# Derivative and Tangential Vector of a Parametrized Curve

Problem
Let $C\subset\mathbb{R}^2$ be a simple closed convex curve with $O$ as its interior point and its regular parametrization is given as $\gamma_C:[0,1]\rightarrow C$ such that $\gamma_C(0)=(1,0)$ and $\gamma_C'(0)=(0,1)$.
Let $L\subset\mathbb{R}^2$ be a simple curve with its regular parametrization given as $\gamma_L:[0,1]\rightarrow L$ such that $\gamma_L(0)=(1,0)$ and $\gamma_L'(0)=(-1,1)$.

Show that there exists $t\in[0,1]$ such that $\gamma_L(t)$ is interior to $C$.

Description
I am trying to prove a lemma in my research regarding topology but I come across this small problem about geometry of a curve. I do not attend a full course on geometry of curves, hence I may lack of ideas to solve this. It is geometrically intuitive but I need an algebraic approach to prove this. I try Taylor's Theorem or Cauchy's Mean Value Theorem but I'm going nowhere.

• It's not true because the original curve $C$ may be turning to the right. Take $C$ to be the circle $(x-2)^2+y^2=1$ (traversed clockwise), for example. – Ted Shifrin May 2 '17 at 20:14
• The closed curve bisects, near it's starting point, any sufficiently small ball centered at that point into two components. In order to ensure that what you claim is true you somehow have to ensure that the component 'to the left' is contained in what you refer to as the interior region, because that's the direction into which the second curve is pointing at its starting point. Convexity of $C$ will obviously allow you to conclude this. – Thomas May 3 '17 at 5:10