# How to compute the derivative for my particular line?

Let me give a short definition then make a question. Let $L$ be any smooth directional line on the XY plane. That is ,for any point $p$ on $L$ , just one tangent line exists. Let the tangent spin speed at $p$ on $L$ be defined as the limit of $function1$ at $p$ when a point $q$ approaches to $p$ along $L$ , against the direction of $L$. Where $function1$ is the ratio = $( tangent(q) - tangent(p) ) / (q.x - p.x)$ .

My question is this. Let $sL$ be an enough short segment of $L$. Where the tangent spin speed on $sL$ is known as constant $c$ ; and the initial tangent at the first end of $sL$ is known as $t$. Can you compute the derivative for $sL$? If yes , how?

Please read the answer by user7530,it almost answers what I want. The remaining question follows. I said that the point $q$ moves along $L$ , but the tangent spin speed concerns the speed of $q$ only along the x-axis, not along $L$.

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The question is a bit hard to read. Is "Exp" a name or an exponential? Also "the ratio as ..." doesn't make any sense -- do you mean just "the ratio", or "as" as in "as $q\to p$"? It would also help with the legibility if you used $\TeX$; you can do so by enclosing things in dollar signs like $this$. – joriki Oct 25 '11 at 13:17

I'm having a lot of trouble understanding your question, but I will do the best I can.

If I understand correctly, what you are trying to capture with "tangent spin speed" is the curvature of the curve: the rate at which the unit tangent vector changes as you move at unit speed along the curve.

Let $\gamma(s)$ be your curve. We'll suppose $\gamma$ is parameterized by arc length, which means that the "speed" of the point $\gamma(s)$ that sweeps out the curve is constant and equal to one: $\|\gamma'(s)\| = 1$. Curvature $\kappa$ is then defined as the magnitude of the second derivative of $\gamma$: $$\kappa = \|\gamma''(s)\| = \left\|\frac{d}{ds} \gamma'(s)\right\|.$$

Curvature captures a notion of "tangent spin speed": straight line segments have 0 curvature, and the more sharply $\gamma$ bends, the higher the curvature at that point. (Curvature, or at least this definition of it, is always positive; you can extend this to signed curvature by e.g. defining curvature as positive if the curve bends clockwise and negative if it bends counterclockwise.) We use arc-length parametrization to guarantee that two curves that look the same have the same curvature.

Your question, then, would be equivalent to the following: given a curve of constant curvature $c$ and initial tangent vector $\gamma'(0) = t$, what is the tangent vector $\gamma'(s)$ at arbitrary $s$?

This question has a particularly simple solution: curves in the plane of constant curvature $c$ are precisely circles of radius $\frac{1}{c}$. Thus $\gamma'(s) = R(\pm cs) t$, where $R(\theta)$ is rotation by $\theta$ radians, e.g. $$R(\theta) = \left[\begin{array}{cc} \cos\theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right].$$ The precise sign will depend on your convention.

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:Thank you for the answer. Quote,"as you move at unit speed along the curve". I think it is where I have failed to let you know what I want. Borrowing your words, you move at unit speed along the x-axis. – seven_swodniw Oct 26 '11 at 12:01