Let $\alpha:I\to {\mathbb R}^3$ be a cylindrical helix with a unit vector $u$ such that $u\cdot T_{\alpha}$ is a constant for all $t\in I$. For $t_0\in I$, the curve $$\gamma(t)=\alpha(t)-((\alpha(t)-\alpha(t_0))\cdot u)u$$ is called a cross-sectional curve of the cyliner on which $\alpha$ lies. Prove that the curvature of $\gamma$ is $\frac{\kappa_{\alpha}}{sin^2\theta}$, where $\kappa_{\alpha}$ is the curvature of $\alpha$, and $\theta$ is the angle between $T_{\alpha}$ and $u$

I've proved that:


so we're trying to show that:


Expressing both sides in terms of their derivatives, we have:


Since $\gamma'=\alpha'-(\alpha'\cdot u)u$ and $\gamma''=\alpha''-(\alpha''\cdot u)u$, we get:

$$\frac{|(\alpha'-(\alpha'\cdot u)u)\times{\alpha''}-(\alpha''\cdot u)\alpha'\times u|}{|\alpha'-(\alpha'\cdot u)u|^3}=\frac{|\alpha'\times\alpha''|}{|\alpha'|^3}+\frac{|\alpha'|^3}{|\alpha'\times\alpha''|}\frac{((\alpha'\times\alpha")\cdot{\alpha'''})^2}{|\alpha'\times\alpha"|^4}$$

Because of the complexity of the equation, I think I should approach to it at some other perspective. I also believe that the conclusion $\sin \theta=\frac{\kappa_{\alpha}}{\sqrt{{\kappa_{\alpha}}^2+\tau_{\alpha}^2}}$ should still be utilized. I hope someone could give me a clue.

  • $\begingroup$ Just want to confirm that t is NOT a arc-length parameter? $\endgroup$ – Xipan Xiao Jan 2 '15 at 16:43
  • $\begingroup$ If it is arc-length parameter, the curvature is just $|\alpha''|$ and $\theta$ is constant. $\endgroup$ – Xipan Xiao Jan 2 '15 at 17:19
  • $\begingroup$ @XipanXiao The fact that $\theta$ is constant is not implied by assuming that $t$ is an arc length parameter, instead, this is implied by the fact that $\alpha$ is a cylindrical helix. Also why is the curvature just $|\alpha''|$ when assuming that $t$ is an arc length paramenter? I tried and get $|\alpha''-(\alpha''\cdot u)u|$ $\endgroup$ – pxc3110 Jan 2 '15 at 17:29
  • $\begingroup$ $const=u\cdot \alpha'=1\cdot 1\cdot \cos\theta=\cos\theta$ and $k_\alpha=|\alpha''|$ $\endgroup$ – Xipan Xiao Jan 2 '15 at 17:36
  • $\begingroup$ And you need to state it explicitly in your post whether t is an arc-length parameter or not. It makes much difference. $\endgroup$ – Xipan Xiao Jan 2 '15 at 17:38

If t is an arc-length parameter, $\gamma'=\alpha'-(\alpha'\cdot u)u=\alpha'-cu$ where $c=\alpha'\cdot u=\cos\theta$ is constant as assumed. So $\gamma''=\alpha''$ and $|\gamma'|^2=|\alpha'|^2+c^2-2c|\alpha'|\cos\theta=|\alpha'|^2-c^2=1-\cos^2\theta=\sin^2\theta$
So let $\beta(s)=\gamma(s/\sin\theta)$, $s$ is an arc-length parameter and the curvature is $|\beta''|=|\gamma''/\sin^2\theta|=|\alpha''/\sin^2\theta|=k_\alpha/\sin^2\theta$

  • $\begingroup$ My bad, it's not an arc length parameter. My English's not good enough :) $\endgroup$ – pxc3110 Jan 2 '15 at 18:36
  • $\begingroup$ By the way, do you know that a curve is called a cylindrical curve if an only if $\frac{\kappa}{\tau}$ is a constant, which in this case equals to $tan\theta$? $\endgroup$ – pxc3110 Jan 3 '15 at 1:08
  • $\begingroup$ @TedShifrin: got it. Didn't know there are so many types of helixes. $\endgroup$ – Xipan Xiao Jan 3 '15 at 1:18

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