I'm taking a course in elementary differential geometry, and I was trying to understand a result.
Let $S$ be a nonempty, compact, oriented regular surface in $\mathbb{R}^{3}$ and let $p \in S$ be a point which maximises the function $f: S \rightarrow \mathbb{R}$ defined by $f(x)=|x|^{2}$. Prove that the normal curvature of any curve $C \subset S$ passing thought $p$ satisfies $\left|k_{n}\right| \geq 1 /|p|$.
Proof:
Let $\gamma:(-\epsilon, \epsilon) \rightarrow S$ be an arbitrary smooth curve parametrised by arc-length and $\gamma(0)=p .$ By the assumptions on $p,$ we must have $ \left.\frac{d}{d t}|\gamma(t)|^{2}\right|_{t=0}=0,\left.\quad \frac{d^{2}}{d t^{2}}|\gamma(t)|^{2}\right|_{t=0} \leq 0. $ Since above equations hold for all such $\gamma$, the first equation implies that the vector $p=\gamma(0)$ is proportional to the unit normal $N(p)=\pm \gamma(0) /|\gamma(0)|$ of $S$ at $\gamma(0)=0$. The second inequality implies that $ \left\langle\gamma^{\prime \prime}(0), \gamma(0)\right\rangle+\left\langle\gamma^{\prime}(0), \gamma^{\prime}(0)\right\rangle \leq 0 $, and…
Why is the normal given by $N(p)=\pm \gamma(0) /|\gamma(0)|$?
Best Answer
The first equation tells you that $p$ (as a vector) is always perpendicular to $\dot{\gamma}(0)$, whatever $\gamma$ you choose. Since all tangent vectors are of the form $\dot{\gamma}(0)$, then $p$ needs to be proportional to the normal vector.