(a) Explanation Given: A tank initially contains 5 lb of salt in 50 ft 3 of solution. Brine begins to enter the tank at the rate of 2 ft 3 / h , C ( t ) = 1 − e − 0.2 t lb / ft 3 Calculation: Let S ( t ) denotes amount of salt in the tank at the end of t min. Because 2 f i 3 / h of brine flows and brine has concentration C ( t ) = 1 − e − a z [ b / f 3 ] so, inflow rate is 2 ( 1 − e − 2 z ) | b / h At time, t there is S ( t ) amount of salt. So, outflow rate is S ( t ) 25 Thus, net rate of change of salt is d S d t = 2 ( 1 − e − 0.2 t ) − S 25 d S d t + S 25 = 2 ( 1 − e − 0.2 t ) P ( t ) = 1 25 and Q ( t ) = 2 ( 1 − e − 2 t ) So, I ( t ) = e ∫ P ( t ) d t = e ∫ 1 25 d t = e t 25 = e 0 (b) To determine To find: the total amount of salt in the tank (c) To determine To find: the maximum amount of salt in the solution

Student's Solution and Survival Manual for Calculus

7th Edition

ISBN: 9781524934040

Author: STRAUSS MONTY J, TODA MAGDALENA DANIELE, SMITH KARL J

Publisher: Kendall Hunt Publishing

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Chapter 7.6, Problem 46PS

(a)

To determine

To setup and solve: the differential equation for the amount of salt in the solution at time t

(b)

To determine

To find: the total amount of salt in the tank

(c)

To determine

To find: the maximum amount of salt in the solution

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Chapter 7.6, Problem 45PS

Chapter 7.6, Problem 47PS

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Student's Solution and Survival Manual for Calculus

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To setup and solve: the differential equation for the amount of salt in the solution at time t

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To setup and solve: the differential equation for the amount of salt in the solution at time t

To find: the total amount of salt in the tank

To find: the maximum amount of salt in the solution

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Chapter 7 Solutions