Ch 17 Review Ex

# Ch 17 Review Ex - 1170 C H A P T E R 17 L I N E A N D S U R...

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1170 CHAPTER 17 LINE AND SURFACE INTEGRALS (ET CHAPTER 16) Therefore the vector h 4 y , x i is tangent to the ellipses. Since F ( x , y ) is a scalar multiple of this vector, i.e, F ( x , y ) = 1 x 2 + 16 y 2 h 4 y , x i , F is parallel to h 4 y , x i hence F is tangent to the ellipses x 2 + 4 y 2 = c 2 .WeFndtha t F can be sketched by Frst sketching ellipses in the family x 2 + 4 y 2 = c 2 , and then sketching the unit tangents at points on the ellipses. The Feld F ( x , y ) is shown in the Fgure: In Exercises 7–14, determine whether or not the vector feld is conservative and, iF so, fnd a potential Function. 7. F ( x , y , z ) = - sin x , e y , z ® SOLUTION We examine the cross partials of F .Since F 1 = sin x , F 2 = e y , F 3 = z we have: F 1 y = 0 F 2 z = 0 F 3 x = 0 F 2 x = 0 F 3 y = 0 F 1 z = 0 F 1 y = F 2 x , F 2 z = F 3 y , F 3 x = F 1 z Since the cross partials are equal, F is conservative. We denote the potential Feld by ϕ ( x , y , z ) .Sowehave: x = sin x y = e y z = z By integrating we get: ( x , y , z ) = Z sin xdx =− cos x + C ( y , z ) y = C y = e y C ( y , z ) = e y + D ( z ) ( x , y , z ) cos x + e y + D ( z ) z = D z = z D ( z ) = z 2 2 We conclude that ( x , y , z ) cos x + e y + z 2 2 . Indeed: = ¿ x , y , z À = - sin x , e y , z ® = F 8. F ( x , y , z ) = - 2 , 4 , e z ® We examine the cross partials of F .Wehave , F 1 = 2, F 2 = 4, F 3 = e z hence: F 1 y = 0 F 2 x = 0 F 1 y = F 2 x F 2 z = 0 F 3 y = 0 F 2 z = F 3 y F 3 x = 0 F 1 z = 0 F 3 x = F 1 z Since the cross-partials are equal, F is conservative. We denote the potential Feld by ( x , y , z ) .Wehave: x = 2 y = 4 z = e z

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Chapter Review Exercises 1171 By integrating we get: ϕ ( x , y , z ) = Z 2 dx = 2 x + C ( y , z ) y = C y = 4 C ( y , z ) = 4 y + D ( z ) ( x , y , z ) = 2 x + 4 y + D ( z ) z = D z = e z D = e z We conclude that ( x , y , z ) = 2 x + 4 y + e z .Weverify: = ¿ x , y , z À = - 2 , 4 , e z ® = F 9. F ( x , y , z ) = - xyz , 1 2 x 2 z , 2 z 2 y ® SOLUTION No. We show that the cross partials for x and z are not equal. Since the equality of the cross partials is a necessary condition for a Feld to be a gradient vector Feld, we conclude that F is not a gradient Feld. We have: F 1 z = z ( ) = xy F 3 x = x ( 2 z 2 y ) = 0 F 1 z 6= F 3 x Therefore the cross partials condition is not satisFed, hence F is not a gradient vector Feld. 10. F ( x , y , z ) = - 2 z , x 2 yz , 1 2 x 2 y 2 ® Yes. We Fnd by guessing that ( x , y ) = 1 2 x 2 y 2 z is a potential function for F . Indeed, = ¿ x , y , z À = ¿ 2 z , x 2 , 1 2 x 2 y 2 À = F The potential function is determined up to an additive constant, therefore the functions 1 2 x 2 y 2 z + C are potential func- tions for F as well. 11. F ( x , y , z ) = ¿ y 1 + x 2 , tan 1 x , 2 z À We examine the cross partials of F .Since F 1 = y 1 + x 2 , F 2 = tan 1 x , F 3 = 2 z we have: F 1 y = 1 1 + x 2 F 2 x = 1 1 + x 2 F 1 y = F 2 x F 2 z = 0 F 3 y = 0 F 2 z = F 3 y F 3 x = 0 F 1 z = 0 F 3 x = F 1 z Since the cross partials are equal, F is conservative. We denote the potential Feld by
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Ch 17 Review Ex - 1170 C H A P T E R 17 L I N E A N D S U R...

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