# Showing another form of a curve $\alpha(s)$ parametrized by arc-length

I have recently come across a question and am looking for some advice as to how to approach it. The question reads:

Let $\alpha(s)$, $s=$ arc length, be a curve whose torsion $\tau$ is non-zero constant, say $\tau=\frac{1}{a}$. Show that $\alpha$ can also be expressed as: $$\alpha(s)=a\int g(s)\wedge g'(s)ds$$ for some vector valued function $g$ satisfying $|g(s)|=1$, and also $(g\wedge g')\cdot g''\neq0$.

I have tried to use the torsion formula for an arc-length parametrized curve: $$-\frac{(\alpha'\wedge\alpha'')\cdot\alpha'''}{|\kappa(s)|^2}$$ Where $\kappa(s)=|\alpha''(s)|$ is the curvature of the curve $\alpha$. I'm confused by where the integral comes from basically. I think the curve $g$ is just the derivative of $\alpha$, but like I said, I'm confused by where the integral comes from or how to derive it. If anyone could offer some advice it would be appreciated!

For this specific curve, figure out what the "$g$" function would be. Draw some pictures and think about what cross products mean.
Then, I expect you'll be able to figure out how to invent a "$g$" function for more general curves.
• I am trying to use $\alpha(s)$ as the helix, and differentiated both sides of the equation to reach: $$\alpha'(s)=a\cdot (g(s)\wedge g'(s))$$ So would I just solve this equation to find $g$? Apr 13 '17 at 16:50