Program Explanation Given information: The pollutant concentration is c ( t ) = 10 ( 1 + cos t ) L / m 3 and it varies between 0 and 20, average depth of water is 2 meters the liquid pollutant flows at the rate of 2 × 10 5 m 3 per month. The period of oscillation is about 6 1 4 months. Explanation: The mount of solute x ( t ) satisfies the below differential equation. d x d t = r i c i − r o V x ....... (1) Here, r i , c i , r o are constants and V is the volume of the solution. The value of r i in millions of liters per second is calculated as, r i = 200 × 10 3 m 3 / month = 0.2 × 10 6 m 3 / month = 1 5 million liters The rate of inflow and outflow is same, so r o = 1 5 and the value of volume in millions of cubic meters is 2 . Substitute 1 5 , r i and 1 5 for r o , 10 ( 1 + cos t ) for c i and 2 for V in equation (1). d x d t = 1 5 ⋅ 10 ( 1 + cos t ) − 1 5 2 x d x d t = 2 ( 1 + cos t ) − 1 10 x d x d t + 1 10 x = 2 ( 1 + cos t ) The above equation is a linear first-order equation and its solution is calculated as, x e 1 10 ∫ d t = 2 ∫ ( 1 + cos t ) e 1 10 ∫ d t d t x e t 10 = 2 ∫ ( 1 + cos t ) e t 10 d t = 2 ∫ e t 10 d t + 2 ∫ cos t ( e t 10 ) d t Simplify the above equation. x e t 10 = 20 e t 10 + 2 ∫ cos t ( e t 10 ) d t + C ....... (2) Consider the integral ∫ cos t ( e t 10 ) d t as I and evaluate. I = ∫ cos t ( e t 10 ) d t The integral I can be solved with the integration by parts and then consider the below substitution. u = cos t ⇒ d u = − sin t d t d v = e t 10 d t ⇒ v = 10 e t 10 The integral I can be written as, I = 10 e t 10 cos t + 10 ∫ e t 10 sin t d t Consider the integral ∫ sin t ( e t 10 ) d t as I 1 and evaluate. I 1 = ∫ sin t ( e t 10 ) d t The integral I can be solved with the integration by parts and then consider the below substitution. u = sin t ⇒ d u = cos t d t d v = e t 10 d t ⇒ v = 10 e t 10 The integral I 1 can be written as, I 1 = 10 e t 10 sin t − 10 ∫ e t 10 cos t d t = 10 e t 10 sin t − 10 I The integration I can be written as, I = 10 e t 10 cos t + 10 I 1 = 10 e t 10 cos t + 10 ( 10 e t 10 sin t − 10 I ) = 10 e t 10 cos t + 100 e t 10 sin t − 100 I 101 I = 10 e t 10 cos t + 100 e t 10 sin t Simplify the above equation

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

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Chapter 1.5, Problem 46P

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Program Description: Purpose of problem is to calculate the amount of pollutant x(t) and show that the graph is common with the steadily rising curve and the time taken pollutant to reach 10L/m3 .

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Chapter 1.5, Problem 45P

Chapter 1.6, Problem 1P

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THIS IS NOT A GRADING ASSIGNMENT: Please only do lab 2.2 (bottom part of the first picture) For that Lab 2.2 do: *Part 1 (do the CODE, that's super important I need it) *Part 2 *Part 3 I also attached Section 2.5.2 which is part of the step 1 so you can read what is it about. Thank you!

THIS IS NOT A GRADING ASSIGNMENT: Please only do lab 2.2 (bottom part of the first picture) For that Lab 2.2 do: *Part 1 *Part 2 *Part 3 I also attached Section 2.5.2 which is part of the step 1 so you can read what is it about. Thank you!

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Chapter 1 Solutions

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Program Description: Purpose of problem is to calculate the amount of pollutant x ( t ) and show that the graph is common with the steadily rising curve and the time taken pollutant to reach 10 L / m 3 .

Solution Summary: The author calculates the amount of pollutant x(t) and shows that the graph is common with the steadily rising curve.

Concept explainers

Program Description: Purpose of problem is to calculate the amount of pollutant x(t) and show that the graph is common with the steadily rising curve and the time taken pollutant to reach 10L/m3 .

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Chapter 1 Solutions