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Math Help - Invariance of Curvature

  1. #1
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    Invariance of Curvature

    Show that when the equation of a curve is given in parametric form x=x(t),y=y(t) that the curvature given by:

    \frac{\dot x \ddot y - \dot y \ddot x}{(\dot x^2 + \dot y^2)^{-3/2}}

    remains invariant under the change of parameter t=t(\bar t)

    I think I have to make the substitutions:

    \dot x = \frac{dx}{d \bar t}\cdot \frac{d \bar t}{dt}

    and

    \ddot x = \frac{d^2x}{d \bar t^2} (\frac{d \bar t}{dt})^2+\frac{dx}{d \bar t} \frac{d^2 \bar t}{dt^2}

    and expect to find that the terms like \frac {d \bar t}{dt} disappear.

    Unfortunately I have not been able to make that work.
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  2. #2
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    Quote Originally Posted by Kiwi_Dave View Post
    Show that when the equation of a curve is given in parametric form x=x(t),y=y(t) that the curvature given by:

    \frac{\dot x \ddot y - \dot y \ddot x}{(\dot x^2 + \dot y^2)^{-3/2}} (**)

    remains invariant under the change of parameter t=t(\bar t)

    I think I have to make the substitutions:

    \dot x = \frac{dx}{d \bar t}\cdot \frac{d \bar t}{dt}

    and

    \ddot x = \frac{d^2x}{d \bar t^2} (\frac{d \bar t}{dt})^2+\frac{dx}{d \bar t} \frac{d^2 \bar t}{dt^2}

    and expect to find that the terms like \frac {d \bar t}{dt} disappear.

    Unfortunately I have not been able to make that work.
    Let me use r's instead of \bar{t}

    So x_t = x_r r_t,\;\;\;x_t = y_r r_t and as you said x_{tt} = x_{rr} r_t^2 + x_r r_{tt},\;\;\;y_{tt} = y_{rr} r_t^2 + y_r r_{tt}, so

    \dot x \ddot y - \dot y \ddot x = x_r r_t \left( y_{rr} r_t^2 + y_r r_{tt}\right) - y_r r_t \left( x_{rr} r_t^2 + x_r r_{tt}\right) = r_t^3\left( x_r y_{rr} - y_r x_{rr} \right)

    Also

     \left( x_t^2 + y_t^2\right)^{3/2} = \left( r_t^2\left( x_r^2 + y_r^2 \right) \right)^{3/2} =r_t^3 \left( x_r^2 + y_r^2 \right)^{3/2}

    and dividing the two gives

    \frac{x_t y_{tt} - y_t x_{tt}}{\left( x_t^2 + y_t^2\right)^{3/2}} = \frac{x_r y_{rr} - y_r x_{rr}}{\left( x_r^2 + y_r^2\right)^{3/2}}

    although I think it's +3/2 in (**)
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