34 The Tractory.

Article 34
The Tractory.

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.

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Let $P T$ , $P' T'$ be two consecutive tangents such that $P T = P' T' = c$ , and let $O T = t$ ; draw $T S$ perpendicular to $P' T'$ ; then if $P P' = d s$ , it is evident that $S T'$ diﬀers from $d s$ by an inﬁnitesimal of a higher order. Let $P T$ make an angle $φ$ with $O A$ , the axis of $y$ ; then (to the ﬁrst order of inﬁnitesimals) $P T d φ = T S = T T' cos φ$ ; that is,

\begin{matrix} c d φ = cos φ d t, t = c {gd}^{- 1} φ, \\ x = t - c sin φ = c ({gd}^{- 1} φ - sin φ), y = c cos φ . \end{matrix}

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

d s = S T' = d t sin φ = c tan φ d φ, s = c {log}_{e} sec φ .

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

\frac{x}{c} = {gd}^{- 1} ({cos}^{- 1} \frac{y}{c}) - sin ({cos}^{- 1} \frac{y}{c}) = {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, $φ = gd u$ , $\frac{x}{c} = u - tanh u$ , $\frac{y}{c} = sech u$ , $\frac{s}{c} = log cosh u .$

Prob. 100.: Given $t = 2 c$ , show that $φ = 7 4^{\circ} 35'$ , $s = 1.3249 c$ , $y = . 2658 c$ , $x = 1.0360 c .$ 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 $φ$ ; and derive the intrinsic equation of the involute.

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Article 34The Tractory.

Article 34
The Tractory.