- #1
- 1,104
- 961
It is well known that a curve in ##\mathbb{R}^3## is uniquely (up to a position in the space) defined by its curvature ##\kappa(s)## and torsion ##\tau(s)##, here ##s## is the arc-length parameter. We will consider ##\kappa(s),\tau(s)\in C[0,\infty)##
Thus a natural problem arises: to restore features of the curve from given functions ##\kappa(s),\tau(s)##. This is not a simple problem: the Frenet-Serret equations are not in general integrable explicitly.A simplest property of the curve is boundedness. We shall say that a curve is bounded iff its radius-vector is bounded: $$\sup_{s\ge 0}|\boldsymbol r(s)|<\infty.$$
For example, if a function ##\kappa(s)/\tau(s)## is monotone and
$$\lim_{s\to\infty}\frac{\kappa(s)}{s\cdot\tau(s)}=0$$ then the curve is unbounded. This is an almost trivial fact, it follows from some another almost trivial theorem, for details see http://www.ma.utexas.edu/mp_arc/c/16/16-63.pdf
The comments are welcome. Particularly is there a geometrical interpretation of brought above proposition
Thus a natural problem arises: to restore features of the curve from given functions ##\kappa(s),\tau(s)##. This is not a simple problem: the Frenet-Serret equations are not in general integrable explicitly.A simplest property of the curve is boundedness. We shall say that a curve is bounded iff its radius-vector is bounded: $$\sup_{s\ge 0}|\boldsymbol r(s)|<\infty.$$
For example, if a function ##\kappa(s)/\tau(s)## is monotone and
$$\lim_{s\to\infty}\frac{\kappa(s)}{s\cdot\tau(s)}=0$$ then the curve is unbounded. This is an almost trivial fact, it follows from some another almost trivial theorem, for details see http://www.ma.utexas.edu/mp_arc/c/16/16-63.pdf
The comments are welcome. Particularly is there a geometrical interpretation of brought above proposition
Last edited: