Proving that the trace of a curve $\alpha(s)$ is symmetric if the curvature is an even function

Let $b >0$ and $\alpha : (-b, b) \rightarrow \mathbb{R^2}$ be a plane curve parametrized by arc length. Suppose that $k(s) = k(-s) \quad \forall s \in (-b,b)$, where $k(s)$ denotes the curvature of $\alpha$ at the point $s$. Show that the trace of $\alpha$ is symmetric with respect to the normal line of $\alpha$ at $0$.

I'll admit that geometry is not my favorite subject but I tried with a differential geometry course nevertheless since many important tools for multivaribale calculus (which I do like) can be studied and developed there so please be patient if I don't fully understand the geometric arguments at first. I think that we can suppose that the normal line of $\alpha$ at $0$ passes through the origin (otherwise we can translate the normal line and the curve so that this happens), so that the normal line of $\alpha$ at $0$ is given by $\beta(t) = N(0)t, \quad t \in \mathbb{R}$. Then I tried to make a reflection of the curve with respect to the normal line, but I didn't get very far. I guess there might be a different and easier approach like defining an auxiliary function involving the curve, the normal vector or the curvature and proving that it's identically zero or something along those lines.

Any hints or ideas will be fully appreciated.

• this is exercise 1.24 of montiel and ros curves and surfaces – Jacques Saliba Apr 11 '18 at 19:07

It has to be seen in the light of the slope $\phi$ , the auxiliary function you are probably seeking, which connects coordinates and even functioned curvature w.r.t. arc $s.$ On integration the slope turns odd. We take arctangent on x-axis at origin, set all constants of integration of integration to zero as implied in your attempt:

Slope or tangent rotation

$\kappa = \frac{d \phi}{ds}$ is even and $\phi = \int \kappa ds$ is odd

x- co-ordinate

$\cos \phi = \frac{d x}{ds},$ is even $\;x =\int \cos \phi\; ds$ is odd

y- co-ordinate

$\sin \phi = \frac{d y}{ds}$ is odd , $y =\int \sin \phi\; ds$ is even

So $y=f(x),$ the trace of $\alpha (s)$ should be symmetric about y-axis.

In general Cornu Spirals we look at last diagram. When curvature is an odd function of arc $s$ the curve ( when boundary conditions are such as shown at origin in the Figure) is antisymmetric and when even it is having $y-$ axis symmetry.