Find integral curve over vector field

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SUMMARY

The discussion focuses on finding the integral curve of a vector field defined in R^3, specifically for the vector field X(p) = (p; X_x(p), X_y(p), X_z(p)), where X_x(p) = -x + y + z, X_y(p) = x - y + z, and X_z(p) = x + y - z. The integral curve α must pass through a specified point q = (a, b, c) in R^3, leading to the equations for partial derivatives of α. The integration process involves treating y and z as constants, resulting in the expression α(x, y, z) = -1/2 x^2 + xy + xz + u(y, z), where u(y, z) is an arbitrary function of y and z.

PREREQUISITES
  • Understanding of Differential Geometry concepts
  • Familiarity with vector fields in R^3
  • Knowledge of partial derivatives and integration techniques
  • Experience with path integrals in Calculus
NEXT STEPS
  • Study the properties of vector fields in R^3
  • Learn about integral curves and their applications in Differential Geometry
  • Explore the concept of path independence in integrals
  • Investigate the role of arbitrary functions in integration results
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Students and professionals in mathematics, particularly those studying Differential Geometry, vector calculus, and anyone interested in understanding integral curves of vector fields.

Alteran
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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.
 
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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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