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Exact differentials

  1. Mar 1, 2016 #1
    Why can't (1.39) be put in the form of an exact differential? Seems like I could and the solution to the first equation is

    ##x-a\phi\sin\theta=c##, where ##c## is an arbitrary constant.

    Let ##x-a\phi\sin\theta## be ##h##.

    By considering ##dh=\frac{\partial h}{\partial x}dx+\frac{\partial h}{\partial \phi}d\phi=0##, we get the first equation of (1.39). So it must be a solution. Isn't it?

    Screen Shot 2016-03-02 at 12.04.16 am.png
    Derivation 4:
    Screen Shot 2016-03-02 at 12.04.37 am.png
     
  2. jcsd
  3. Mar 1, 2016 #2

    TeethWhitener

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    I don't know for sure, but I think it's because ##h = h(x,y,\theta,\phi)##, so
    $$dh=\frac{\partial x}{\partial h}dx+\frac{\partial y}{\partial h}dy+\frac{\partial \theta}{\partial h}d\theta+\frac{\partial \phi}{\partial h}d\phi$$
    and ##\frac{\partial \theta}{\partial h}## is nonzero, so you don't actually get back equation 1.39.
     
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