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

  1. Jan 10, 2014 #1
    Let [itex] P(x,y) [/itex] be a point on a unit circle that is centered at (0,0). How to compute exactly the function
    [itex] \frac{\partial^2 x}{\partial s^2} [/itex]​
    where [itex] x [/itex] is the x-coordinate of the point [itex] P(x,y) [/itex] and [itex]s[/itex] is the tangent at point [itex] P(x,y) [/itex]. Clearly,
    [itex] \frac{\partial x}{\partial s} = t_x = -n_y [/itex] ​
    where [itex] t_x [/itex] is the x-component of the tangent at point [itex] P(x,y) [/itex] and [itex] n_y [/itex] is the y-component of the normal to circle boundary at point [itex] P(x,y) [/itex]. I have verified above equation with finite difference. Now how do I obtain an exact expression for
    [itex] \frac{\partial }{\partial s }\left(\frac{\partial x}{\partial s}\right) [/itex] ​
    to get [itex] \frac{\partial^2 x}{\partial s^2} [/itex]? Thanks for help.
     
  2. jcsd
  3. Jan 11, 2014 #2

    vanhees71

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

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

    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 [itex]\varphi[/itex] of position vector [itex]\vec{r}[/itex] to the [itex]x[/itex] axis. Then the circle is parametrized as
    [tex]\vec{r}(\varphi)=\cos \varphi \vec{e}_x+\sin \varphi \vec{e}_y, \quad \varphi \in [0, 2\pi[.[/tex]
    The path length, measured from the point [itex](1,0)[/itex] is then given by
    [tex]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.[/tex]
    So you simply have [itex]s=\varphi[/itex].

    Now you can take derivatives. The tangent vectors at the points along the cirlce are given by
    [tex]\vec{t}(\varphi)=\frac{\mathrm{d} \vec{r}(\varphi)}{\mathrm{d} \varphi}=-\sin \varphi \vec{e}_x+\cos \varphi \vec{e}_y,[/tex]
    and the second derivative gives
    [tex]\frac{\mathrm{d}^2 \vec{r}(\varphi)}{\mathrm{d} \varphi^2}=-\vec{r}(\varphi).[/tex]
    It's clear that, because of [itex]s=\varphi[/itex] you can as well write [itex]s[/itex] instead of [itex]\varphi[/itex] everywhere.
     
  4. Jan 11, 2014 #3
    @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.
     
  5. Jan 11, 2014 #4

    vanhees71

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