# Prove that for any $\kappa$ > 0 and $\tau$ there is an helix with such numbers as curvature and torsion respectively

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

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.