Invariance of Curvature

**Kiwi_Dave** · February 17th 2009, 02:13 PM

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.

**Danny** · February 17th 2009, 03:16 PM

Originally Posted by Kiwi_Dave

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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