Explanation Given: The tank can hold 1,000 gallons of water. Initially the tank contains 200 gallons of water where each gallon of water contains two pounds of pollutants. Additional polluted water containing 3 pounds of pollutants per gallon is pumped into the tank at the rate of 150 gallons of water per hour. The uniformly mixed water in the tank is released into the sewage system at a rate of 50 gallons per hour. Formula used: Integrating factor: The integrating factor of a function f ( x ) is given by I ( x ) = e ∫ f ( x ) d x . Calculation: Let p ( t ) be the total amount of pollutants in the tank t hours after this process begins. Since the tank initially contains 200 gallons of water and each gallon of water contains 2 pound of pollutants, p ( 0 ) = 2 × 200 = 400 Since water containing 3 pounds of pollutants per gallon is entering the tank at a constant rate of 150 gallons per hour, pollutants are entering the tank at a constant rate of 3 × 150 = 450 pounds per hour . Note that 150 gallons of water enter the tank each hour and 50 gallons leave each hour. That means the amount of water in the tank increases at the rate of 100 gallons per hour. Therefore, the amount of water in the tank at time t is 200 + 100 t . The amount of pollutants in each gallon of water at any time t is obtained by dividing the total amount of pollutants by the total amount of water. That is, p ( t ) 200 + 100 t . Since water is leaving the tank at the rate of 50 gallons per hour, the rate at which the pollutants are leaving the tank is 50 p ( t ) 200 + 100 t = p ( t ) 4 + 2 t . So the rate of change of P ( t ) must satisfy the equation p ′ ( t ) = 450 − p ( t ) 4 + 2 t where p ( 0 ) = 400 . That is p ′ ( t ) + p ( t ) 4 + 2 t = 450 . Now find the integrating factor of this first order linear equation where f ( t ) = 1 4 + 2 t and g ( t ) = 450 . I ( t ) = e ∫ 1 4 + 2 t d t = e ln ( 4 + 2 t ) 2 = e ln ( 4 + 2 t ) 1 2 = ( 4 + 2 t ) 1 2 Multiply both sides of the equation by the integrating factor. ( 4 + 2 t ) 1 2 p ′ ( t ) + ( 4 + 2 t ) − 1 2 p ( t ) = 450 ( 4 + 2 t ) 1 2 ( ( 4 + 2 t ) 1 2 p ( t ) ) ′ = 450 ( 4 + 2 t ) 1 2 [ ∵ ( x y ) ′ = x y ′ +

Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

14th Edition

ISBN: 9780134668574

Author: Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen, Christopher J. Stocker

Publisher: PEARSON

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Chapter 9.3, Problem 74E

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To find: The time at which the tank will contain 1,500 pounds of pollutants.

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Chapter 9.3, Problem 73E

Chapter 9.3, Problem 75E

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Chapter 9 Solutions

Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

Ch. 9.1 - Show that y=Cx+1 is the general solution of the...Ch. 9.1 - Prob. 2MP Ch. 9.1 - Prob. 3MP Ch. 9.1 - Prob. 4MP Ch. 9.1 - Prob. 5MP Ch. 9.1 - Prob. 1ED Ch. 9.1 - Prob. 2ED Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E

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The time at which the tank will contain 1,500 pounds of pollutants.

Solution Summary: The author calculates the time at which the tank will contain 1,500 pounds of pollutants after 3.4 hours. The integrating factor of a function f(x) is given.

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To find: The time at which the tank will contain 1,500 pounds of pollutants.

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Chapter 9 Solutions