# How to prove this curve lies in the plane through $\alpha(0)$ orthogonal to $u$?

Let $\alpha$ be cylindrical helix with unit vector $u$, angle $\theta$, and arc length $s$ (measured from $0$). The only curve $\gamma$ such that $$\alpha(t)=\gamma(t)+s(t)\cos(\theta)u$$ is called the cross-section curve of the cylinder where $\alpha$ lies.

$\bf (a)$ How to show $\gamma(t)$ lies on the plane through $\alpha(0)$ orthogonal to $u$? I know I must show $$\langle \gamma(t)-\alpha(0), u\rangle=0$$ for every $t$, however I wasn't able to do that yet, maybe I'm missing some property of this kind of curve.

Obs A cylindrical helix is a curve $\alpha$ such that $\langle \alpha^{'}(t), u\rangle=\cos(\theta)$ for every $t$, where $u$ is fixed unit vector.

$\bf (b)$ How to show the curvature of $\gamma$ is $\kappa/\sin^2(\theta)$ where $\kappa$ is the curvature of $\alpha$?

• We have to show that $\langle \gamma'(t)-\alpha(0),u\rangle=0$. – DiegoMath Apr 26 '14 at 21:07

Suppose $\alpha$ parametrized by the arc length, remember that $$s(t)=\int_0^t|\alpha'(s)|\,ds,$$ hence $s'(t)=|\alpha'(t)|=1$.
Now $$\begin{matrix} \langle\gamma'(t)-\alpha(0),u\rangle&=&\langle\alpha'(t)-s'(t)\cos(\theta) u-\alpha(0),u\rangle\\ &=&\langle\alpha'(t),u\rangle-\langle s'(t)\cos(\theta)u,u\rangle-\langle\alpha(0),u\rangle\\ &=&\cos(\theta)-\cos(\theta)\langle u,u\rangle-0=0. \end{matrix}$$ We use $s'(t)=1$ and the fact that the vector $\alpha(0)$ lies in the plane ortogonal to $u$.
As to second part I did a proof. Since $\gamma(t)=\alpha(t)-s(t)\cos(t)u$,$$\gamma^{'}(t)=\alpha^{'}(t)-\cos(\theta)u\quad \textrm{and}\quad \gamma^{''}(t)=\alpha^{''}(t).$$ Hence:$$\gamma^{'}(t)\times \gamma^{''}(t)=\alpha^{'}(t)\times \alpha^{''}(t)-\cos(\theta)(u\times \alpha^{''}(t)).$$ Then $$\|\gamma^{'}(t)\times \gamma^{''}(t)\|^2=\|\alpha^{'}\times \alpha^{''}\|^2-2\cos(\theta)\langle \alpha^{'}\times \alpha^{''}, u\times \alpha^{''}\rangle+\cos(\theta)^2\|u\times \alpha^{''}\|^2.$$ But $$\|\alpha^{'}\times \alpha^{''}\|^2=\|\alpha^{'}\|^2\|\alpha^{''}\|^2\sin(90)=\kappa^2,$$ and analogously, $$\|u\times \alpha^{''}\|^2=\kappa^2.$$ Also, notice, $$\langle \alpha^{'}\times \alpha^{''}, u\times \alpha^{''}\rangle=\cos(\theta)\kappa^2.$$ Hence $$\|\gamma^{'}(t)\times \gamma^{''}(t)\|^2=\kappa^2(1-\cos^2(\theta))=\kappa^2\sin^2(\theta).$$ Finally, it is easy to see, $$\|\gamma^{'}\|=\sin^2(t).$$ Therefore, $$\kappa_\gamma=\frac{\|\gamma^{'}\times \gamma^{''}\|}{\|\gamma^{'}\|^3}=\frac{\kappa \sin(\theta)}{\sin^3(\theta)}=\frac{\kappa}{\sin^2(\theta)}.$$