# Prove that the curvature of $\gamma$ is $\frac{\kappa_{\alpha}}{\sin^2\theta}$

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.

• Just want to confirm that t is NOT a arc-length parameter? – Xipan Xiao Jan 2 '15 at 16:43
• If it is arc-length parameter, the curvature is just $|\alpha''|$ and $\theta$ is constant. – Xipan Xiao Jan 2 '15 at 17:19
• @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 $|\alpha''-(\alpha''\cdot u)u|$ – pxc3110 Jan 2 '15 at 17:29
• $const=u\cdot \alpha'=1\cdot 1\cdot \cos\theta=\cos\theta$ and $k_\alpha=|\alpha''|$ – Xipan Xiao Jan 2 '15 at 17:36
• And you need to state it explicitly in your post whether t is an arc-length parameter or not. It makes much difference. – 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 $|\beta''|=|\gamma''/\sin^2\theta|=|\alpha''/\sin^2\theta|=k_\alpha/\sin^2\theta$
• 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 $tan\theta$? – pxc3110 Jan 3 '15 at 1:08