Non-Vanishing Derivative Functions for Vector Fields X and Y in R^3

daishin
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Homework Statement


On R^3 with the usual coordinates (x,y,z), consider the pairs of vector fields X,Y given below. For each pair, determine if there is a function f:R^3-->R with non-vanishing derivative df satisfying Xf=Yf=0, and either find such a function or prove that there is none.
(a) X=(e^x)d/dx - ((e^x)z + 2y)d/dz, Y=(e^x)d/dy - (2y)d/dz
(b) X=(e^x)d/dx - ((e^x)z + 2x)d/dz, Y=(e^x)d/dy - (2y)d/dz

Homework Equations



Could you help me start this problem.

The Attempt at a Solution



Sorry I don't know how to start this problem.
 
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You'll want to start by looking up Frobenius integrability conditions.
 
More hint or idea.

OK I looked at Frobenius integrability condition and still have no clue.
How can I use the integrability condition??
 
What is the integrability condition stated in terms of vector fields? You are talking about stuff I haven't looked at for a long time, but isn't the existence of a solution to the PDE's corresponding to these vector fields related to the commutator of the vector fields?
 
Frobenius' Theorem

Yes. A distribution D={E(x)} on a manifold M is integrable iff for all vector fiels V,W on M with V(x),W(x) in E(x) for all x in M, [v,w](x) in E(x) for all x in M. But how can I relate this with the problem I asked?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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