# The Tractrix

The tractrix is the curve defined by the property that the distance from a point on the curve along its tangent to a fixed straight line is a constant independent of the point.

We first show how this curve can be described by a first-order ordinary differential equation. We will then solve this equation to obtain a parametrization of the curve. To begin, let’s suppose that the distance along the tangent at a point on the tractrix to the $$y$$-axis is equal to $$1$$. Suppose, then, that the curve is given by $$\alpha(t) = (x(t),\, y(t))$$ and that the corresponding point on the $$y$$-axis is given by $$\beta(t) = (0,\, v(t))$$. Furthermore, for a fixed value of $$t$$, let $$\gamma(s)$$ denote the parametrization of the tangent line given by

\begin{aligned}
\gamma(s)
&= \alpha(t) + s\,\alpha'(t)\\
&= (x(t) + s\,x'(t),\, y(t) + s\,y'(t)).
\end{aligned}

We seek $$s$$ such that $$x(t) + s\,x'(t) = 0$$ or $$s = -x(t)/x'(t)$$. We now take the parameter $$t$$ to be the angle that the tangent line makes with the $$y$$-axis. With respect to this parametrization, we have that $$x(t) = \sin t$$, hence

s = -\frac{\sin t}{\cos t} = -\tan t.

Now note that

v(t) = y(t) + s\, y'(t) \Rightarrow y(t) – v(t) = y'(t) \tan t.

Since $$\| \alpha – \beta \|^2 = x^2 +(y – v)^2 = 1$$, we obtain

\sin^2 t + y'(t)^2 \tan^2 t = 1,

Taking the square root and noting that $$\sin t > 0$$ for $$0 < t < \pi$$, we obtain $$\label{eqYprime} y' = \frac{\cos^2 t}{\sin t} = \frac{1 - \sin^2 t}{\sin t} = \frac{1}{\sin t} - \sin t, \quad 0 < t < \pi.$$ To find $$\alpha$$, we must integrate (\ref{eqYprime}). Noting that $$y(\pi/2) = 0$$, integration of (\ref{eqYprime}) yields $$y(t) = \int_{\pi/2}^t \frac{du}{\sin u} - \int_{\pi/2}^t\sin u\,du.$$ The second integral is easy to evaluate. Namely, we have $$\int_{\pi/2}^t\sin u\,du = \left[ -\cos u \right]_{\pi/2}^t = -\cos t.$$ The first integral is a little trickier. Note first that the identity $$\sin u = 2 \sin(u/2) \cos(u/2)$$ leads to $$\int\frac{du}{\sin u} = \int \frac{dv}{\sin v \cos v}, \quad v = u/2.$$ Now observe that $$\frac{1}{\sin v \cos v} = \frac{\cos v}{\sin v} + \frac{\sin v}{\cos v}.$$ Substituting this into the previous integral yields \begin{aligned} \int \frac{dv}{\sin v \cos v} &= \int\frac{\cos v}{\sin v}dv + \int \frac{\sin v}{\cos v}dv\\ &= \log(\sin v)- \log(\cos v)\\ &= \log(\tan v)\\ &= \log(\tan(u/2)). \end{aligned} It follows that \begin{aligned} \int_{\pi/2}^t \frac{du}{\sin u} &= \left[ \log(\tan(u/2) \right]_{\pi/2}^t\\ &= \log(\tan(t/2)) - \log(\tan(\pi/4))\\ &= \log(\tan(t/2)), \end{aligned} since $$\tan(\pi/4) = 1$$ and $$\log 1 = 0$$. This yields the formula $$y(t) = \cos t + \log(\tan(t/2)).$$ Hence the tractrix is given by the system of equations $$\boxed{ x(t) = \sin t,\quad y(t) = \cos t + \log(\tan(t/2)),\quad 0 < t < \pi. }$$