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Conservative Fields. a Vector Field Is Called Gradient If It Is a Gradient F = Grad Φ of a Scalar Potential. It Is

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Conservative Fields. a Vector Field Is Called Gradient If It Is a Gradient F = Grad Φ of a Scalar Potential. It Is Lecture 22: Conservative Fields. A vector ¯eld is called gradient if it is a gradient F = grad Á of a scalar potential. It is called path independent if the line integral dependsZ only onZ the endpoints, i.e. if c1 and c2 are any two paths from P to Q then F ¢ ds = F ¢ ds. c1 c2 This is equivalent to that the line integral along any closed path or loop vanishes. Th A vector ¯eld F in a domain D is gradient if and only if it is path independent. In that case we say that it is conservative Zand the integral is the di®erence in potential of the endpoint minus initial point: F ¢ ds = Á(Q) ¡ Á(P ). Z c Ex. Evaluate F¢ds, where F= yi + xj and c= cos ti + sin tj, 0·t·¼=4. c Sol. We want to ¯nd Á so that @Á=@x = y, @Á=@y = x and @Á=@z = 0. Integration of @Á=@x = y gives Á = xy + g(y; z), where g is any function of y and z. With this Á it follows that @Á=@y = x and @Á=@z = 0 if g(y; z) = C is a constant. Hence Á = xy + C, for any constant C, satis¯es grad Á = F. Hence Z p p 1 F ¢ ds = Á(1= 2; 1= 2; 0) ¡ Á(1; 0; 0) = c 2 Note that F = grad Á is perpendicular to the level surfaces of Á and hence the level surfaces of the potential function are perpendicular to the flow lines of F. Proof of Th: If F = grad Á then for any curve c from P to Q: Z Z Z b @Á dx @Á dy @Á dz b d F¢ds = + + dt = Á(x(t); y(t); z(t)) dt = Á(Q)¡Á(P ); c a @x dt @y dt @z dt a dt If the integral is independent of the way then we de¯ne Z (x;y;z) Á(x; y; z) = F ¢ ds; (x0;y0;z0) where the integral is along any curve from a ¯xed point (x0; y0; z0) to (x; y; z). This is well-de¯ned since it does not depend on which path we integrate along. In particular we can pick a curve going from (x0; y0; z0) to (a; y; z) and then along a straight line segment (t; y; z), a · t · x, to (x; y; z): Z (a;y;z) Z (x;y;z) Z (a;y;z) Z x Á(x; y; z) = F ¢ ds + F ¢ ds = F ¢ ds + F1(t; y; z)dt: (x0;y0;z0) (a;y;z) (x0;y0;z0) a The ¯rst integral is independent of x and by the fundamental theorem of calculus Z @Á @ x (x; y; z) = F1(t; y; z)dt = F1(x; y; z): @x @x a The proof of that @Á=@y =F2 and @Á=@z =F3 is similar. Ex. Show that F = xy2i + x3yj is not conservative. Sol. If @Á=@x = xy2 then @2Á=@y@x = 2xy but if @Á=@y = x3y then @2Á=@x@y = 3x2y which is a contradiction. 1 2 Conservative ¯elds-Irrotational ¯elds. We have just seen an example of a vector ¯eld F = F1i + F2j + F3k that could not be conservative because if there was a potential F = grad Á then @2Á=@x@y = @2Á=@y@x etc. so we must have @F1=@y = @F2=@x etc. Hence for a vector ¯eld to be conservative we must have ³@F @F ´ ³@F @F ´ ³@F @F ´ curl F = 2 ¡ 3 i + 3 ¡ 1 j + 1 ¡ 2 k = 0 @z @y @x @z @y @x A vector ¯eld satisfying this is called irrotational. We have Theorem. A vector ¯eld F de¯ned and continuously di®erentiable throughout a simply connected domain D is conservative if and only if it is irrotational in D. The physical interpretation of this is that the flow lines for a gradient vector ¯eld can not curl around in a closed orbit since Á increases in the direction of the gradient. Ex. Show that F = xi + yj is conservative in all space. Sol. By the previous theorem it su±ces to show that it is irrotational: curl F=0. Ex. Is F = (¡yi + xj)=(x2 + y2) conservative in D =f(x; y; z); (x; y) 6= (0; 0)g. Sol. r £ F = ::: = 0 but D is not simply connected so it does not follow that it is conservative. In fact, it is not since the line line integral along a circle around the z-axisR is nonvanishingR as we shall see. If x = cos t, y = sinR t and z = 0, 0 · t · 2¼ 2¼ 2¼ 2 2 then C F ¢ ds = 0 (F1dx=dt + F2dy=dt + F3dz=dt)dt = 0 sin t + cos t dt = 2¼. Ex. Show that the vector ¯eld above in the domain D of space where x > 0 is conservative and ¯nd a potential there. Sol. That it is conservative follows from that r £ F = 0 and that the domain D is simply connected. Á = tan¡1(y=x) is a potential in D, since x 6= 0 in D. Proof of the theorem. That conservative implies irrotational is just the calculation above that r £ rÁ = 0. We shall prove that irrotational implies conservative if the domain is all of space or a rectangular box containing the origin. We de¯ne Z x Z y Z z Á(x; y; z) = F1(t; 0; 0) dt + F2(x; t; 0) dt + F3(x; y; t) dt 0 0 0 and we will show that rÁ=F if r£F=0. By the Fundamental Theorem of Calculus ¯ Z z @Á d ¯ d (x; y; z) = ¯ Á(x; y; z) = F3(x; y; t) dt = F3(x; y; z) @z dz (x;y)¡fixed dz 0 Similarly, also using that we can di®erentiate below an integral sign: Z z @Á @F3 (x; y; z) = F2(x; y; 0) + (x; y; t) dt @y 0 @y If we now also use the fact that @F3=@y = @F2=@z we obtain Z z ¯ @Á d ¯ (x; y; z) = F2(x; y; 0) + ¯ F2(x; y; t) dt @y 0 dt (x;y)¡fixed = F2(x; y; 0) + F2(x; y; z) ¡ F2(x; y; 0) = F2(x; y; z) by the Fundamental Theorem of Calculus. Finally by the same argument Z Z @Á y @F z @F (x; y; z) = F (x; 0; 0) + 2 (x; t; 0) dt + 3 (x; y; t) dt @x 1 @x @x Z0 Z0 y d z d = F1(x; 0; 0) + F1(x; t; 0) dt + F1(x; y; t) dt 0 dt 0 dt = F1(x; 0; 0) + F1(x; y; 0) ¡ F1(x; 0; 0) + F1(x; y; z) ¡ F1(x; y; 0) = F1(x; y; z).
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