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 $\alpha$?
Best Answer
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 $u$.