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Math Help - Exact differential

  1. #1
    Member Ruun's Avatar
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    Exact differential

    Hi. Given equations of constraint for the motion of a particle in the form:

    \sum_{i=1}^{n} g_{i}(x_{1},...,x_{n} ) dx_{i} = 0

    will be holonomic only if exists a function f(x_{1},..,x_{n}) such that:

    \frac{\partial fg_{i}}{\partial x_{j}} = \frac{\partial fg_{j}}{\partial x_{i}} for all  i \neq j.

    In the particular case of:

     dx - a sin(\theta) d\phi =0
    dy + a cos (\theta) d\phi = 0

    I should find an integrating factor of the form f(x,y,\theta,\phi), but I don't know wich are exactly the g_{i} functions. There are four g_{i}.

    Thanks for your time.
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  2. #2
    MHF Contributor

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    Well, your problem is that those two equations are NOT of the form \sum_{i=1}^{n} g_{i}(x_{1},...,x_{n} ) dx_{i} = 0 and can be put in that form in a number of different ways. For example, just adding them gives dx+ dy+ a(cos(\theta)- sin(\theta))d\phi= 0. In that case g_x= 1, g_y= 1, g_\theta= 0, and g_\phi= a(cos(\theta)- sin(\theta)).

    If you were to subtract the two equations rather than add, you would get dx- dy- a(sin(\theta)+ cos(\theta))d\phi and that would a different f.

    Personally, what I would do is integrate dx= a sin(\theta)d\phi and dy= -acos(\theta)d\phi, treating \theta as a constant. That will give x and y as functions of \theta and \phi.
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