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

[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.

 

[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.

 

[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.

 

[vanhees71]

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