Equation of Tractrix/X-Axis Asymptote/Cartesian Form

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 $y$-axis of a cartesian plane with $T$ at the origin and $P$ therefore at the point $\tuple {0, a}$.

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

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

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

Proof

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

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

Thus from Pythagoras's Theorem:

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

Hence:

\(\ds \int \rd x\)

\(=\)

\(\ds \int -\frac {\sqrt {a^2 - y^2} } y \rd y\)

Solution to Separable Differential Equation

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds a \map \ln {\frac {a + \sqrt {a^2 - y^2} } y} - \sqrt {a^2 - y^2} + C\)

Primitive of $\dfrac {\sqrt {a^2 - y^2} } y$

\(\ds \)

\(=\)

\(\ds a \paren {\map \ln {a + \sqrt {a^2 - y^2} } - \ln y} - \sqrt {a^2 - y^2} + C\)

after algebra

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

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.

This needs considerable tedious hard slog to complete it. In particular: This covers only the part where $x > 0$. The left hand half needs doing too. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Linguistic Note

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

The plural is tractrices.