# Prove that this class of curves has constant speed and curvature

Let $$\gamma: (a,b) \rightarrow \mathbb{R}^2$$ be a smooth regular curve such that $$\forall s,t \in (a,b)$$ , $$||\gamma(s)-\gamma(t)||$$ is a non-negative real valued function which depends only on $$|t-s|$$.

Show that $$\gamma(t)$$ has speed and curvature both constant.

Here is my attempt: I somehow have to use the constraint given of the function $$||\gamma(t)- \gamma(s)||$$. Since this acts on $$(a,b)^2$$, I want to handle a nicer funcion defined as follows: $$f(h):= ||\gamma(s+h)-\gamma(s)||^2$$ for $$s\in(a,b)$$ fixed. Taking the derivative with respect to $$h$$, denoting with $$\langle \cdot , \cdot \rangle$$ the Euclidean inner product on $$\mathbb{R}^2$$

$$f'(h)= \frac{d}{dh}\langle \gamma(s+h) - \gamma(s) , \gamma(s+h) - \gamma(s) \rangle = \frac{d}{dh} [\langle \gamma(s+h), \gamma(s+h) \rangle + \langle \gamma(s), \gamma(s) \rangle -2 \langle \gamma(s+h), \gamma(s) \rangle ] = 2[\langle \frac{d}{dh}\gamma(s+h), \gamma(s+h) \rangle - \langle \frac{d}{dh}\gamma(s+h), \gamma(s) \rangle] = 2[\langle \frac{d}{dh}\gamma(s+h), \gamma(s+h) - \gamma(s) \rangle ]$$

So $$f'(0)=0 \quad \forall s$$

I think this is somehow useful, but I don't see how to continue from here.

For the speed it is sufficient to remark that $$|\gamma'(s)|=\lim_{h\to0+}{|\gamma(s+h)-\gamma(s)|\over h}\ .$$ Here the RHS is independent of $$s$$, by assumption on $$\gamma$$.
For the curvature we have to extract the value of $$\kappa(s)$$ from distance measurements. We thereby may assume that $$|\gamma'(s)|\equiv1$$.
Consider the following model situation: $$s\mapsto\gamma(s)=\bigl(x(s),y(s)\bigr)$$ is a smooth arclength parametrized curve with $$x(0)=y(0)=0,\quad \dot x(0)=1,\quad \dot y(0)=0,\quad \theta(0)=0\ ,$$ where $$\theta(s):={\rm arg}\bigl(\dot x(s),\dot y(s)\bigr)$$ is the argument of the tangent direction. Then one has \eqalign{\dot x&=\cos\theta,\quad\ddot x=-\sin\theta\cdot\dot\theta,\quad x^{...}=-\cos\theta\cdot\dot\theta^2-\sin\theta\cdot\ddot\theta,\cr\dot y&=\sin\theta,\quad \ddot y=\cos\theta\cdot\dot\theta\cr} identically in $$s$$, and therefore $$\ddot x(0)=0,\quad x^{...}(0)=-\kappa^2,\qquad \ddot y(0)=\kappa\ ,$$ where $$\kappa:=\dot\theta(0)$$ is the curvature of $$\gamma$$ at $$(0,0)$$. Taylor's theorem then gives $$x(s)=s-{\kappa^2\over6}s^3+O(s^4),\quad y(s)={\kappa\over2}s^2+O(s^3)\qquad(s\to0)\ .$$ It follows that $$|\gamma(s)|^2=\left(s^2-{\kappa^2\over3}s^4+O(s^5)\right)+\left({\kappa^2\over4}s^4+O(s^5)\right)=s^2-{\kappa^2\over12}s^4+O(s^5)\qquad(s\to0)\ ,$$ so that $$\lim_{s\to0}{|\gamma(s)|^2-s^2\over s^4}=-{\kappa^2\over12}\ .$$ Back to our given special curve $$\gamma$$ this means that $$-{\kappa^2(s)\over12}=\lim_{h\to0}{\bigl|\gamma(s+h)-\gamma(s)\bigr|^2-h^2\over h^4}\ .$$ Here the RHS is independent of $$s$$, hence $$s\mapsto\kappa^2(s)$$ is constant. By continuity this implies that $$s\mapsto\kappa(s)$$ is constant.
• Thank you, this helped a lot. What if $\gamma$ is not parametrized by arc-length? Does this line of reasoning still work with more tedious calculations, or do we have to try something else? – Siupa Nov 1 '18 at 17:54
• Part (i) shows that $\gamma$ is parametrized by arc length, up to a constant factor. Assuming $|\gamma'(s)|\equiv1$ instead of $|\gamma'(s)|\equiv c$ is no essential restriction in what follows. – Christian Blatter Nov 1 '18 at 19:26
Your derivative is right but this is not what you should look at. Fro the problem you want $$\parallel \gamma^\prime(s) \parallel = cst$$. So you should calculate $$\frac{\langle \gamma(s+h)-\gamma(s), \gamma(s+h)-\gamma(s) \rangle}{h^2}$$ as $$h \to 0$$.