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!