Show that a curve in the plane, ${\bf r}(t) = (x(t),y(t),0)$, has curvature $\kappa(t)=\frac{|\dot x \ddot y−\dot y \ddot x|}{{(\dot x^2+\dot y^2)}^{3/2}}$.
My attempt: I have taken some curve in the plane given as ${\bf r}(t)=(x(t),y(t),0)$. I have it written down that ${\bf t}(s)=\frac{d{\bf r}}{ds}, {\bf t'}=\kappa {\bf n}$ and $\frac{ds}{dt}=|\frac{d{\bf r}}{ds}|$ where $s$ is the arc length. Since $\bf n$ is a unit vector, $\kappa=|{\bf t'}|$. By the chain rule ${\bf t}=\frac{d{\bf r}}{ds}=(\frac{d{\bf r}}{dt})(\frac{dt}{ds})=\frac{(\dot x, \dot y, 0)}{\sqrt{\dot x^2+\dot y^2}}$. Similarly, ${\bf t'}=(\frac{d{\bf t}}{dt})(\frac{dt}{ds})=\frac{(\ddot x,\ddot y,0)}{\sqrt{\dot x^2+\dot y^2}\sqrt{\ddot x^2+\ddot y^2}}$. So $\kappa=\frac{\sqrt{\ddot x^2+\ddot y^2}}{\sqrt{\dot x^2+\dot y^2}\sqrt{\ddot x^2+\ddot y^2}}=\frac 1{\sqrt{\dot x^2+\dot y^2}}$, which is clearly wrong.
As I'm writing this right now I've noticed at least one mistake (I think)-$\frac{ds}{dt}=|\frac{d{\bf r}}{ds}|$, I have taken $\frac{ds}{dt}=|\frac{d{\bf t}}{ds}|$ which is incorrect as $s$ is the arclength of $\bf r$. To fix this I think it may be necessary to get ${\bf r}(u(s))$ at the very start, but I'm not sure how to do this, so it may be useful if someone could explain whether or not that's right and if so how to do it. As you can probably tell I'm not very confident on this topic.
To be honest I could be wrong on any one of these steps (or probably all of them), if not that then it's entirely possible I've fundamentally misunderstood everything.
Apologies if any of that (the fractions) is hard to read. Any help is appreciated.
Thank you
For the plane curve, curvature $\kappa$ is given by $$\kappa=\frac{d\psi}{ds}$$ where the tangential angle is $\psi=\arctan\left(\frac{dy}{dx}\right)$ and $s$ is arc length.
So $$\kappa=\frac{d\psi}{dt}\frac{dt}{ds}$$
$$=\frac{d}{dt}\left(\arctan\left(\frac{\dot{y}}{\dot{x}}\right)\right)\frac{1}{\sqrt{\dot{x}^2+\dot{y}^2}}$$
$$=\frac{1}{1+\left(\frac{\dot{y}^2}{\dot{x}^2}\right)}\frac{d}{dt}\left(\frac{\dot{y}}{\dot{x}}\right)\frac{1}{\sqrt{\dot{x}^2+\dot{y}^2}}$$
$$=\frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{(\dot{x}^2+\dot{y}^2)^{\frac 32}}$$