Since the cyclist is stationary, the rate of change of temperature observed by the cyclist is 0 mph * 0°F/mile = 0°F/hour. Condition (c): The cyclist arrives at the beach at 1 P.M.. At 1 P.M., the temperature at home is 60°F + 4*5 = 80°F. The temperature at the cyclist's location is 60°F. Because the cyclist pedals at 20 mph, the rate of change of temperature observed by the cyclist is 20*mile/hour * 0°F/mile = 0°F/hour.