Equation of Tractrix/Cartesian Form

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Definition

Let $S$ be a cord of length $a$ situated as a (straight) line segment whose endpoints are $P$ and $T$.

Let $S$ be aligned along the $x$-axis of a cartesian plane with $T$ at the origin and $P$ therefore at the point $\tuple {a, 0}$.

Let $T$ be dragged along the $y$-axis.


The equation of the tractrix along which $P$ travels is:

$y = a \map \ln {\dfrac {a \pm \sqrt {a^2 - x^2} } x} \mp \sqrt {a^2 - x^2}$


Proof

Consider $P$ when it is at the point $\tuple {x, y}$.


Tractrix.png


The cord $S$ is tangent to the locus of $P$.

Thus from Pythagoras's Theorem:

$\dfrac {\d y} {\d x} = -\dfrac {\sqrt {a^2 - x^2} } x$


Hence:

\(\ds \int \rd y\) \(=\) \(\ds \int \frac {\sqrt {a^2 - x^2} } x \rd x\) Solution to Separable Differential Equation
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds -a \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + \sqrt {a^2 - x^2} + C\) Primitive of $\dfrac {\sqrt {a^2 - x^2} } x$
\(\ds \) \(=\) \(\ds a \map \ln {\frac {a - \sqrt {a^2 - x^2} } x} + \sqrt {a^2 - x^2} + C\) after algebra

Taking the negative square root:

\(\ds y\) \(=\) \(\ds -a \map \ln {\frac {a - \sqrt {a^2 - x^2} } x} - \sqrt {a^2 - x^2} + C\)
\(\ds \) \(=\) \(\ds a \paren {\map \ln {a + \sqrt {a^2 - x^2} } - \ln x} - \sqrt {a^2 - x^2} + C\) after algebra


When $y = 0$ we have $x = a$.

Thus:

\(\ds 0\) \(=\) \(\ds a \map \ln {\frac {a + \sqrt {a^2 - a^2} } a} - \sqrt {a^2 - a^2} + C\)
\(\ds \leadsto \ \ \) \(\ds C\) \(=\) \(\ds -a \map \ln {\frac a a}\)
\(\ds \leadsto \ \ \) \(\ds C\) \(=\) \(\ds -a \ln 1\)
\(\ds \) \(=\) \(\ds 0\)


Hence the result.

$\blacksquare$


Linguistic Note

The word tractrix derives from the Latin traho (trahere, traxi, tractum) meaning to pull or to drag.

The plural is tractrices.


Sources