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. The process continues for 2 hours. 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 1 2 ln ( 4 + 2 t ) = e ln ( 4 + 2 t ) 1 2 = ( 4 + 2 t ) 1 2 Multiply both sides of the equation p ′ ( t ) + p ( t ) 4 + 2 t = 450 by the integrating factor. ( 4 + 2 t ) 1 2 p ′ ( t ) + p ( t ) ( 4 + 2 t ) 1 2 = 450 ( 4 + 2 t ) 1 2 ( ( 4 + 2 t ) 1 2 p ( t ) ) ′ = 450 ( 4 + 2 t ) 1 2

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Chapter 9.3, Problem 72E

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To find: The total number of pollutants in the tank and the rate at which pollutants are being released after 2 hours.

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Chapter 9.3, Problem 71E

Chapter 9.3, Problem 73E

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The total number of pollutants in the tank and the rate at which pollutants are being released after 2 hours.

Solution Summary: The author calculates the total number of pollutants in the tank and the rate at which pollutants are being released after 2 hours.

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To find: The total number of pollutants in the tank and the rate at which pollutants are being released after 2 hours.

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