2
$\begingroup$

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:

$$\sin\theta=\frac{\kappa_{\alpha}}{\sqrt{{\kappa_{\alpha}}^2+\tau_{\alpha}^2}}$$

so we're trying to show that:

$$\kappa_{\gamma}=\kappa_{\alpha}+\frac{\tau_{\alpha}^2}{\kappa_{\alpha}}$$

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

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

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $|\beta''|=|\gamma''/\sin^2\theta|=|\alpha''/\sin^2\theta|=k_\alpha/\sin^2\theta$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.