Greenwood press
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book-20600
38
INTEGRATION The area between the graph of f (t) = 6.37e −0.04t and the t axis determined by 2 0 6.37e −0.04t dt. graph is much lower in this interval (the dark solid region), than from 0 to 2 hours of work (the light shaded region). This information can help managers determine when employees should take breaks so that they can optimize their performance, because they would likely feel more productive when they returned to work. A definite integral can help heating and cooling companies estimate the amount of costs needed to send power or gas to each household. On any given day, the temperature can be modeled with a sinusoidal function, because tem- perature increases during the day, decreases at night, and then repeats the cycle throughout the year. For example, suppose the temperature reached a low of 50° Fahrenheit at 2 AM and a high of 90° at 2 PM . If x represents the number of hours that have passed during the day, then the temperature in degrees Fahrenheit, T , can be represented by the equation T = 20 cos 2π(x−14) 24 + 70. Suppose that the thermostat in the house is set to 80° so that the air conditioning will turn on once the temperature is greater than or equal to that setting. The amount of energy used for the air conditioner is proportional to the temperature outside. That means that the air conditioner will use more energy to keep the house cool when it is closer to 90° than when it is near 80°. The price to cool the house might be five cents per hour for every degree above 80°. If the temperature were 83° for the entire hour, then the cost to run the air conditioner would be fifteen cents. However, since temperature varies according to a sinusoidal function, the cost per minute would actually change. Therefore, a definite integral bounded by the time when the temperature is above 80° will help predict the cooling costs. The temperature should be 80° at x = 10 (10 AM ) and x = 18 (6 PM ), so the cool- ing costs per day for days like this would be approximately $2.62 based on an evaluation of the expression $0.05 18 10 (20 cos 2π(x−14) 24 + 70 − 80)dx = $2.62. Notice that the answer is also represented by 0.05 times the area of the curve between T = 20 cos 2π(x−14) 24 + 70 and T = 80, as shown in the following diagram. Download 1.81 Mb. Do'stlaringiz bilan baham: |
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