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Find integral curve over vector field

  1. Apr 15, 2006 #1
    The question should be very easy, its from topics of Differential Geometry, I just want to make sure that I understands it right :shy: . My question is:

    in [tex]R^3[/tex] we have vector field [tex]X[/tex] and for every point [tex]p(x,y,z)[/tex] in [tex]R^3[/tex] space, vector field [tex]X(p) = (p; X_x(p), X_y(p), X_z(p))[/tex] has:
    [tex]X_x(p) = -x+y+z[/tex]
    [tex]X_y(p) = x-y+z[/tex]
    [tex]X_z(p) = x+y-z[/tex]

    Find integral curve [tex]\alpha[/tex] of this vector field and it should pass through point [tex]$q=(a,b,c) \in R^3[/tex] so that [tex]\alpha(0)=q[/tex].

    Thanks for any advise.
  2. jcsd
  3. Apr 15, 2006 #2


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    Staff Emeritus
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    In other words, you are looking for [itex]\alpha (x,y,z)[/itex] such that
    [tex]\frac{\partial \alpha}{\partial x}= -x+ y+ z[/tex]
    [tex]\frac{\partial \alpha}{\partial y}= x- y+ z[/tex]
    [tex]\frac{\partial \alpha}{\partial z}= x+ y- z[/tex]

    Integrating the first with respect to x (treating y and z as constants) gives [itex]\alpha(x,y,z)= -\frac{1}{2}x^2+ xy+ xz+ u(y,z)[/itex].
    Note that the "constant of integration", since we are treating y and z as constants, may be a function of x and y. Do you remember dealing with "path integrals independent of the path" in Calculus? You should be able to finish this yourself.

    We can't tell you whether you are "understanding it right" or not since you haven't told us how you understand it.
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