With the help of the existing answers I have been able to put together this clearer animation, and I asked this question to discover the shape is called a cochleoid. What I am really trying to find out at this point is the following:

It seems like this curve should be perfectly smooth through the point at $(0, 1)$, but because the curve is based on $sin(x)/x$ it is technically undefined at this point, and if I were programming a function to evaluate this curve I would have to add a special case for values near this singularity. I'm curious if there is any way to re-phrase this equation to remove the singularity and make evaluation of it near zero more numerically stable.

--- Original text of question preserved below ----------------------------------------------------

Imagine I have a unit line segment going from $(0, 0)$ to $(0, 1)$. Over time, I want to bend this segment such that it always forms a circular arc (the initial configuration can be considered an arc on a circle with infinite radius). What shape will the end point trace out, and how can I get the coordinates of that point if I am given as input the angle that should be spanned by the arc length?

Edit: Thanks to John Bales below, this animation accurately depicts what I am trying to describe, although I'm not sure how to render the bent segment itself, which would always connect back to the origin. Is this shape a cartioid? Is there a way to rephrase the equation so it doesn't become undefined when the input is zero?

  • $\begingroup$ Your question is not clear to me. How are you bending the segment? Does the segment maintain its unit length? What exactly is the circle that this segment is the arc of? Oh and what is the end point you are talking about? $\endgroup$ – Mike Pierce Jul 16 '16 at 3:50
  • $\begingroup$ So are you assuming the point 0,0 always has tangent 0? $\endgroup$ – fleablood Jul 16 '16 at 3:54
  • $\begingroup$ I think they mean that for each radius from 1/2 through infinity, there is one circle (say, whose center lies in the left half-plane on $y=1/2$) passing through the two points. Maybe the end point they're referring to is actually the mid point? $\endgroup$ – Mark S. Jul 16 '16 at 3:56
  • $\begingroup$ Suppose the radius of the imaginary circles go from infinite (line is straight) to 1/2pi (line is bent to a circle with circumference 1. What is the formula for the end point of the arch. Can you figure what is codomain/image of this formula? $\endgroup$ – fleablood Jul 16 '16 at 3:59
  • $\begingroup$ @Jason-- You need to edit your question to specify that the segment is always mapped homeomorphically onto a circular arc of unit length along a circle tangent to the $y$-axis and with center on the positive $x$-axis. $\endgroup$ – John Wayland Bales Jul 19 '16 at 19:40

The angle in radians subtended by an arc of length $s$ on the circumference of a circle of radius $r$ is given by

\begin{equation} \theta=\dfrac{s}{r} \end{equation}

In this instance $s=1$ and $r\ge\tfrac{1}{\pi}$. The circle has center $(r,0)$ and radius $r$. The arc $s$ of length $1$ extends upward along the circumference with one end fixed at $(0,0)$ and the other end ends at the point

\begin{equation} (x,y)=\left(r-r\cos\left(\frac{1}{r}\right),r\sin\left(\frac{1}{r}\right)\right) \end{equation}

These are parametric equations of the curve.

Here is a desmos.com animation of the curve along which the point moves.


This curve has the following polar equation:

\begin{equation} r=\dfrac{\sin\left(\frac{\pi}{2}-\theta\right)}{\frac{\pi}{2}-\theta} \end{equation}

Here is a desmos.com graph of the polar curve with the particle moving along it. You want just the portion of the curve in the first quadrant.


I think it is interesting that this turns out to be a $\dfrac{\sin x}{x}$ curve but in the polar coordinate system.

If instead of the interval from $(0,0)$ to $(0,1)$ we bend the unit interval on the horizontal axis along circles with centers $\left(0,\frac{1}{2\theta}\right)$ and radii $\frac{1}{2\theta}$ as $0\le\theta\le\frac{\pi}{2}$ then the ends of the unit arcs along the circles from the origin will trace out the portion of the polar curve

\begin{equation} r=\dfrac{\sin\theta}{\theta} \end{equation}

in quadrant I.

The following is a link to a GeoGebra animation illustrating this variation of the problem.


Still image of GeoGebra animation

  • $\begingroup$ @Jason I found the polar equation of the curve and posted a link to an additional graph. $\endgroup$ – John Wayland Bales Jul 16 '16 at 6:10
  • $\begingroup$ @fleablood I found the polar equation and graph of Jason's 'bending a line segment' problem and it turns out to be a $\dfrac{\sin x}{x}$ graph but in the polar plane. $\endgroup$ – John Wayland Bales Jul 16 '16 at 6:14
  • $\begingroup$ For some strange reason GeoGebra is speaking Portuguese in the app sidebar today. sen means sin $\endgroup$ – John Wayland Bales Jul 16 '16 at 21:24
  • $\begingroup$ how to calculate the arc angle in above equation. $\endgroup$ – Ajay Singh Thakur May 21 at 10:48

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