# What is a geometric explanation to the coincidence of two tangential derivatives

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 $$\dfrac{\partial u}{\partial\xi}\left(z_{0}\right)=\dfrac{\partial v}{\partial\xi}\left(z_{0}\right)?$$

## 1 Answer

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 $$u\vert_{\partial\Omega}$$.