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Is it true that if I have a planar curve parametrized by its arc length, then the second derivative points towards to the center of the osculating circle?

I can´t see it, but the book says that it´s true. I don´t see why it's true geometrically or algebraically. Can someone explain it to me?

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Second derivative of what ? –  Sasha Aug 7 '11 at 20:36
    
of the function s --> ( x(s) , y(s) where s is the arcoparameter –  Daniel Aug 7 '11 at 20:38
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3 Answers 3

I'll give you the intuition. And I'll tell you what you could prove to see it.

Firstly, note that the 'speed' or a particle going along a curve parametrized by arc length is constant. The first derivative relates to velocity, and so will be parallel to the curve. The second derivative relates to acceleration, but since the speed is constant, the acceleration vector will be perpendicular to the curve. Thus it's perpendicular to the line tangent to the curve, and therefore to the circle kissing the tangent line. Thus it points to the center of the circle.

To prove it rigorously, you should prove the following. The velocity and acceleration of a particle on a curve parametrized by arc length are perpendicular. And if you don't believe it, any straight line perpendicular to the circumference of a circle passes through the center of that circle.

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Thanks mixedwatch but i know that, i can derivate the dot product and deduce that, but one thing is that the second derivate towards to the center of the osculating circle, and other it´s that intersect , the difference is that any parallel of that vector can intersect the center, but my question is in the sense that if i multiply that vector by a positive constant, then i´ll intersect the center. –  Daniel Aug 7 '11 at 21:02
    
@Daniel: I'm sorry - I don't understand what you're saying. Can you restate your question? –  mixedmath Aug 8 '11 at 5:00
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There is a problem of orientation here; therefore it is not enough to consider orthogonality only.

Let $\gamma:\ s\mapsto z(s)=\bigl(x(s),y(s)\bigr)$ be our curve parametrized by arc length and consider the point $z_0=z(s_0)$ of $\gamma$. At $z_0$ we set up a new orthonormal frame given by $e_1:=(\dot x,\dot y)$ and $e_2:=(-\dot y, \dot x)$, where I have suppressed the reference to $s_0$. Note that this system is again positively oriented, i.e., you get $e_2$ by rotating $e_1$ ninety degrees counterclockwise.

As $\dot x^2 +\dot y^2\equiv1$ we have $\langle e_1,\ddot z\rangle=\dot x\ddot x+\dot y\ddot y=0$ (this we knew before) which implies $\ddot z=\kappa e_2$ for some $\kappa\in{\mathbb R}$. What is at stake here is the sign of $\kappa=\langle e_2,\ddot z\rangle$.

To clear this question we shall look at the polar angle $\theta$ of $\dot z$ which is defined by $\theta:=\arg(\dot x,\dot y)$. As $$\nabla\arg(\xi,\eta)=\Bigl({-\eta\over \xi^2+\eta^2},{\xi\over \xi^2+\eta^2}\Bigr)$$ we obtain by the chain rule $$\dot\theta ={-\dot y\over \dot x^2 +\dot y^2}\ \ddot x\ +\ {\dot x\over \dot x^2+\dot y^2}\ \ddot y =\langle e_2,\ddot z\rangle =\kappa\ .\qquad\qquad(*)$$ This equation is the key to our problem. Number one, it tells us that the quantity $\kappa$ is nothing else but the (signed!) angular velocity by which the tangent vector $\dot z$ turns along $\gamma$ as $s$ increases, so we are allowed to call $\kappa$ the curvature of $\gamma$ at the point $z_0$.

Number two, we now look at the sign of $\kappa$, and this brings us to the solution of your original problem. If the curve $\gamma$ turns to the left at $z_0$ then its polar angle $\theta$ is increasing there, and the center $c$ of the oscillating circle at $z_0$ will lie to the left of $\gamma$, i.e. in the direction of $e_2$. On the other hand the equation $(*)$ tells us that $\kappa$ is positive, whence $\ddot z=\kappa e_2$ points in the same direction as $e_2$, which means in the direction where $c$ lies.

Similarly, if the curve $\gamma$ turns to the right at $z_0$ then its polar angle $\theta$ is decreasing there, and the center $c$ of the oscillating circle at $z_0$ will lie to the right of $\gamma$, i.e. in the direction of $-e_2$. On the other hand the equation $(*)$ tells us that $\kappa$ is negative, whence $\ddot z=\kappa e_2$ points in the same direction as $-e_2$ which again is in the direction where $c$ lies.

If, however the curve $\gamma$ is parametrized by "time" $t$ in the form $t\mapsto \tilde z(t)$ then its arc length $s$ is given by $$s(t)\ =\ \int_0^t\sqrt{x'^2(t)+y'^2(t)}\ dt\ ,$$ and the two functions $s\mapsto z(s)$ and $t\mapsto \tilde z(t)$ are related by $$\tilde z(t)\ =z\bigl(s(t)\bigr)\qquad(0\leq t\leq T)\ .$$ Differentiating with respect to $t$ (denoted by $'$) gives, using the chain rule, $$\tilde z'\ =\ \dot z\ s' = v\ e_1\qquad\qquad(3)$$ where $v:=s'=\sqrt{x'^2+y'^2}>0$ is the "velocity" along $\gamma$ measured by a speedometer. Differentiating (3) with respect to $t$ once more we get $$\tilde z''=\ddot z\ s'^2 + \dot z\ s''\ .$$ We can write this in the following form: $$\tilde z''=v^2 \kappa e_2 + a e_1\ ,$$ where $s''=v'=:a$ is the speed gain along $\gamma$. The last equation tells us that the vector $\tilde z''$ has two components: a component parallel to $\tilde z'=v e_1$ coming from the speed gain and a component orthogonal to $z'$ which is the centripetal acceleration. The word says it: This component is directed towards the center $c$ of the osculating circle, because ${\rm sgn}(\langle \tilde z'', e_2\rangle) ={\rm sgn}(\kappa)$.

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I think the Frenet-Serre equations:

http://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas

Explain it: by construction, N has length 1, i.e., ||T||:=$\frac{1}{\kappa}T$=1.

Since N is a vector field of unit length, $T(s). T(s)$=1, so that $\frac{d}{ds}T.T=0$=

$2T.T'=0$ , so (since N is defined as $\frac{dT}{ds}$) $T.N=0$, so that the normal points towards the osculating circle. (I think the fact that the curve is planar here guarantees that the direction points towards the center of the osculating circle.

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