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## Article 34The Tractory.

[Note.16]

To ﬁnd the equation of the curve which possesses the property that the length of the tangent from the point of contact to the axis of $x$ is constant.

Let $PT$, $P\prime T\prime$ be two consecutive tangents such that $PT=P\prime T\prime =c$, and let $OT=t$; draw $TS$ perpendicular to $P\prime T\prime$; then if $PP\prime =ds$, it is evident that $ST\prime$ diﬀers from $ds$ by an inﬁnitesimal of a higher order. Let $PT$ make an angle $\phi$ with $OA$, the axis of $y$; then (to the ﬁrst order of inﬁnitesimals) $PTd\phi =TS=TT\prime cos\phi$; that is,

This is a convenient single-parameter form, which gives all values of $x$, $y$ as $\phi$ increases from $0$ to $\frac{1}{2}\pi$. The value of $s$, expressed in the same form, is found from the relation

 $ds=ST\prime =dt\phantom{\rule{0.3em}{0ex}}sin\phi =ctan\phantom{\rule{0.3em}{0ex}}\phi \phantom{\rule{0.3em}{0ex}}d\phi ,\phantom{\rule{1em}{0ex}}s=c\phantom{\rule{0.3em}{0ex}}{log}_{e}sec\phi .$

At the point $A$, $\phi =0$, $x=0$, $s=0$, $t=0$, $y=c$. The Cartesian equation, obtained by eliminating $\phi$, is

 $\frac{x}{c}={gd}^{-1}\left({cos}^{-1}\frac{y}{c}\right)-sin\left({cos}^{-1}\frac{y}{c}\right)={cosh}^{-1}\frac{c}{y}-\sqrt{1-\frac{{y}^{2}}{{c}^{2}}}.$

If $u$ be put for $\frac{t}{c}$, and be taken as independent variable, $\phi =gdu$, $\frac{x}{c}=u-tanhu$, $\frac{y}{c}=sechu$, $\frac{s}{c}=logcoshu.$

Prob. 100.
Given $t=2c$, show that $\phi =7{4}^{\circ }\phantom{\rule{0.3em}{0ex}}35\prime$, $s=1.3249c$, $y=.2658c$, $x=1.0360c.$ At what point is $t=c$?
Prob. 101.
Show that the evolute of the tractory is the catenary. (See Prob. 92.)
Prob. 102.
Find the radius of curvature of the tractory in terms of $\phi$; and derive the intrinsic equation of the involute.

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