I'm trying to show that a closed, smooth plane curve with curvature 1 is a circle.
The Frenet equations are:
\begin{align} t' &= kn \\ n' &= -kt - \tau b \\ b' &= \tau n. \end{align}
Now, I've shown that if $\alpha(t)$ is a plane curve, then $\tau = 0$.
Thus, our equations give:
$$t'' = -t$$ and $$b'=0.$$
From the first equation we have that,
\begin{align} t'' &= t'\\ \implies \alpha'''(s)&=-\alpha'(s)\\ \implies \alpha(s) &= c_1sinx+c_2cosx+c_3. \end{align}
It seems like I may be on the right track... though I have not yet used the fact that $\alpha$ is a closed curve. Or is there a different approach I should take?
*Also, I have moved the other direction: I have previously shown that a circle possesses curvature $\frac{1}{r}$ where $r$ is its radius. A circle of radius one, then has curvature 1. This proof doesn't work the other direction.
*For reference, I'm almost exclusively using Do Carmo's Differential Geometry of Curves and Surfaces
Thx!
Update:
No one has answered with anything yet, but here's another idea:
Do Carmo asks that we prove the existence of an Osculating Circle. Here's an idea for that:
Let there be two points $s_0$ and $s_1$ on $\alpha(s)$. Let their corresponding normals intersect at point which we will define as the center of a circle. Let ${s_0}$ approach ${s_1}$. Now show the limiting position of the intersection of normals becomes the center of an osculating circle (i.e., a circle with a tangent that coincides with the tangent of our curve at $s_1$) and the limiting position of the circle become the osculating circle. We may then prove that the radius of this circle is $\frac{1}{k(s)}$ Is this correct and how may I approach this?
