# Proving a couple things about $\beta(t) = \alpha(t) - s(t)\cos(\theta)u$, where $\alpha$ is a helix. Is this correct?

Let $$\alpha: I \mapsto \mathbb{R^3}$$ be a helix and $$u$$ be the constant unit vector whose angle $$\theta$$ with $$\alpha'(t)$$ is constant. Let $$s(t)$$ be the arclength function of $$\alpha$$ starting at $$t = 0$$. Consider the curve $$\beta(t) = \alpha(t) - s(t)\cos(\theta)u$$ and prove that:

a) $$\beta$$ is contained in the plane that goes through $$\alpha(0)$$ and is orthogonal to $$u$$

b) Prove that $$K_\beta = \cfrac{K_\alpha}{\sin^2(\theta)}$$

b) is easy, but I'm having some trouble with a). Everything I tried got hairy pretty fast. I think there might be a typo and I should consider $$\beta'$$ as the curve contained in the mentioned plane, 'cause I've already proven it in that case and it makes much more sense, but I'm not sure.

Progress: I think I got it. Assuming, WLOG, that $$\alpha$$ is a unit speed curve, since $$\beta'(t) \cdot u =( \alpha'(t) - s'(t)\cos(\theta)u)\cdot u = \cos(\theta)-\cos(\theta) = 0$$, we can integrate and find that $$\beta(t) \cdot u = c$$... and for anyone seeing this who wants the answer, see the discussion in the comments, because the previous update had an error.

Your computation isn't quite right, as $\alpha'(t)\cdot u = s'(t)\cos\theta$. (Recall that $s'(t)$ is the speed with which the particle moves.) [You also never defined $\theta$ in your question!]
Note that $s(t)$ is the arclength starting at $t=0$, so $s(0)=0$. This shows that $\beta(0)=\alpha(0)$. Now, can you show that $\big(\beta(t)-\alpha(0)\big)\cdot u = 0$ for all $t$?
• Here I assumed the curve $\alpha$ is parameterized by arclength, so $s(t) = \int_0^t ||\alpha'(s)|| \ ds$, whence $s'(t) = 1$, so $\alpha'(t) \cdot u = \cos(\theta)$. I'll fix the question, sorry, $\theta$ is the angle between the unit vector $v$ and $\alpha'(t)$. I thought I already showed that $(\beta(t) - \alpha(0)) \cdot u = 0$ in my original computation, since $\beta'(t) \cdot u = 0$ , so $\beta(t) \cdot u$ is constant, and since it is $0$ at $t = 0$ it must be zero everywhere, from which follows $(\beta(t) - \alpha(0))\cdot u = 0, \forall t \in I$ Are there any mistakes here? Feb 5, 2018 at 17:35
• You're assuming things that are wrong!! Your argument only works if $\alpha(0)$ lies in the plane through the origin with normal $u$. Feb 5, 2018 at 18:06