I will start with an interpretation for curvature sign of a plane curve:
Let $\alpha : I \to \mathbb{R}^2$ a curve parametrized by arc length and $s_0 \in I$ such that $\kappa (s_0) \neq 0$ and its implie $\alpha ''(s_0) \neq 0$.
Let $r(\lambda) = \alpha (s_0) + \lambda T(s_0)$, $\lambda \in \mathbb{R}$ the tangent line to $\alpha$ in $s_0$.
Because $\alpha''(s_0)$ is orthogonal $\alpha'(s_0)=T(s_0)$ we can write the tangent line to $\alpha$ in $s_0$ like
$$p \in r \Leftrightarrow \langle p – \alpha (s_0), \alpha'' (s_0) \rangle =0$$
Note that $r$ cut the plane in two half planes $H^+$ and $H^-$
$$H^+ = \{ p \in \mathbb{R}^2 : \langle P – \alpha (s_0), \alpha ''(s_0) \rangle >0 \}$$
and
$$H^- = \{ p \in \mathbb{R}^2 : \langle P – \alpha (s_0), \alpha ''(s_0) \rangle <0 \}.$$
We will show that, for a $s$ sufficiently close to $s_0$, $\alpha(s) \in H^+$, i.e., $\alpha (s)$ belongs to the same half plane of $\alpha''(s_0)$.
Expanding $\alpha(s)$ using Taylor we have
$$\alpha (s) = \alpha(s_0) + (s-s_0) \alpha'(s_0) + \frac{(s-s_0)^2}{2} \alpha''(s_0) + R(s)$$
where $ \lim_{s \to s_0} \frac{R(s)}{(s-s_0)^2} =0$.
And
$$\langle \alpha(s) – \alpha(s_0), \alpha''(s_0)\rangle = (s-s_0) \langle \alpha'(s_0), \alpha''(s_0)\rangle +\frac{(s-s_0)^2}{2} \langle \alpha''(s_0), \alpha''(s_0) \rangle + \langle R(s), \alpha''(s_0) \rangle$$
$$=k(s_0)^2 \frac{(s-s_0)^2}{2} + \langle R(s), \alpha''(s_0) \rangle.$$
So
$$\lim_{s \to s_0} \frac{\langle \alpha(s) – \alpha(s_0), \alpha''(s_0)\rangle}{(s-s_0)^2} = \frac{\kappa (s_0)^2}{2} >0$$
And taking $\varepsilon = \frac{\kappa (s_0)^2}{4}$ exists $\delta>0$ such that $s \in I$, $0 < |s-s_0|<\delta$
$$\Rightarrow \left| \frac{\langle \alpha(s) – \alpha(s_0), \alpha''(s_0)\rangle}{(s-s_0)^2} – \frac{\kappa (s_0)^2}{2} \right| < \frac{\kappa (s_0)^2}{4}$$
$$\Rightarrow \frac{\langle \alpha(s) – \alpha(s_0), \alpha''(s_0)\rangle}{(s-s_0)^2} > \frac{\kappa (s_0)^2}{2} – \frac{\kappa (s_0)^2}{4} = \frac{\kappa (s_0)^2}{2} >0$$
I.e., $\langle \alpha(s) – \alpha(s_0), \alpha''(s_0) \rangle >0$, $\forall s \in (s_0 – \delta, s_0 + \delta) \cap I$, $s \neq s_0$, and by this $\alpha(s) \in H^+$.
Therefore $\alpha''(s_0)$ points in the direction of concavity of curve and the trace of $\alpha$ do not cross the tangent line $r$.
After this interpretation of sign of curvature of a plane curve, I think about the point where $\kappa (s_0)=0$ and $\kappa'(s_0) \neq 0$. That case is in the figure above when $\alpha''(s_0)=0$ and $\kappa (s_0)=0$.
My question is:
For $\alpha: I \subset \mathbb{R} \to \mathbb{R}^2$ parametrized by arc length, if
$s_0 \in I$ such that $\kappa (s_0)=0$ and $\kappa'(s_0) \neq 0$.
How can I show that for all the neighborhood of $s_0$ there are points of $\alpha$ in
each one of the half planes determined by the tangent line to $\alpha$ in $s_0$?
I try to use something like the resolution before, constructing $r$ that cut the plane in two vertical planes. But can not handle with $\alpha''(s_0)=0$.
Is this, thanks for any help 😀
Best Answer
I followed the comments and got the following answer:
We have that $\alpha$ is parametrized by arch lenght with $\kappa (s_0)=0$ and $\kappa'(s_0) \neq 0$.
Observations:
If $\kappa(s_0) \neq 0$ then the curve do not intersect $\alpha$'s tangent line.
If $\kappa(s_0) = 0$ we have that $\alpha$ not necessarily cross the tangent line. For example, take $\alpha(t) = (t^3, t^6)$ at $s_0=0$, we have $\kappa (s_0)=0$ and the curve cross the tangent line.
Thus, the conditions for the curve to cross the tangent line - case $\kappa(s_0) = 0$ -, is that $\kappa'(s_0) \neq 0$.
Supose $\alpha$ parametrized by arc lenght. If we want that curve cross the line $$\alpha(s) - \alpha(s_0) = \alpha(s) T(s_0) + B(s)N(s_0)$$
were $T=\alpha'$, $B$ is the binormal and $N$ the normal vector of $\alpha$, we need to study the sign of $B(S)$ were $$B(s) = \langle \alpha(s)-\alpha(s_0), N(s_0) \rangle.$$
By Taylor we have $$\alpha(s) = \alpha(s_0) + (s-s_0) \alpha'(s_0) + \frac{(s-s_0)^2}{2} \alpha''(s_0) + \frac{(s-s_0)^3}{3!} \alpha'''(s_0) + R(s_0).$$ How $\|\kappa (s_0) \|=\|\alpha''(s_0)\|=0$ then $$\alpha(s) - \alpha(s_0) = \alpha'(s_0)(s-s_0) + \frac{(s-s_0)^3}{3!} \alpha'''(s_0) + R(s_0)$$ with $$\lim_{s \to s_0} \frac{R(s_0)}{(s-s_0)^3}=0.$$
Replacing in $B(s) = \langle \alpha(s)-\alpha(s_0), N(s_0) \rangle$ we have $$B(s) = (s-s_0) \underbrace{\langle \alpha'(s_0), N(s_0)\rangle}_{=0} + \frac{(s-s_0)^3}{3!} \langle \alpha'''(s_0), N(s_0)\rangle + \langle R(s_0), N(s_0)\rangle$$ and $$B(s) = \frac{(s-s_0)^3}{3!} \langle \alpha'''(s_0), N(s_0)\rangle + \langle R(s_0), N(s_0)\rangle.$$
By Frenet formula $T(s) = \kappa(s) N(s) \Rightarrow T'(s) = \kappa'(s)N(s) + \kappa(s) N'(s)$ then $$T'(s_0) = \kappa'(s_0) N(s_0)$$ and $$B(s) = \frac{(s-s_0)^3}{3!} \langle \kappa'(s_0) N(s_0), N(s_0)\rangle + \langle R(s_0), N(s_0)\rangle.$$ Let $$\frac{B(s)}{(s-s_0)^3} = \frac{1}{3!} \langle \kappa'(s_0) N(s_0), N(s_0)\rangle + \left \langle \frac{R(s_0)}{(s-s_0)^3}, N(s_0)\right\rangle \xrightarrow[s\to s_0]{} \underbrace{\frac{1}{3!}}_{>0} \cdot \kappa'(s_0) \underbrace{\|N(s_0)\|}_{>0}$$ then the signal of $B(s)$ depends only of $\kappa'(s_0)$.
Case 1: If $\kappa'(s_0) >0$ then $B(s)>0$ and so there is a neighborhood of $s_0$ such that $\langle \alpha(s) - \alpha(s_0), N(s_0)\rangle > 0$.
Case 2: If $\kappa'(s_0) <0$ then $B(s)<0$ and so there is a neighborhood of $s_0$ such that $\langle \alpha(s) - \alpha(s_0), N(s_0)\rangle < 0$.
Therefore, for all neighborhood of $s_0$ there are points in both half planes determined by the tangent line to $\alpha$ at $s_0$.