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Partial Derivatives and Implicit Function Thm.

  1. Oct 12, 2009 #1

    So I'm reading a physics book and I come across the following passage:

    Ok, up to this point I'm fairly confident I'm following along. But then they do the following:

    and I have no idea where this comes from. I am guessing here that [itex]p_i=\phi _i(q)[/itex] is only in some sufficiently small region and not the entire space, since they do not explicitly say anything in this regard, so that [itex]p_i-\phi _i[/itex] is not identically zero on the whole space, so I do not know where the above comes from. This is probably something really simple and I'm just being stupid, but any help is greatly appreciated.
    Last edited: Oct 13, 2009
  2. jcsd
  3. Oct 13, 2009 #2
    Anyone :frown:
  4. Oct 13, 2009 #3


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    What is the implication of

    [tex]\frac{\partial p_i}{\partial q_h}\,=\,0[/tex]
  5. Oct 13, 2009 #4
    [tex]\frac{\partial p_i}{\partial q_h} = 0[/tex]
    [tex]\frac{\partial \phi _i}{\partial q_h} = 0[/tex]
    in the region where [tex]p_i=\phi _i[/tex], and outside this region, well, it's identically zero since [tex]p_i[/tex] is not explicitly a function of the [tex]q_i[/tex]'s. Am I on the right track? I don't see where these train-tracks are leading.
  6. Oct 13, 2009 #5
    Aha! Does this mean that, regardless of where we are in the space, if we have the [tex]p_i[/tex] as a function of the [tex]q_i[/tex] or not, the adding of the term [tex]p_i[/tex] has no effect of the partial derivative because:
    If we are in the region where [tex]p_i-\phi _i=0[/tex] the partial derivatives
    [tex]\frac{\partial g_s}{\partial q_h} = 0[/tex]
    anyway, and if we are outside this region then
    [tex]\frac{\partial p_i}{\partial q_h} = 0[/tex]
    Is this right?
  7. Oct 14, 2009 #6
    Ok, anyone please? Even if it's just a " yea I don't get that either"...
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