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Let $r(t)$ be a unit speed curve such that for all $t$, $\frac{\tau(t)}{\kappa(t)}=\cot(\theta)$ for some $0 < \theta < \pi$. Show that there is a constant vector $a$ satisfying $T(t) * a = \cos(\theta)$ for all $t$

(Where $\tau$ is torsion, $\kappa$ is curvature, and T is the tangent in the sense of Frenet Formula)

I noticed that if given $T(t)*a$ equal to some constant, then taking the derivitive gives $N(t)*a=0$ (Where N is the normal in Frenet). On another problem similar to this, I saw $a$ constructed such that if $Ta=\alpha$ and $||a||^2=\alpha^2+\beta^2$

$a=\alpha T(t) + \beta B(t)$ (B(t) being the Binormal) and thus, taking the derivitive $0=\alpha \kappa N(t) - \beta \tau N(t)$

From here, $\frac{\beta}{\alpha} \frac{\tau}{\kappa} N(t)=N(t)$. Taking the norm of both sides gives $\frac{\beta}{\alpha} \cot(\theta)=1$ So $\alpha$ could be $\cos(\theta)$ if $\beta$ was $\sin(\theta)$ but that is not necessarily the unique case. Obviously I'm missing some logic in here, but after playing around with it for a while I can't seem to get on the right track. If anyone can help me out doing it like this, or even a completely different way if that is simpler and I'm totally on the wrong track.

Edit: Maybe if we somehow knew that $||a||^2$=1? Just trying to think of something.

James Snyder
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  • If you're proving existence of something, it suffices to construct an example (e.g., using your formula with $\alpha$ and $\beta$ suitably-chosen). You needn't prove uniqueness (i.e., you needn't show that if some choice of $a$ "works", then it's necessarily some particular choice). – Andrew D. Hwang Mar 04 '14 at 14:25
  • Alright, that makes sense. So otherwise my methodology is correct? – James Snyder Mar 05 '14 at 00:22
  • I didn't check carefully, but on a quick skim your work looks reasonable. – Andrew D. Hwang Mar 05 '14 at 00:39
  • Though looking at my work now, it seems that the a I made is not necessarily $a$ constant vector. – James Snyder Mar 05 '14 at 05:05
  • Your calculation that $da/dt = 0$ looks good on closer inspection, and certainly $T \cdot a = \cos\theta$; are there particular doubts you have? (For a complete argument, you do want to mention that since $\tau/\kappa = \cot\theta$ is real, you have $\kappa > 0$, so the Frenet trihedron is uniquely defined and satisfies the Frenet-Serret equations.) – Andrew D. Hwang Mar 05 '14 at 13:39
  • After going over it again I do see now that $a$ is indeed a constant vector, I had written something out wrong when verifying $a'=0$ so I'm definitely done, thanks for the help. – James Snyder Mar 05 '14 at 14:32
  • I think you did not prove it since you assumed first that $T(t)*a$ is constant which is not given. Also, the ratio in the second paragraph supposed that $\cos \theta=\frac{\alpha}{\alpha^2+\beta^2}$ not $\alpha$ – Semsem Mar 06 '14 at 04:30

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