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i'm trying this problem, but the approaches i find are a little bit tedious. First, i wrote the parametric equation of a helix, and the i used it to find the curvature and the torsion according to the well-known formulas for those numbers without the arc length parametrization. I find those and for me will be suffice to show that those functions are surjective. The other approach is using Frenet-Serret formulas but this is indeed more tedious than the former. Is there an intuitive way to figure out this?

Thank you

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I suppose that you know that for a helix parametrized as $(r\sin t,r\cos t, ct)$, you have $$ \kappa = \frac{r}{r^2+c^2}\\ \tau = \frac{c}{r^2+c^2} $$ All we need to do is show that one can solve for $c,r$ as functions of $\kappa,\tau$ given the above simultaneous equations.

One way to do so is to go through a sort of change of variables. Define $$ \theta=\arctan\left(\frac{c}{r}\right)\\ \rho=\sqrt{c^2+r^2} $$ Knowing that this change is invertible, we see that our original set of equations is rendered as $$ \kappa = \frac{\cos\theta}{\rho}\\ \tau = \frac{\sin\theta}{\rho} $$ Solving this, we have $$ \theta=\arctan\left(\frac{\tau}{\kappa}\right)\\ \rho=\frac{1}{\sqrt{\tau^2+\kappa^2}} $$


In retrospect, the substitution was entirely unnecessary. Starting with the system $$ \kappa = \frac{r}{r^2+c^2}\\ \tau = \frac{c}{r^2+c^2} $$ We divide the two equations to find $$ \frac{\kappa}{\tau}=\frac{r}{c} $$ And add the squares to find $$ \kappa^2+\tau^2=\frac{1}{(r^2+c^2)^2} $$ That is, we have the system $$ r=c\frac{\kappa}{\tau}\\ r^2+c^2=\frac1{\sqrt{\kappa^2+\tau^2}} $$ Which can be solved directly via substitution.

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  • $\begingroup$ Uhh, thank you very much, i was missing the change of variables, is there any book you can recommend me on that? $\endgroup$ – Sebastian Valencia Aug 27 '13 at 20:26
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    $\begingroup$ Sorry, I don't really have one to recommend; that was simply the first thing that occurred to me when I saw the problem. Here's a nice free textbook on differential geometry by Shiffrin at UGA. $\endgroup$ – Omnomnomnom Aug 27 '13 at 20:59

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