I've gotten as far as $\mathbf{n''}+(\kappa^2+\tau^2)\mathbf{n} = 0$, which suggests or at least permits trigonometric expressions for every component of $\mathbf{n}$ -- not something that seems to lead to a standard expression for a helix. Am I just supposed to prove that a helix has constant nonzero curvature and torsion and invoke the fundamental theorem of curves?

ETA: I was able to get further than the point I described, but the equation for the curve I came up with was...


...and I have no idea how to prove that that's a helix, plus I suspect it isn't even right.


You can use the Frenet-Serret formulas to find the parametric equations of the curve with constant nonzero curvature and torsion. This gives you a system of three linear differential equations. It's straightforward (albeit a bit tedious) to solve.

However, if you have the fundamental theorem of curves established, all what you need is to show that for every constant nonzero curvature and torsion, there is a helix that has such values. From there, you know that all other curves with the same curvature and torsion are the same helix (up to an isometry).

  • $\begingroup$ Trying to solve the equations led me to an equation for the curve with no fewer than 12 unknown constants and trigonometric expressions for all three dimensions. I think the big problem for me might be proving that this equation does represent a helix, but I suspect I took a wrong turn a few steps back, which was why I said I was at the point I did. The teacher gave us the solution for y''+(c^2)y=0 as part of the problem, which suggests my track was the right one. $\endgroup$ – Shay Guy Sep 16 '12 at 21:57
  • $\begingroup$ @ShayGuy I solved the equations when I learned the Frenet-Serret formulas as an exercise. It was a while back but I remember that it was quite tedious. If you keep all of integration constants, you'll end up with a general parametrization that covers all possible isometries. Not a pleasant one! $\endgroup$ – Ayman Hourieh Sep 16 '12 at 22:05
  • $\begingroup$ Like I said, I've got a parametrization. I just don't see how to prove that it's a helix. Do I need to define a new orthonormal basis? $\endgroup$ – Shay Guy Sep 16 '12 at 22:33
  • $\begingroup$ @ShayGuy Apply an isometry to the helix parametrization and compare with yours. $\endgroup$ – Ayman Hourieh Sep 16 '12 at 22:37
  • $\begingroup$ I don't think we've covered isometries. At least, they're not in the pseudo-textbook we're using yet. $\endgroup$ – Shay Guy Sep 16 '12 at 22:46

Take a look at section 1.18 on page 2/5 of this PDF.


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