Program Explanation Given information: The acceleration is a = 0.5 mi , v 0 = 9 mi/h and v s = 3 mi/h . Calculation: The velocity of the river by the fourth-degree function is v R = v 0 ( 1 − x 4 a 4 ) . The swimmer’s direction angle θ can be expressed as, tan θ = v R v S = d y d x The value of d y d x can be expressed as, d y d x = v 0 v S ( 1 − x 4 a 4 ) . Substitute 0.5 for a , 9 for v 0 and 3 for v s in the velocity function as, d y d x = 9 3 ( 1 − x 4 0.5 4 ) = 3 ( 1 − 16 x 4 ) Now, integrate the above equation on both sides of the equation to obtain the value of y ( x )

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- For Differential Equations and Boundary Value Problems: Computing and Modeling Tech Update

5th Edition

ISBN: 9780134872971

Author: Edwards, C., Penney, David, Calvis

Publisher: PEARSON

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Question

Chapter 1.2, Problem 40P

Program Plan Intro

To calculate: The distance when the swimmer goes in the downstream if the velocity of the river by the fourth-degree function is vR=v0(1−x4a4) .

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Chapter 1.2, Problem 39P

Chapter 1.2, Problem 41P

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Chapter 1 Solutions

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- For Differential Equations and Boundary Value Problems: Computing and Modeling Tech Update

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The distance when the swimmer goes in the downstream if the velocity of the river by the fourth-degree function is v R = v 0 ( 1 − x 4 a 4 ) .

Solution Summary: The author calculates the distance when the swimmer goes in the downstream if the velocity of the river by the fourth-degree function is v_0(1-x4a

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To calculate: The distance when the swimmer goes in the downstream if the velocity of the river by the fourth-degree function is vR=v0(1−x4a4) .

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Chapter 1 Solutions