Are the tangent lines to a helix disjoint?

128 Views Asked by At

Let $H$ be the helix parametrized by $t \mapsto (\cos(t), \sin(t), t) : \mathbb{R} \to \mathbb{R}^3$. It seems like any two distinct tangent lines to $H$ will be disjoint – in fact I thought this was so clear that I wrote a problem on a worksheet asking students to find a curve whose distinct tangent lines are disjoint and just assumed this would be a solution. Now, as I'm writing the solutions, I'm having trouble actually showing that the tangent lines to $H$ are pairwise-disjoint.

Is there an easy way to see that any two distinct tangent lines to $H$ must be disjoint? Or, if this is false, is it easy to see why?

1

There are 1 best solutions below

4
On BEST ANSWER

Updated answer based on comments by @diracdeltafunk: the tangent lines are not all disjoint.

By symmetry, it suffices to show that there are distinct tangent lines that intersect the line tangent to $(1,0,0)$.

The tangent line at $(1,0,0)$ can be parametrized as $\mathbf r_0(a) = (1,a,a)$. The tangent line at any other point $(\cos(t),\sin(t),t)$ can be parametrized as $\mathbf r_t(b) = (\cos(t) - b\sin(t),\sin(t)+b\cos(t),t+b)$. For these two lines to intersect, we require there to exist $a,b$ so that $$ \begin{cases} \cos(t) - b\sin(t) = 1 \\ \sin(t) + b\cos(t) = a \\ t + b = a. \end{cases}\tag{$\ast$} $$ The last equation reduces us to asking $$ \begin{cases} \cos(a-b) - b\sin(a-b) = 1 \\ \sin(a-b) + b\cos(a-b) = a, \\ \end{cases} $$ or equivalently that $$ \begin{pmatrix} \cos(a-b) & -\!\sin(a-b) \\ \sin(a-b) & \cos(a-b)\end{pmatrix}\begin{pmatrix}1 \\ b\end{pmatrix} = \begin{pmatrix} 1 \\ a\end{pmatrix}. $$

Since the matrix appearing in the last equation is a rotation matrix, it must be the case that $|b| = |a|$. Taking $a = b$ in ($\ast$) corresponds to the solution $\mathbf r_a = \mathbf r_0$. If we take $a = -b$ and let $a$ be a real number such that $a = \tan(a)$ (there are infinitely many such $a$), then we can show that $\mathbf r_0(a) = \mathbf r_{2a}(-a)$. One way to see this is to use the double-angle identities and the fact that for such an $a$, $a-\arctan(a)$ is an integer-multiple of $\pi$. This exhibits infinitely many distinct tangent lines (the $\mathbf r_{2a}$) that intersect any given tangent line ($\mathbf r_0$).