226
Chapter 3 | Derivatives
v ( t ) − v (19.96)
v ( t ) −337.19 t −19.96
t
t −19.96 =
0.0
16.89
3.05
14.74
5.88
13.46
14.51 8.05
Table 3.3 Average acceleration
The rate at which the car is accelerating is decreasing as its velocity approaches 229.9 mph (337.19 ft/s).
Example 3.9 Rate of Change of Temperature
A homeowner sets the thermostat so that the temperature in the house begins to drop from 70°F at 9 p.m., reaches a low of 60° during the night, and rises back to 70° by 7 a.m. the next morning. Suppose that the temperature in the house is given by T ( t ) =0.4 t 2 −4 t +70 for 0≤ t ≤10, where t is the number of hours past 9 p.m. Find the instantaneous rate of change of the temperature at midnight. Solution Since midnight is 3 hours past 9 p.m., we want to compute T ′(3). Refer to Equation 3.5 . T ′(3) = lim t →3 T ( t )− T (3) t −3 Apply the definition. = lim t →3 0.4 t 2 −4 t +70−61.6 t −3 Substitute T ( t ) =0.4 t 2 −4 t +70and T (3) =61.6. = lim t →3 0.4 t 2 −4 t +8.4 t −3 Simplify. = lim t →3 0.4( t −3)( t −7) t −3 = lim t →3 0.4( t −3)( t −7) t −3 = lim t →3 0.4( t −7) Cancel. =−1.6 Evaluate the limit. The instantaneous rate of change of the temperature at midnight is −1.6°F per hour.
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