Multivariable calculus: constant curvature

[Nikitin]

If the curvature is always constant and >0 for a parametrized curve C, does it automatically mean the curve is a circle?

 

[WannabeNewton]

No. Let ##\gamma:\mathbb{R}\rightarrow \mathbb{R}^{3}## be the regular curve given by ##\gamma(t) = (r\cos t,r\sin t, ct)## where ##r## is a positive constant and ##c## is another constant. The extrinsic curvature of this curve as embedded in ##\mathbb{R}^{3}## is given by ##\kappa = \frac{r}{r^{2} + c^{2}}## which is always positive and constant. ##\gamma## is called a helix; ##r## is the radius of the helix and ##2\pi c## is a constant giving the vertical separation of the successive loops of the helix.

EDIT: just to be complete, if the torsion of the curve ##\gamma: J \rightarrow \mathbb{R}^{3}## is zero i.e. ##\gamma## lies on a plane, and ##\gamma## has constant positive curvature then it must be a circle. This is easily proven. If the torsion is zero then the Frenet-Serret formulas give us ##\frac{\mathrm{d} N}{\mathrm{d} s} = -\kappa T##. Recall that the center of the osculating circle to ##\gamma## at some ##s\in J## is given by ##p(s) = \gamma(s) + \frac{1}{\kappa}N(s)##. Note that ##\frac{\mathrm{d} p}{\mathrm{d} s} = \frac{\mathrm{d} \gamma}{\mathrm{d} s} + \frac{1}{\kappa}\frac{\mathrm{d} N}{\mathrm{d} s} = T + \frac{1}{\kappa}(-\kappa T) = 0 ## and that ##\left \| p(s) - \gamma(s) \right \| = \frac{1}{\kappa} = \text{const.} \forall s\in J## hence ##\gamma## must be a circle because the center of the osculating circle to ##\gamma## never changes and the distance from this fixed center to the curve is always constant and is given by ##1/\kappa## i.e. the osculating circle is the curve itself.

 

[Nikitin]

Ah yes, you're completely right. Thanks!