Explanation No work done in displacement from a point A to another point in the field and back to A and the mechanical energy is conserved. i.e., ∮ F ⋅ d R = 0 , the field is said to be conservative. Every irrotational field is conservative. c u r l F = 0 , then vector field is irrotational. The field is irrotational the vector F can be expressed as the gradient of a scalar potential ϕ i.e., F = g r a d ϕ ⇒ F = ∇ ϕ Given: F ( x , y , z ) = ( 2 x y x 2 + z , l n ( x 2 + z ) , y x 2 + z ) is the vector point function. Formula used: F = F 1 i + F 2 j + F 3 k Curl of a continuously differentiable vector point function F is c u r l F = ∇ × F = | i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z F 1 F 2 F 3 | Calculation: The components of the vector F are f 1 = 2 x y x 2 + z , f 2 = l n ( x 2 + z ) , f 3 = y x 2 + z c u r l F = ∇ × F = | i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z 2 x y x 2 + z l n ( x 2 + z ) y x 2 + z | , by expanding 3 × 3 determinant = i ( ∂ ∂ y ( y x 2 + z ) − ∂ ∂ z ( l n ( x 2 + z ) ) ) − j ( ∂ ∂ x ( y x 2 + z ) − ∂ ∂ z ( 2 x y x 2 + z ) ) + k ( ∂ ∂ x ( l n ( x 2 + z ) ) − ∂ ∂ y ( 2 x y x 2 + z ) ) = i ( ∂ ∂ y ( y ) ( 1 x 2 + z ) − 1 x 2 + z ∂ ∂ z ( x 2 + z ) ) − j ( y ∂ ∂ x ( 1 x 2 + z ) − 2 x y ∂ ∂ z ( 1 x 2 + z ) ) + k ( 1 x 2 + z ∂ ∂ x ( x 2 + z ) − 2 x x 2 + z ∂ ∂ y ( y ) ) = i ( ( 1 ) ( 1 x 2 + z ) − 1 x 2 + z ( 0 + 1 ) ) − j ( y ( − 1 ( x 2 + z ) 2 ) ∂ ∂ x ( x 2 + z ) − 2 x y ( − 1 ( x 2 + z ) 2 ) ∂ ∂ z ( x 2 + z ) ) + k ( 1 x 2 + z ( 2 x + 0 ) − 2 x x 2 + z ( 1 ) ) = i ( 1 x 2 + z − 1 x 2 + z ) − j ( − y ( x 2 + z ) 2 ( 2 x 2 − 1 + 0 ) − 2 x y ( − 1 ( x 2 + z ) 2 ) ( 0 + 1 ) ) + k ( 2 x x 2 + z − 2 x x 2 + z ) = i ( 0 ) − j ( − y ( x 2 + z ) 2 ( 2 x ) + 2 x y ( x 2 + z ) 2 ) + k ( 0 ) = i ( 0 ) − j ( 0 ) + k ( 0 ) = 0 c u r l F = 0 Therefore the field is irrotational...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17 Solutions

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Whether the vector field is conservative, if so find potential function.

Solution Summary: The author explains that the vector field is conservative and the scalar potential function is = y l n ( x 2 + z)

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Chapter 17 Solutions