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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.

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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 $t$?

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  • $\begingroup$ 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? $\endgroup$
    – Matheus Andrade
    Feb 5, 2018 at 17:35
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    $\begingroup$ You're assuming things that are wrong!! Your argument only works if $\alpha(0)$ lies in the plane through the origin with normal $u$. $\endgroup$
    – Ted Shifrin
    Feb 5, 2018 at 18:06
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    $\begingroup$ Just make the mild correction to do what I said in the second paragraph? You need to review what the equation of a plane through a given point is. $\endgroup$
    – Ted Shifrin
    Feb 5, 2018 at 18:11
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    $\begingroup$ Righto!! Sometimes you need to absorb what I write to you before you respond. :) $\endgroup$
    – Ted Shifrin
    Feb 5, 2018 at 18:23
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    $\begingroup$ I do tend to speak/type faster than I think. I'll try to stop that habit, sorry :). Thanks for your patience. $\endgroup$
    – Matheus Andrade
    Feb 5, 2018 at 18:24

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