Explanation Given information: r ( t ) be the rate at which the water flowed out of pipe after it’s sprang. Calculation: Let r ( t ) be the rate at which the water flowed out of pipe after it’s sprang. The leak of water started from 10 am and worsened throughout the afternoon reaching 8 gallons per hour before it shutoff. As the rate of flow eventually reaches 8 gallons per hour from 10 am 6 : 40 pm , the rate is less than 8 gallons per hour. That is. r ( t ) < 8 gallon/hour The integral ∫ 0 8 r ( t ) d t represents the total water our flowed during the 8 hours. As the maximum rate of outflow is 8 gallons per hour, during the 8 hours the total outflow is rate of outflow in 8 hours. Total amount of water leaked in total hours is less than the product of rate of 8 gallon per hour to the time elapsed

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Author: DA

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Chapter 5.5, Problem 25P

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Give a reasonable overestimate of ∫08r(t)dt.

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Chapter 5.5, Problem 24P

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Give a reasonable overestimate of ∫ 0 8 r ( t ) d t .

Solution Summary: The author calculates the rate at which the water flowed out of pipe after it’s sprang.

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Give a reasonable overestimate of ∫08r(t)dt.

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Chapter 5 Solutions