Computing tangential derivative d2x/ds2 at a point on a circle.

1. Jan 10, 2014

nawidgc

Let $P(x,y)$ be a point on a unit circle that is centered at (0,0). How to compute exactly the function
$\frac{\partial^2 x}{\partial s^2}$​
where $x$ is the x-coordinate of the point $P(x,y)$ and $s$ is the tangent at point $P(x,y)$. Clearly,
$\frac{\partial x}{\partial s} = t_x = -n_y$ ​
where $t_x$ is the x-component of the tangent at point $P(x,y)$ and $n_y$ is the y-component of the normal to circle boundary at point $P(x,y)$. I have verified above equation with finite difference. Now how do I obtain an exact expression for
$\frac{\partial }{\partial s }\left(\frac{\partial x}{\partial s}\right)$ ​
to get $\frac{\partial^2 x}{\partial s^2}$? Thanks for help.

2. Jan 11, 2014

vanhees71

I don't now what you mean by "s is the tangent at point $P(x,y)$". Then the derivative doesn't make sense.

I guess, what you are doing here is differential geometry of a line in the Euclidean plane $\mathbb{R}^2$.

The "natural" way to parametrize a curve is to use the path length measured from an arbitrary starting point as parameter.

For your unit circle you can go as follows: You start with an arbitrary parametrization. Take the angle $\varphi$ of position vector $\vec{r}$ to the $x$ axis. Then the circle is parametrized as
$$\vec{r}(\varphi)=\cos \varphi \vec{e}_x+\sin \varphi \vec{e}_y, \quad \varphi \in [0, 2\pi[.$$
The path length, measured from the point $(1,0)$ is then given by
$$s(\varphi)=\int_0^{\varphi} \mathrm{d} \varphi' \left|\frac{\mathrm{d}\vec{r}(\varphi')}{\mathrm{d} \varphi'} \right|= \int_0^{\varphi} \mathrm{d} \varphi' 1=\varphi.$$
So you simply have $s=\varphi$.

Now you can take derivatives. The tangent vectors at the points along the cirlce are given by
$$\vec{t}(\varphi)=\frac{\mathrm{d} \vec{r}(\varphi)}{\mathrm{d} \varphi}=-\sin \varphi \vec{e}_x+\cos \varphi \vec{e}_y,$$
and the second derivative gives
$$\frac{\mathrm{d}^2 \vec{r}(\varphi)}{\mathrm{d} \varphi^2}=-\vec{r}(\varphi).$$
It's clear that, because of $s=\varphi$ you can as well write $s$ instead of $\varphi$ everywhere.

3. Jan 11, 2014

nawidgc

@vanhees71: I should have probably said that s is the unit tangent at point P. Physically, what I need is the second derivative of x coordinate at point P with respect to the unit tangent s at P (i.e. d2x/ds2). This can also be interpreted as the rate of change of x-component of unit tangent s with respect to s, i.e. d/ds (dx/ds) which I think will be a scalar quantity. Appreciate your help.

4. Jan 11, 2014

vanhees71

I have no clue what you mean by taking the derivative or a coordinate with respect to a tangent vector.