# Prove that a SRAN curve is part of a circle

• Nov 6th 2012, 07:49 PM
dykesa
Prove that a SRAN curve is part of a circle
Let alpha(s): I -> R^3 be smooth, regular, non-singular curve parameterized by arc length with the property that all normal lines to alpha(s) pass through the origin. We can assume that alpha(s) is non-zero for any s in I.
Prove that alpha is part of a circle. (i.e. the trace is contained in a plane and the distance from alpha to a fixed point is constant). Hint: explain why alpha(s)+f(s)n(s) = 0 for some f:I -> R, and use this together with the Frenet equations to prove the result.

From the hint, I see that since n(s) is the normal vector to alpha at s, f can be -alpha(s) to give us the zero. But what does this have to do with the Frenet equations? Do we want to prove that curvature is 1/r, thus the trace of alpha is a circle at that point?
• Feb 4th 2015, 05:04 PM
Rebesques
Re: Prove that a SRAN curve is part of a circle
The normal line to $a(s)$ is given by

$\epsilon(u)=a(s)+u\eta(s),u\in \mathbb{R}$.

Since every normal line passes through the origin, for every $s\in I$ there exists an $u=u(s)$ such that

$0=a(s)+u(s)\eta(s),s\in I$.

The function $u(s)$ is differentiable (why?). Differentiating the last relation gives

$0=a'(s)+u'(s)\eta(s)+u(s)\eta'(s) \Rightarrow$
$(1-u(s)k(s))t(s)+u'(s)\eta(s)+u(s)\tau(s)b(s)=0$

where $\{t,\eta,b\}$ is the Serret-Frenet trihedron. Linear independence now implies

$u'(s)=0\Rightarrow u(s)=u_0\neq 0$
$u_0\tau(s)=0\Rightarrow \tau(s)=0$, so the curve is planar and
$1-u_0k(s)=0\Rightarrow k(s)=u_0^{-1}$, so the curvature is constant;
Thus $a$ is part of a circle.