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Hometractrix

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# tractrix

Tractrix (from the Latin verb trahere ‘pull, drag’) is the curve along which a small object (tractens) moves when pulled on a horizontal plane with a piece of thread by a puller (tractendus) which moves rectilinearly.

Let the object initially be in the $xy$-plane on the $x$-axis in the point $(a,\,0)$ and the puller in the origin; $a$ is the length of the pulling thread. Then the puller begins to move along the $y$-axis in the positive direction. The object follows drawing the path curve $y=y(x)$ so that the line determined by the thread is at every moment the tangent of the curve. This condition gives in the point $(x,\,y)$ the simple differential equation

$\frac{dy}{dx}=-\frac{\sqrt{a^{2}-x^{2}}}{x}$ |

with the initial condition $y(a)=0$. The solution is

$y=\int_{x}^{a}\frac{\sqrt{a^{2}-x^{2}}}{x}\,dx$ |

or

$y=\pm(a\ln{\frac{a+\sqrt{a^{2}-x^{2}}}{x}}-\sqrt{a^{2}-x^{2}}).$ |

Here the minus alternative is for the case that the puller moves in the negative direction from the origin. In fact, both branches, corresponding to both signs, belong to the tractrix. The branches meet in the cusp point $(a,\,0)$.

The substitution $x:=a\cos{t}$ gives for the tractrix the parametric presentation

$x=a\cos{t},\quad y=\pm a(\ln\frac{1+\sin{t}}{\cos{t}}-\sin{t}).$ |

Another one is

$x=\frac{a}{\cosh{u}},\quad y=\pm a(u-\tanh{u}),$ |

where $\cosh$ and $\tanh$ are the hyperbolic functions cosinus hyperbolicus and tangens hyperbolica.

Remarks

1. 2. The differential equation of the orthogonal curves of the tractrix is

$\frac{dy}{dx}=\frac{x}{\sqrt{a^{2}-x^{2}}},$ whence they are the circles $x^{2}+(y-C)^{2}=a^{2}$.

3. The arc length of one branch on the interval $[b,\,a]$ is simply

$\int_{b}^{a}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx=a\int_{b}^{a}\frac{dx}% {x}=a\ln\frac{a}{b}.$ 4. The area $A$ between the tractrix and its asymptote is $\frac{\pi a^{2}}{2}$. This may be calculated ordinarily as

$A=2\int_{0}^{a}(a\ln{\frac{a+\sqrt{a^{2}-x^{2}}}{x}}-\sqrt{a^{2}-x^{2}})\,dx;$ integrating by parts and using the area of a quarter-circle yield

$A=2\left[a\operatornamewithlimits{\Big/}_{{\!\!\!x=0}}^{{\,\quad a}}x\ln\frac{% a+\sqrt{a^{2}-x^{2}}}{x}-a\int_{0}^{a}x\frac{d}{dx}\left(\ln\frac{a+\sqrt{a^{2% }-x^{2}}}{x}\right)\,dx-\frac{\pi a^{2}}{4}\right]$ and moreover

$A=2a\operatornamewithlimits{\Big/}_{{\!\!\!x=0}}^{{\,\quad a}}\left[x\ln(a+% \sqrt{a^{2}-x^{2}})-x\ln{x}+a\arcsin\frac{x}{a}\right]-\frac{\pi a^{2}}{2}=2a% \left(0-0+a\cdot\frac{\pi}{2}\right)-\frac{\pi a^{2}}{2}=\frac{\pi a^{2}}{2}$ (see this entry for $\lim_{{x\to 0+}}x\ln{x}=0$). Another way to determine $A$ is differential-geometric: as the object draws the entire tractrix from above to down, the thread turns $180^{{\mathrm{o}}}$ and thus sweeps an area equal to a half-circle.

5. The envelope of the normal lines of the tractrix, i.e. the evolute of the tractrix is the catenary (or “chain curve”) $x=a\cosh{\frac{y}{a}}$.

## Mathematics Subject Classification

51N05*no label found*

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## Comments

## Sign errors in tractrix equations

N.B., I have fixed my errors -- the equations are now right.

Jussi