Let $\alpha: I \to \mathbb R^n (n = 2, 3)$ be a regular curve of any parameter $r \in I$. Let $ \beta: J \to \mathbb R^n$ be a reparameterization of $\alpha$ by arc length, ie $\beta(s(r)) = \alpha(r)$. If $t(s), n(s),$ and $b(s)$ form the Frenet frame of reference of $\beta, k(s),$ and $\tau (s)$ is curvature and the torsion of $\beta$, then we will say that $t(r) = t(s(r)), n(r) = n(s(r)) and b(s(r))$ are the Frenet reference of $\ alpha$, and $k(r) = k(s(r))$ and $\tau(r) = \tau(s(r))$ is the curvature and twist of $\alpha$.
Let $\alpha(r) = (x(r), y(r)), r \in I$, be a flat regular curve (not necessarily Parameterized by arc length). Show that: $$t(r) = \frac{(x', y')}{\sqrt{(x')^2 + (y')^2}}$$ $$n(r) = \frac{(-y', x')}{\sqrt{(x')^2 + (y')^2}}$$ $$k(r) = \frac{-x''y' + x' y''}{((x')^2 + (y')^2)^{\frac{3}{2}} }$$
Let $\beta$ a reparameterization of $\alpha$ by arc length.
Deriving $\beta(s(r)) = \alpha(r)$, we have
$$\frac{d\beta}{ds}\frac{ds}{dr}=\alpha’(r)\tag1 $$
and
$$\frac{d^2\beta}{ds^2}(\frac{ds}{dr})^2 + \frac{d\beta}{ds}\frac{d^2s}{dr^2} =\alpha''(r)\tag2$$
Where
$$\frac{ds}{dr} = \vert \alpha’(r)\vert\tag3$$
And, therefore,
$$\frac{d^2s}{dr^2}=\frac{\langle\alpha'(r), \alpha''(r)\rangle}{\vert\alpha'(r)\vert}\tag4$$
Considering that $\alpha(r) = (x(r),y(r))$, it follows from (1) and (3) that
$$t(r) = \frac{(x',y')}{\sqrt{(x')^2 + (y')^2}}$$.
For the definition of normal vector, we have
$$n(r) = \frac{(-y',x')}{\sqrt{(x')^2 + (y')^2}}$$.
How
$$k(s(r)) = \langle \frac{d^2\beta}{ds^2}(s(r),n(r)) \rangle$$.
we conclude using (1) to (4) that
$$k(r) = \frac{-x''y' + x'y'}{((x'^2) + (y')^2)^{3/2}}$$
I don't know if I understand correctly, I'm not able to fill in the missing steps.
Thanks for any help.
The curve $\alpha(r)=(x(r),y(r))=(x,y)$ is parametrized by $r$. We will show it simply by $\alpha$ when it is parametrized by its arclenght $s$. The vector functions $t(r)$, $n(r)$ and the curvature $\kappa(r)$ are to be calculated. Apologize us for that we don't use arrows or bold notation for vector functions. When they are parametrized by $s$, we show them simply by $t$, $n$ and $\kappa$. Definitions are given in here: https://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas.
Note that $ds=\sqrt{(x')^{2}+(y')^{2}}dr=vdr$ where $v=\sqrt{(x')^{2}+(y')^{2}}$. First one is easy. Recall that $t=\frac{d\alpha}{ds}$. So in $r$, $$t(r)=\frac{d\alpha}{ds}(r)=\frac{d\alpha}{dr}\frac{dr}{ds}=\frac{(x',y')}{v}=\frac{(x',y')}{\sqrt{(x')^{2}+(y')^{2}}}.$$
The second one is defined by, $n=\frac{dt/ds}{||dt/ds||}$. So lets calculate first, $$\frac{dt}{ds}(r)=\frac{dt}{dr}\frac{dr}{ds}=\frac{(x'',y'')v-(x',y')v'}{v^2}\frac{1}{v}=\frac{2v^2(x'',y'')-(x',y')(2x'x''+2y'y'')}{2v^4}$$ $$=\frac{((y')^2x''-x'y'y'', (x')^2y''-x'y'x'')}{v^4}=\frac{x'y''-y'x''}{v^4}(-y',x').$$ And from this, assuming that $x'y''-y'x''\neq 0$, after normalization we get $$n(r)=\frac{(-y',x')}{\sqrt{(x')^{2}+(y')^{2}}}.$$
And the last one $\kappa$. We already found it! We found $\frac{dt}{ds}=\frac{x'y''-y'x''}{v^3}\frac{(-y',x')}{v}$. But, $\frac{dt}{ds}=\kappa n$ and so $$\kappa=\frac{x'y''-y'x''}{((x')^{2}+(y')^{2})^{\frac{3}{2}}}.$$
Note: For dimension $n=3$, the formulas are ugly, except the tangent vector $t(r)=\frac{(x',y',z')}{v}$ where $v=\sqrt{(x')^2+(y')^2+(z')^2}$.