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A colleague asked me a question and I got stumped. Suppose that $f:\mathbb{R}^2\rightarrow \mathbb{R}$, $g:\mathbb{R}^2\rightarrow\mathbb{R}^2$, and that $\Gamma$ is a closed $C^\infty$ curve in $\mathbb{R}^2$. Further let $\tau_x$ be a unit tangent at each point $x\in\Gamma$ and let $\nu_x$ be a unit normal. If we have

$ \nabla f\cdot\nu_x=g\cdot \tau_x $

at each point of $\Gamma$, does that imply that we have

$ \nabla f\cdot\tau_x=-g\cdot\nu_x? $

I did an example with the circle $x^2+y^2=1$. If we work in polar, a unit normal is $\langle \cos\theta,\sin\theta\rangle$ and a unit tangent is $\langle\sin\theta,-\cos\theta\rangle$. The examples I tried all worked out.

What I'd like to say is that, given an $x\in\Gamma$, we can change coordinates to make the curve at $x$ tangent to some circle and then use the above argument. Clearly we can't do this for all $x\in\Gamma$ simultaneously, which sort of bothers me.

What I want to know is, is the above true? Can my argument be made more rigorous? If either of those questions turn out negative, what is a counterexample? Or, what is a valid argument?

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    @Hans Lundmark: Agreed. Did you want to post that as an answer, so I can give you credit and close this silly question?2011-11-15

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Given $f$, the first equation only determines the tangential component of $g$. The normal component (the right-hand side of the second equation) can be anything, and therefore the second equation need not hold.