# Partial Derivatives and Implicit Function Thm.

1. Oct 12, 2009

### qspeechc

Hi.

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 $p_i=\phi _i(q)$ is only in some sufficiently small region and not the entire space, since they do not explicitly say anything in this regard, so that $p_i-\phi _i$ 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. Oct 13, 2009

### qspeechc

Anyone

3. Oct 13, 2009

### Staff: Mentor

What is the implication of

$$\frac{\partial p_i}{\partial q_h}\,=\,0$$

4. Oct 13, 2009

### qspeechc

If
$$\frac{\partial p_i}{\partial q_h} = 0$$
then
$$\frac{\partial \phi _i}{\partial q_h} = 0$$
in the region where $$p_i=\phi _i$$, and outside this region, well, it's identically zero since $$p_i$$ is not explicitly a function of the $$q_i$$'s. Am I on the right track? I don't see where these train-tracks are leading.

5. Oct 13, 2009

### qspeechc

Aha! Does this mean that, regardless of where we are in the space, if we have the $$p_i$$ as a function of the $$q_i$$ or not, the adding of the term $$p_i$$ has no effect of the partial derivative because:
If we are in the region where $$p_i-\phi _i=0$$ the partial derivatives
$$\frac{\partial g_s}{\partial q_h} = 0$$
anyway, and if we are outside this region then
$$\frac{\partial p_i}{\partial q_h} = 0$$
Is this right?

6. Oct 14, 2009

### qspeechc

Ok, anyone please? Even if it's just a " yea I don't get that either"...