Explanation Given The given force field F = 4 y 2 i + ( 3 x + y ) j from (1,0) to (-1,0) along the top half of the ellipse x 2 + y 2 k 2 = 1 . Concept used: Line integral of a vector field: If F ( x , y , z ) = u ( x , y , z ) i + v ( x , y , z ) j + w ( x , y , z ) k is a vecor field and C be a piecewise smooth curve with parametric equation R ( t ) = x ( t ) i + y ( t ) j + z ( t ) k for a ≤ t ≤ b then the line integral of F along C is given by ∫ C F ⋅ d R = ∫ C ( u d x + v d y + w d z ) = ∫ a b F [ R ( t ) ] ⋅ R ′ ( t ) d t Where d R = d x i + d y j + d z k Formula used: Fundamental theorem of Calculus I: ∫ a b f ( x ) d x = F ( b ) − F ( a ) Where F ′ ( x ) = f ( x ) Calculation: The parametric equation of the ellipse x = cos t y = k sin t As the objects moves from (1,0) to (-1,0) hence the parameter t vary from 0 to π Therefore consider, r = ( cos t ) i + ( k sin t ) j That is, d r = ( − sin t ) i + ( k cos t ) j Along the curve, F = 4 ( cos t ) 2 i + ( 3 cos t + k sin t ) j F = 4 ( cos 2 t ) i + ( 3 cos t + k sin t ) j Now consider , F ⋅ d r = { 4 ( cos t ) 2 i + ( 3 cos t + k sin t ) j } ⋅ { ( − sin t ) i + ( k cos t ) j } = − 4 cos 2 t sin

6th Edition

ISBN: 9781465208880

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

Publisher: Kendall Hunt Publishing

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Chapter 13, Problem 88SP

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To find : the work done when an object moves against the given force field.The value of k the minimizes work.

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Chapter 13, Problem 87SP

Chapter 13, Problem 89SP

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