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A curve is said to be helix if its tangent line have a constant angle with a fixed direction. i.e. $\langle T(t),u\rangle$ is constant for some unit vector $u$.

I am trying to prove: a regular curve with $\kappa(t)>0$ is helix if and only if $\frac{\kappa}{\tau}$ is constant.

I can show the sufficient part： if $\alpha(t)$ is helix, then $\langle T'\!, u\rangle=0$ and therefore $\langle N,u\rangle=0$, which implies $\langle-\kappa(t)T(t)+\tau(t)B(t), u\rangle=0$. We then have $$ \frac{\kappa(t)}{\tau(t)} = \frac{\langle T(t),u\rangle}{\langle B(t),u\rangle}, $$ which is constant, since $\langle T(t),u\rangle,\langle B(t),u\rangle$ are constant.

But I have no idea in proving the reverse direction.

Sammy Black
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xyz
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1 Answers1

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This is a proof for the reverse direction, and is largely inspired from "Differential Geometry" by Martin M. Lipschutz. For the reverse direction, we're essentially trying to show that the angle made is constant, i.e. trying to backwards solve for our unit vector $u$. So, we start off by writing that $$\frac{k}{\tau} = \tan \alpha = \frac{\sin \alpha}{\cos \alpha}.$$ Note that using tangent as the value for our constant is allowed because $\operatorname{ran} \tan = \mathbb{R}$. Next, we multiply out and see $$k \cos \alpha - \tau \sin \alpha = 0,$$ and then, by the Frenet Equations, $$|t'|\cos \alpha - |b'|\sin \alpha = 0.$$ We now integrate and see that $$|t|\sin \alpha - |b| \cos \alpha = u.$$ Note that in the last line I implicitly assumed that $t$ parameterized $\alpha$ by arc length, which is allowed. We now want to know what the magnitude of $u$ is, so we compute it to be \begin{align*} |u| &= \sqrt{|t|^2 \sin^2 \alpha + |b|^2 \cos^2 \alpha} \\ &= \sqrt{\sin^2 \alpha + \cos^2 \alpha} \\ &= 1. \end{align*} So, this means that $u$ is a constant unit vector. Thus, $$\langle t, u \rangle = |t|\cdot |u| \cos \alpha = \cos \alpha$$ which is a constant angle. Thus, it's a helix.

sidj
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