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.


gTractrix

The tractrix. In this figure, the horizontal axis is the \( y \)-axis and the vertical axis is the \( x \)-axis.

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

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