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Let $\Omega\subset\mathbb{R}^{3}$ be a bounded domain with $C^{\infty}$ boundary. Assume that $u$ and $v$ are smooth functions on a neighbourhood of $\Omega$ such that $u=v$ on the boundary $\partial\Omega$. Let $\xi$ be a tangent vector to $\partial\Omega$ at $z_{0}\in\partial\Omega$. What is the geometric intuition of the conclusion $$ \dfrac{\partial u}{\partial\xi}\left(z_{0}\right)=\dfrac{\partial v}{\partial\xi}\left(z_{0}\right)? $$

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Look at one dimension lower and straighten the boundary with a chart, this is simply every (smooth) extension $f(x,y)$ of $f(x,0)=g(x)$ to the upper half space have the same $x$-partial derivative at the boundary $(\partial_1f)(x,0)=g'(x)$, which should be obvious.

If you are using the equivalence class of curves definition of tangent vector, this has an intuitive picture: there is a representative curve $\gamma\colon(-\varepsilon,\epsilon)\to\mathbb{R}^3$ $\gamma(0)=z_0$, $\gamma'(0)=\xi$ that is completely contained in $\partial\Omega$. Therefore $\dfrac{\partial u}{\partial\xi}(z_0)$ depends only on $u\vert_{\partial\Omega}$.

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