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Is there a common name for curves, obtained from dragging a point along another curve, similar to how tractrix is obtained by dragging a point along a line?

What is a parametric equation of such curve given the parametric equation of a curve along which the master goes?

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Generalized tractrices (tractories) exist, see this or this or this. (the last two are in French, but with slightly more detail than the first one.)

This old book ought to be of interest as well.

This book describes Euler's treatment of the problem of the tractory.

As a note, the problem of finding the generating curve, given the tractory, is a much easier problem (hint: use the tangent vector of a curve) than finding the tractory corresponding to a generating curve.


For those who have difficulty reading French, the third link I mentioned gives the prescription for generating the corresponding tractory from a generating curve with parametric equations $(f(t)\quad g(t))$.

The parametric equations for the tractory of $(f(t)\quad g(t))$ (in vector form) is

$$\begin{pmatrix}f(t)\\g(t)\end{pmatrix}-\frac{a}{\sqrt{f^{\prime}(t)^2+g^{\prime}(t)^2}}\begin{pmatrix}\cos\;\alpha(t)&\sin\;\alpha(t)\\-\sin\;\alpha(t)&\cos\;\alpha(t)\end{pmatrix}\cdot\begin{pmatrix}f^{\prime}(t)\\g^{\prime}(t)\end{pmatrix}$$

or explicitly

$$\begin{align*}x&=f(t)-\frac{a}{\sqrt{f^{\prime}(t)^2+g^{\prime}(t)^2}}(f^{\prime}(t)\cos\;\alpha(t)+g^{\prime}(t)\sin\;\alpha(t))\\y&=g(t)-\frac{a}{\sqrt{f^{\prime}(t)^2+g^{\prime}(t)^2}}(g^{\prime}(t)\cos\;\alpha(t)-f^{\prime}(t)\sin\;\alpha(t))\end{align*}$$

where the function $\alpha(t)$ satisfies the differential equation

$$\frac{\mathrm d\alpha}{\mathrm dt}=\frac{f^{\prime}(t)g^{\prime\prime}(t)-g^{\prime}(t)f^{\prime\prime}(t)}{f^{\prime}(t)^2+g^{\prime}(t)^2}-\frac{\sin\alpha}{a}\sqrt{f^{\prime}(t)^2+g^{\prime}(t)^2}$$

and $a$ is the length of the segment running through the generating curve.

As an example, here is an animation showing the curve $(3\cos\;t-2\cos^3 t+\cos 2t\quad 2\sin^3 t+\sin 2t)$ and its tractory with segment length 1:

nephroid tractory

(This is a less fancy version of the last bicycle animation in the third French link.)

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  • $\begingroup$ Probably there is a typo in the formula because it is not equation as supposed to be $\endgroup$ – Anixx Dec 31 '10 at 12:32
  • $\begingroup$ Which one do you think has the typo, @Anixx? $\endgroup$ – J. M. is a poor mathematician Dec 31 '10 at 12:44
  • $\begingroup$ "The parametric equations for the tractory is" is not equation... $\endgroup$ – Anixx Dec 31 '10 at 17:27
  • $\begingroup$ @Anixx: Multiply out the (rotation) matrix and the vector, and add up the vectors. The components yield the parametric equations. I only chose matrix-vector representations here because it's more intuitive (at least, for me). $\endgroup$ – J. M. is a poor mathematician Apr 7 '11 at 6:20
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    $\begingroup$ @Anixx: $\alpha$ is a scalar function. I guess you aren't accustomed to vector notation... I'll edit this answer to give the explicit parametric equations. $\endgroup$ – J. M. is a poor mathematician Apr 10 '11 at 15:34
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Some of them are called pursuit curves.

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    $\begingroup$ They're similar, but the tractrix is not identical to the pursuit curve for a straight line. For one thing, the pursuit curve has no cusp. $\endgroup$ – Rahul Nov 14 '10 at 16:21

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