Definition of curvature and intuition I am trying to derive the formula for curvature by finding the radius of best fitting circle. So let $$ \vec{r} \left(t \right)$$
be the parametrization of the curve in question. Let's define tangent lines at points $$ \vec{r} \left(t_0 \right) \ \mathrm{and} \ \vec{r} \left(t_0+\Delta t \right).$$
Then when we let $$ \Delta t \longrightarrow 0,$$
the intersection of normal lines should approach the center of the circle best fitting to point $$ \vec{r} (t_0).$$ And then we could find the radius of curvature R and define curvature as $$ \kappa= 1/R.$$
The problem is that computing this seems a bit harder than I thought. Most of my calculators can't do it. The reason why I use this geometric approach is that it sort of gives intuition to what curvature means. Instead of simply postulating the formula as definition of curvature this  has nice intuitive meaning.
So could someone help me derive the formula? I have seen some other derivations as well but those don't really consider the best fitting circles. And they mostly use something called arc length parameter which I am not familiar with.
 A: Very quickly since I have done it previously on this site: Suppose we have a curve $(x(t),y(t))$ on the plane.  The normal at parameter $t$ is
$$
x'(t)(x-x(t))+y'(t)(y-y(t))=0
$$
and similarly the normal at parameter $t+\Delta t$ is
$$
x'(t+\Delta t)(x-x(t+\Delta t))+y'(t+\Delta t)(y-y(t+\Delta t))=0
$$
Solving, we have
$$
x=\frac{\mbox{$x'(t)y'(t+\Delta t)x(t)+y'(t)y'(t+\Delta t)y(t)$}
-\mbox{$y'(t)x'(t+\Delta t)x(t+\Delta t)-y'(t)y'(t+\Delta t)y(t+\Delta t)$}}{x'(t)y'(t+\Delta t)-y'(t)x'(t+\Delta t)}
$$
and similarly for $y$.
Letting $\Delta t\to 0$, both numerator and denominator $\to 0$.  The first order term in denominator is $[x'(t)y''(t)-y'(t)x''(t)]\Delta t$ and the numerator is $[[x'(t)y''(t)-y'(t)x''(t)]x(t)-y'(t)[x'(t)^2+y'(t)^2]]\Delta t$, similarly for $y$.
So the center of curvature is
$$
\begin{bmatrix}x(t)\\y(t)\end{bmatrix}+\frac{x'(t)^2+y'(t)^2}{x'(t)y''(t)-y'(t)x''(t)}
\begin{bmatrix}-y'(t)\\x'(t)\end{bmatrix}
$$
and we can read off $R=\frac{(\dot{x}^2+\dot{y}^2)^{3/2}}{\dot x\ddot y-\ddot x\dot y}$.
