Showing whether a regular curve in R3 with constant positive curvature is always in a sphere I'm having trouble with showing:
Let $\alpha$ be a regular curve in $R^3$ parametrized by arc length with constant positive curvature. Is the curve always in a sphere?
I already know that such curve is a planar curve iff it is (a part of) a circle
But I don't know how to start proving that the curve is in a sphere
Thanks a lot :)
 A: The curve needn't lie in a sphere.
Consider
$\gamma(t) = (\alpha \cos kt, \alpha \sin kt, \beta t); \tag 1$
then
$\dot \gamma(t) = (-\alpha k \sin kt, \alpha k \cos kt, \beta); \tag 2$
denoting arc-length by $s$, the speed is then
$\dfrac{ds}{dt} = \sqrt{\dot \gamma(t) \cdot \dot \gamma(t)} = \sqrt{\alpha^2 k^2 \sin^2 kt + \alpha^2 k^2 \cos^2 kt + \beta^2} = \sqrt{\alpha^2 k^2 + \beta^2}, \tag 3$
a constant independent of $t$.  Thus, if we intitialize $s = 0$ at $t = 0$, we have
$s = \sqrt{\alpha^2 k^2 + \beta^2} t, \tag 4$
whence
$t = \dfrac{s}{\sqrt{\alpha^2 k^2 + \beta^2}}, \tag 5$
and we may express the tangent vector $\dot \gamma$ in terms of $s$ as
$\dot \gamma(s) = (-\alpha k \sin \dfrac{ks}{\sqrt{\alpha^2 k^2 + \beta^2}}, \alpha k \cos \dfrac{ks}{\sqrt{\alpha^2 k^2 + \beta^2}}, \beta); \tag 6$
now by (3) the unit tangent vector $T(s)$ is
$T(s)$$
= (-\dfrac{\alpha k}{\sqrt{\alpha^2 k^2 + \beta^2}} \sin  \dfrac{ks}{\sqrt{\alpha^2 k^2 + \beta^2}}, \dfrac{\alpha k}{\sqrt{\alpha^2 k^2 + \beta^2}} \cos  \dfrac{ks}{\sqrt{\alpha^2 k^2 + \beta^2}}, \dfrac{\beta}{\sqrt{\alpha^2 k^2 + \beta^2}}); \tag 7$
then 
$\kappa(s) N(s) = \dot T(s) = (-\dfrac{\alpha k^2}{\alpha^2 k^2 + \beta^2} \cos  \dfrac{ks}{\sqrt{\alpha^2 k^2 + \beta^2}}, -\dfrac{\alpha k^2}{\alpha^2 k^2 + \beta^2} \sin  \dfrac{ks}{\sqrt{\alpha^2 k^2 + \beta^2}}, 0); \tag 8$
it follows that
$\kappa^2 = (\kappa N) \cdot (\kappa N) = \dfrac{\alpha^2 k^4}{(\alpha^2 k^2 + \beta^2)^2}, \tag 9$
whence
$\kappa = \dfrac {\vert \alpha \vert k^2}{\alpha^2 k^2 + \beta^2}, \tag{10}$
is the constant curvature of the curve $\gamma(t)$ given in (1).  But $\gamma(t)$ cannot be contained in any sphere, since it becomes undbounded as $t \to \infty$.
$\gamma(t)$ is in fact, a helix, an example Lord Shark the Unknown pointed out in his comment to the question itself.
