A tricky tractrix question about vertical tangents

Question

This is raised by a recent question occurring in combinatorial geometry. It is about a sort of tractrix, but instead of a line, the pulling end moves along a circle of radius $r>\frac12$ (typically around $0.8$), starting with a chord if length 1. Imagine this chord horizontally on the bottom and take its center as the origine of a cartesian coordinate system. If the circle has center $(0,h)$, we have $h^2+\frac14=r^2$, so only one of $h$ and $r$ is needed to define the tractrix.

The tractrix is approximately the green line in the picture, where all blue segments (the tangents) have unit length.
Here is what I have got so far :

For a point $(x,y)$ on the tractrix where it has slope $f’$, we have 3 equations for the point $(u,v)$ where the tangent intersects the circle on the right:
$(i)\ \ \ \ v-y=(u-x)f’\ $ (equation of the tangent)
$(ii) \ \ \ u^2+(h-v)^2=r^2\ $ (intersection with the circle), equivalently $u^2+v^2-2hv-\frac14=0$. $(iii)\ \ (u-x)^2+(v-y)^2=1\ $ (constant length of the tangent between tractrix and circle).

Eliminating $u$ and $v$ yields the following differential equation for the tractrix: $$x+\sqrt{1-s^2}=\sqrt{2h(y+s)-(y+s)^2-\frac14}$$ where $$s :=\frac{f'}{\sqrt{1+f’^2}}.$$ (Thus $s$ is the sine of the slope angle.)

I don’t think there is a closed form of the tractrix equation. But is there a way to determine at which point $(x,y)$ the tangent is vertical? I'd expect it to be a not-too-involved function of $h$.

Note that, unless $r$ is too small for reaching the vertical position at all, the tractrix will carry on winding beyond the vertical for a finite time. That is, under the assumption that at each moment, the segment is tangent to the tractrix at its (the segment's) endpoint. (Imagine $r$ a bit smaller than in the picture, such that the vertical tangent of the tractrix coincides with the $y$-axis. From there on, the top of the blue segment cannot move further to the left without "breaking the smoothness" of the tractrix. I think that for each $r$, the movement will eventually arrive at such a point, which we will naturally consider as the endpoint of the tractrix.)

But where is that point? I have no idea how far it can go if $r$ is big, but I don't think it will go further than becoming horizontal again. All this is easier to perceive if we keep $r$ constant and require the 'rotating' segment of length $\epsilon$ instead of unit length. So:

Where does the tractrix stop if $\epsilon\to0$?

Answer 1

In fact, using the moving frame, it is easy explicitly to solve the equations and get the formula for the slope $\tan\bigl(\theta(s)\bigr)$ as a function of arc-length along the curve. However, one sees immediately that it is a transcendental function, so explicitly solving for the value of $s$ for which $\theta(s) = \tfrac12\pi$ is not going to be easy.

Here are the steps: Let $$ X(s) = \pmatrix{x(s)\\y(s)},\qquad \text{with}\qquad X(0) = \pmatrix{-\tfrac12\\-\sqrt{r^2-\tfrac14}} $$ be the arclength parametrization of the curve, and set $$ E_1 = X'(s) = \pmatrix{\cos\bigl(\theta(s)\bigr)\\\sin\bigl(\theta(s)\bigr)} \quad\text{and}\quad E_2 = \pmatrix{-\sin\bigl(\theta(s)\bigr)\\\cos\bigl(\theta(s)\bigr)}. $$ We then have structure equations $X'(s) = E_1(s)$, $E_1'(s) = \kappa(s)E_2(s)$, and $E_2'(s) = -\kappa(s)E_1(s)$, where $\kappa(s) = \theta'(s)$. We also have $\theta(0) = 0$, by the above normalization.

We are requiring that $\bigl|X(s)+E_1(s)\bigr| = r$. (This is the tractrix equation.)

Let us set $$ X(s) = u(s)\ E_1(s) + v(s)\ E_2(s), $$ where the initial conditions imply $u(0) = -\tfrac12$ and $v(0) = -\sqrt{r^2-\tfrac14}$.

Then the above structure equations imply, since $X'(s) = E_1(s)$, that $$ u'(s) = 1 + \kappa(s) v(s)\qquad\text{and}\qquad v'(s) = -\kappa(s) u(s). $$

Moreover, the tractrix equation implies that $(u(s)+1)^2 + v(s)^2 = r^2$. Differentiating this equation and using the above differential equations then yields $$ 1 + u(s) + \kappa(s)v(s) = 0, $$ which implies $u'(s) = -u(s)$, which, with the initial condition above implies $$ u(s) = -\tfrac12\,e^{-s} $$ and then, consequently, that $$ v(s) = -\sqrt{r^2 - \bigl( 1-\tfrac12\,e^{-s} \bigr)^2}. $$ Moreover, we have $$ \theta'(s) = \kappa(s) = -\frac{(1+u(s))}{v(s)} = \frac{1-\tfrac12\,e^{-s}}{\sqrt{r^2 - \bigl( 1-\tfrac12\,e^{-s} \bigr)^2}}, $$ i.e., $$ \theta(s) = \int_0^s \frac{2-e^{-\sigma}}{\sqrt{4r^2-\bigl( 2-e^{-\sigma}\bigr)^2}}\,\mathrm{d}\sigma. $$ (Note, by the way, that this is an elementary integral, but the integral is a transcendental function.)

Armed with the knowledge of $u(s)$, $v(s)$, and $\theta(s)$, the above formulae give an explicit arc-length parametrization of the tractrix. Note, that, of course, we must have $r>\tfrac12$ or the equations (and the condition) don't make sense.

When $r\ge1$, these formulae exist for all $s\ge0$ and, of course, $u(s)$ goes to zero as $s\to\infty$, while $v(s)$ goes to $-\sqrt{r^2-1}$, i.e., the tractrix asymptotically approaches a circle of radius $\sqrt{r^2-1}$. Moreover, not surprisingly, $\theta(s)$ increases without bound as $s\to\infty$, so there is a first value of $s$ for which $\theta(s) = \tfrac12\pi$.

When $\tfrac12<r<1$, the curvature of the curve goes to infinity in finite arclength, in fact, when $s = -\log\bigl(2(1{-}r)\bigr)>0$. At this point, the curve will have a cusp, and the tangent line will point directly towards the origin. In fact, when $r < 0.724651057$, the (strictly increasing) function $\theta$ will not actually reach $\tfrac12\pi$ in the interval $0\le s\le -\log\bigl(2(1{-}r)\bigr)$, so the curve does not turn vertical in this range before its curvature goes to infinity.

Addendum: Continuing the curve past the cusps

It turns out that there is a reasonable way to continue the curve past the cusps, and this gives a more reasonable picture. The way to do this is to realize that one can reparametrize the curve smoothly (not by arclength) in a smooth way as follows: Consider the equations $$ \mathrm{d}u = -u\,\mathrm{d}s,\quad \mathrm{d}v = \frac{u(u+1)}{v}\,\mathrm{d}s,\quad \mathrm{d}\theta = -\frac{(u+1)}{v}\,\mathrm{d}s\,. $$ Setting $\mathrm{d}s = v\,\mathrm{d}t$, these can be rewritten as $$ \mathrm{d}u = -uv\,\mathrm{d}t,\quad \mathrm{d}v = u(u+1)\,\mathrm{d}t,\quad \mathrm{d}\theta = -(u+1)\,\mathrm{d}t\,. $$ Now, regarding $(u,v,\theta)$ as functions of $t$, we get the formula for $X(t)$ in the form $$ X(t) = u(t)\ \pmatrix{\cos\bigl(\theta(t)\bigr)\\\sin\bigl(\theta(t)\bigr)} + v(t)\ \pmatrix{-\sin\bigl(\theta(t)\bigr)\\\cos\bigl(\theta(t)\bigr)}. $$ Note that $(\dot u,\dot v) = \bigl(-uv,u(u{+}1)\bigr)$ defines a vector field on the circle $(u{+}1)^2+ v^2 = r^2$, and, when $\tfrac12<r<1$, this vector field has no zeros. Hence $u$ and $v$ are smooth periodic functions of $t$ of some period $T(r)$, and $\theta$ satisfies $\theta(t+T(r)) = \theta(t) + P(r)$, where $$ P(r) = \int_0^{T(r)} (1+u(t))\,\mathrm{d}t. $$ (The cusps occur where $v$ vanishes.) In fact, it is easy to compute these period integrals when $r^2<1$. One finds that $$ T(r) = \int_{1-r}^{1+r}\frac{2\,\mathrm{d}\rho}{\rho\sqrt{r^2-(1{-}\rho)^2}} = \frac{2\pi}{\sqrt{1-r^2}} $$ and $$ P(r) = \int_{1-r}^{1+r}\frac{2(1{-}\rho)\,\mathrm{d}\rho}{\rho\sqrt{r^2-(1{-}\rho)^2}} = \frac{2\pi\bigl(1-\sqrt{1-r^2}\bigr)}{\sqrt{1-r^2}}\,. $$ In particular, note that, when $\sqrt{1-r^2}$ is rational, the curve closes periodically, with a finite number of cusps. The length of the curve between consecutive cusps is $$ S = \log\left(\frac{1+r}{1-r}\right)\,. $$

A further remark on integration of the tractrix

In fact, by a change of variable, one can integrate the equations completely when $r^2<1$. The equation $(u{+}1)^2+v^2=r^2$ together with the differential equations $$ \mathrm{d}u = -uv\,\mathrm{d}t,\quad \mathrm{d}v = u(u+1)\,\mathrm{d}t,\quad \mathrm{d}\theta = -(u+1)\,\mathrm{d}t\,. $$ show that we can actually write $$ (u,v) = \bigl(r\,\cos 2\phi -1,\ r\,\sin 2\phi) $$ for some function $\phi$ on the curve. From the first two differential equations, this implies $\mathrm{d}t = -2\,\mathrm{d}\phi/(1-r\,\cos 2\phi)$, so $$ \mathrm{d}\theta = -(u+1)\,\mathrm{d}t = \frac{2r\,\cos 2\phi\,\mathrm{d}\phi}{(1-r\,\cos 2\phi)} $$ and this integrates to give $$ \theta(\phi) = 2\left( \frac{\arctan\bigl(\sqrt{\frac{1{+}r}{1{-}r}}\,\tan\phi\bigr)}{\sqrt{1-r^2}}-\phi\right). $$ Hence, $$ X(\phi) = \pmatrix{r\,\cos\bigl(2\phi{+}\theta(\phi)\bigr)-\cos(\theta(\phi))\\ r\,\sin\bigl(2\phi{+}\theta(\phi)\bigr)-\sin(\theta(\phi))}, $$ As a result, it follows that, when $\sqrt{1-r^2}$ is rational, the curve not only closes periodically, but it is algebraic. I suspect that this was known classically.