Explicit curve with unbounded torsion

Question

I have been trying to experiment with curves. Recently I found a curve (in $\mathbb{R}^2$) with unbounded curvature.

However, I am not able to find one curve with unbounded torsion, for various reasons:

• It has to be in $\mathbb{R}^3$, which makes things more complicated
• Torsion is not as easy to calculate as curvature, I'm aware of the formula $\displaystyle\frac{(\alpha''\times\alpha')\cdot \alpha'''}{k^2_{\alpha}}$

I know the existence of this curves because of the Fundamental Theorem, but I would like to find one explicitly.

Can anyone guide/help me?

Answer 1

Take $\gamma\colon (0,+\infty) \to \Bbb R^3$ defined by $$ \gamma(t) = \left(\frac{1}{t}\cos t^2, \frac{1}{t}\sin t^2, t\right). $$ Then for all $t >0$, \begin{align} \gamma'(t) &= \left(-2 \sin t^2 -\frac{1}{t^2}\cos t^2, 2\cos t^2 - \frac{1}{t^2}\sin t^2, 1\right) ,\\ \gamma''(t) &= \left(-4t\cos t^2 + \frac{2}{t} \sin t^2 + \frac{2}{t^3} \cos t^2, -4t\sin t^2 - \frac{2}{t} \cos t^2 + \frac{2}{t^3} \sin t^2, 0\right),\\ \gamma'''(t) &= \left(8t^2\sin t^2 -\frac{6}{t^2}\sin t^2 -\frac{6}{t^4} \cos t^2, -8t^2\cos t^2 + \frac{6}{t^2}\cos t^2 - \frac{6}{t^4} \sin t^2, 0 \right), \end{align} and you can check that, whenever it makes sense, the torsion is $$ \tau(t) = \frac{8t^9 - 10 t^5}{20t^8 - 11t^4 + 2} \sim_{t\to\infty} \frac{2}{5}t, $$ and thus is unbounded.

Answer 2

If we plot Eqn={u Cos[t],u Sin[t], a t} in 3d we see a linear helicoid whose $u=0$ central line has a constant torsion. The spine twists as can be seen by the rotation of binormal ( not shown).

However, if the third parametric coordinate $z$ red line is kept as any monotonously increasing function then its torsion also increases monotonously as evident in this particular parametrized example:

Eq={u Cos[t],u Sin[t],1+Sinh[t]/5}
    ParametricPlot3D[Eq,{u,0,6},{t,0,1.8Pi},Mesh->{3,55},
    PlotStyle->{Yellow},Axes->None,Boxed->False]