Curvature of a planar curve by reversing its direction

After the last exercise this afternoon, I will revise what is the curvature of a planar curve when we reverse its orientation. In order to simplify the notation, we will assume that the domain $I$ of the planar curve is $I=\mathbb{R}$, so we have $\alpha:\mathbb{R}\rightarrow{\mathbb R}^2$, which we suppose parametrized by the length-arc. In classroom we have said "when we reverse the orientation, then the curvature change the sign at the point". Firstly, we explanation of our intuition.

When we change the orientation on the curve, we are taking $-s$ instead of $s$ as the new parameter the curve, and as we increase $s$, we are going in the initial curve on the contrary direction. Then the result that we want to prove is that when we pass by a point of the curve, its curvature is the opposite if we go along the curve in the other direction. That is, we say that at $s_0$  the curvature of $\alpha$ when we reverse the sense of the parameter is the opposite with respect to the other direction.

The computations are clear and were made in the classroom. Define $\beta(s)=\alpha(-s)$ the curve $\alpha$ parameterized in the opposite direction. Then $T_\beta(s)=-T_\alpha(-s)$, $N_\beta(s)=JT_\beta(s)=-JT_\alpha(-s)=-N_\alpha(-s)$ and $T_\beta'(s)=T'_\alpha(-s)$. Thus $$\kappa_\beta(s)=\langle T'_\beta(s),N_\beta(s)\rangle=-\langle T'_\alpha(-s),N_\alpha(-s)\rangle=-\kappa_\alpha(-s). (*)$$

The problem appears when we utilize the words to describe the above result. The symbol $\kappa_\beta(s)$ is the curvature of $\alpha$ with the reverse direction at the point $s$. The right hand side says that it coincides with the opposite of the curvature of $\alpha$ at $-s$.

An example. Consider the curve $\alpha(t)=(t,t^2+t^3)$. This curve is not parameterized by the length-arc, but the above argument (which it involves only the curvature) holds for any curve. The curvature of $\alpha$ is $$\alpha'(t)=(1,2t+3t^2), \alpha''(t)=(0,2+6t)\Rightarrow\kappa_\alpha(t)=\frac{2+6t}{(1+(2t+3t^2)^2)^{3/2}}.$$ We fix the point $t=1$, that is, $\alpha(1)=(1,2)$. Then the curvature at $t=1$ is $\kappa_\alpha(1)=8/26^{3/2}$.  We reverse the orientation of $\alpha$ by doing $\beta(t)=(-t,t^2-t^3)$. Now the point $t=1$ for $\alpha$ corresponds with $t=-1$. The curvature of $\beta$ is $$\kappa_\beta(t)=\frac{-2+6t}{(1+(2t-3t^2)^2)^{3/2}}.$$ Then $\kappa_\beta(-1)=-8/26^{3/2}$, which corresponds with the formula  (*).

To finish, I propose a question. Suppose now that the curve is the graph of a function $y=f(x)$, where we know that the curvature of the curve is $f''(x)/(1+f'(x)^2)^{3/2}$. If we move in the opposite direction, then the curve is the graph of the function $y=f(-x)$. Using the above formula,  the curvature would be $f''(-x)/(1+f'(-x)^2)^{3/2}$ because the first derivative gives $-f'(-x)$, but the second one yields $f''(-x)$. But according to (*), this should give $-f''(-x)/(1+f'(-x)^2)^{3/2}$, the oppossite sign.

Question: what is it happening?

Curvature and symmetry about a point

Motivated by one of the proposed exercises today, we observe that the curve $y=x^3$, which is symmetric about the origin of the plane, satisfies that the curvature at the point $-x$ is the opposite in $x$: look in which side lies the graph of $y=x^3$ around $x$. Thus we prove the next result:

Result. If $\alpha:I\rightarrow{\mathbb R}^2$ is a planar curve which is symmetric about  a point $p_0\in\alpha(I)$, then $\kappa(-s)=-\kappa(s)$. Here we are assuming that $\alpha(0)=p$.

After a translation and without loss of generality, we suppose that $p_0=(0,0)$. Say that $\alpha$ is symmetric about the origin means $\alpha(-s)= M\alpha(s)$, where $M$ is the symmetry, that is, $Mx=-x$. Thus $\alpha(-s)=-\alpha(s)$.  Let $\beta(s)=\alpha(-s)$. Then $\kappa_\beta(s)=-\kappa_\alpha(-s)$. On the other hand, the curve $\gamma(s)=-\alpha(s)$ is a direct rigid motion of $\alpha$, so $\kappa_\gamma(s)=\kappa_\alpha(s)$. Since $\kappa_\gamma(s)=\kappa_\beta(s)$, we conclude $$-\kappa_\alpha(-s)=\kappa_\alpha(s).$$
This can be checked for curves that are graphs of $y=f(x)$ where we know that $$\kappa_\alpha(x)=\frac{f''(x)}{(1+f'(x)^2)^{3/2}}.$$ Denote $\alpha$ the curve $y=f(x)$. As $f$ is symmetric about the origin, then $f(-x)=-f(x)$. The curvature of $y=g(x)$, where $g(x)=f(-x)$ is
$$g'(x)=-f'(-x), g''(x)=f''(-x)\Rightarrow \kappa_g(x)=\frac{f''(-x)}{(1+f'(-x)^2)^{3/2}}.$$
The curvature of $y=-h(x)$ is $$h'(x)=-f'(x), h''(x)=-f''(x)\Rightarrow \kappa_h(x)=\frac{-f''(x)}{(1+f'(x)^2)^{3/2}}.$$ Since the curvatures of both graphs coincide, we have $$\kappa_f(-x)=\frac{f''(-x)}{(1+f'(-x)^2)^{3/2}}=\frac{-f''(x)}{(1+f'(x)^2)^{3/2}}=-\kappa_f(x).$$
The converse is also true:

Theorem. If $\alpha:I=(-a,a)\rightarrow{\mathbb R}^2$ is a planar curve  such that $\kappa(-s)=-\kappa(s)$, then the graphic of $\alpha$ is symmetric about  the point $p=\alpha(0)$.

After a translation, we suppose $\alpha(0)=(0,0)$. Define two curves: $\beta(s)=\alpha(-s)$ and $\gamma(s)=-\alpha(s)$. We compute their curvatures. For $\beta$, $\kappa_\beta(s)=-\kappa(-s)$ and for $\gamma$, $\kappa_\gamma(s)=\kappa(s)$ because the symmetry is a direct rigid motion.  By hypothesis, $\kappa_\gamma(s)=\kappa_\beta(s)$ so there is a direct rigid motion $M$ such that $\gamma(s)=M\beta(s)$. Because $\gamma(0)=\alpha(0)=\beta(0)$ and $\gamma'(0)=\beta'(0)$ (check!), then their normal vector also coincide at $s=0$. Then $M$ must be the identity, proving the result.