If two closed simple curves have the same trace, same starting point and same starting derivative, then they are equal.

Question

This is not a textbook problem, at least that I'm aware of. I was doodling around with some curves and came to this reasonable conjecture: Let $A\subseteq\mathbb{R}^n$ and $\gamma, \bar\gamma:[a, b]\rightarrow\mathbb{R}^n$ be two regular simple closed curves (so, both are of class $C^{\infty}$ with non vanishing derivative, injective in $[a, b)$, and $\gamma^{(k)}(a)=\gamma^{(k)}(b)$, $\bar\gamma^{(k)}(a)=\bar\gamma^{(k)}(b)$, $\forall k\geq 0$) such that $\gamma([a, b])=A=\bar\gamma([a, b])$, $||\gamma '(t)||=||\bar\gamma '(t)||$ $\forall t\in [a, b]$, $\gamma (a)=\bar\gamma(a)$ and $\gamma ' (a)=\bar\gamma ' (a)$, then $\gamma=\bar\gamma$. I put the last hypothesis so that $\gamma$ and $\bar\gamma$ travel $A$ in the same direction. I still don't know how to attack this problem or even if these hypothesis are enough, so I would appreciate any help with this. Thank you.

Answer 1

HINT:

Like Ted Shifrin suggested, the idea would be that the length of the curve from $\gamma(a)$ to $\gamma(x)$ equals $$\int_{a}^x \|\gamma'(t)\| dt$$ (this because the function is injective) so it equals the length from $\gamma(a)= \bar\gamma(a)$ to $\bar\gamma(x)$. Now if they start in the same direction, it follows that $\gamma(x) = \bar\gamma(x)$, for all $x$.