Partial Derivatives and Implicit Function Thm.

  • Thread starter qspeechc
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  • #1
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Hi.

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

Let [itex]p=(p_1,p_2,...,p_n)[/itex] and [itex]q=(q_1,q_2,...,q_n)[/itex] and the system of n algebraic equations
[itex]g_r(p,q)=0, \quad r\in \{1,2,...,n\}[/itex]
can be solved with respect to the [itex]p_i[/itex] in the form [itex]p_i=\phi _i(q)[/itex] for [itex]i\in \{1,2,...,n\}[/itex]
Then we have
[itex]\frac{d}{dq_h}g_r(\phi ,q)=0[/itex]
where [itex]\phi =(\phi _1,...,\phi _n)[/itex]. If we denote [itex]G_r(q)=g_r(\phi ,q)[/itex] then the last relation gives

[itex]0=\frac{\partial}{\partial q_h}G_s = \Big[\frac{\partial g_s}{\partial q_h} + \sum_{i=1}^{n}\frac{\partial g_s}{\partial p_i}\frac{\partial \phi _i}{\partial q_h} \Big]_{p=\phi (q)}[/itex]
Ok, up to this point I'm fairly confident I'm following along. But then they do the following:

Therefore the partial derivatives can be expressed in the form
[itex]\frac{\partial g_s}{\partial q_h} = \sum_{i=1}^n\frac{\partial g_s}{\partial p_i}\frac{\partial (p_i-\phi _i)}{\partial q_h}[/itex]
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.
 
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Answers and Replies

  • #2
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Anyone :frown:
Please?
 
  • #3
Astronuc
Staff Emeritus
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What is the implication of

[tex]\frac{\partial p_i}{\partial q_h}\,=\,0[/tex]
 
  • #4
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If
[tex]\frac{\partial p_i}{\partial q_h} = 0[/tex]
then
[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.
 
  • #5
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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?
 
  • #6
836
13
Ok, anyone please? Even if it's just a " yea I don't get that either"...
 

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