A curve is defined as y=f(x) and has curvature:

$\displaystyle \frac {y''} {(1+y'^2)^{3/2}}$

show that the curvature is invariant under rotation of the axes.

Hint: calculate $\displaystyle \frac{dy}{dx} =\frac{dy}{d \bar x}\frac{d \bar x}{dx}$

OK so we can define the transformation:

$\displaystyle u=xCos(\alpha)-ySin(\alpha)$
$\displaystyle v=xSin(\alpha)+yCos(\alpha)$

Then

$\displaystyle \frac {dy} {dx} = \frac {dy} {du} \frac {du} {dx} + \frac {dy} {dv} \frac {dv} {dx} = y_u Cos(\alpha) + y_v Sin(\alpha)$

And

$\displaystyle \frac {d^2y} {dx^2} = y_{uu} \frac {du} {dx} Cos(\alpha) + y_{uv} \frac {dv} {dx} Sin(\alpha) + y_{vu} \frac {du} {dx} Sin(\alpha) + y_{vv} \frac {dv} {dx} Sin(\alpha)$

$\displaystyle \frac {d^2y} {dx^2} = y_{uu} Cos^2(\alpha) + 2y_{uv} Cos(\alpha) Sin(\alpha) + y_{vv} Sin^2(\alpha)$

Now substituting for $\displaystyle \frac{dy}{dx}$ and $\displaystyle \frac{d^2y}{d^2x}$ into $\displaystyle \frac {y''} {(1+y'^2)^{3/2}}$ dosen't seem to get me anywhere.

I would have thought that the aim of the exercise is to show that:

$\displaystyle \frac {y''} {(1+y'^2)^{3/2}} = \frac {v_{uu}} {(1+v_u^2)^{3/2}}$