Vtyjkyuk

I am trying to understand the curvature of a plane curve. The Curvature is given by: $$kappa = fraci+f'(x)j(1+f'(x)^2)^frac12 times fracf''(x)j(1+f'(x)^2)^frac12 div sqrt1+f'(x)^2 =frac(1+f'(x)^2)^frac32 $$

and also by: $$kappa = fracdthetads$$

I can see that $sqrt1+f'(x)^2$ is an arc length $ds$ and equivalent to $|r'(t)|$.

$fraci+f'(x)j(1+f'(x)^2)^frac12$ is equivalent to T, $fracr'$ and $fracf''(x)(1+f'(x)^2)^frac12$ is equivalent to $fracr''(t)$ in 3D.

But where in this equation is $dtheta$ equivalent to?

In the plane we have the argument function $rm arg:>dotmathbb R^2tomathbb R/(2pi)$, also known as polar angle. When $x>0$ then $rm arg(x,y)=arctanyover x$. This $rm arg$ function has a well defined gradient
$$nablarm arg(x,y)=left(-yover x^2+y^2, xover x^2+y^2right) .tag1$$

The $theta$ in the above formulas is the argument of the tangent vector $bigl(1,f'(x)bigr)$, resp. $bigl(x'(t),y'(t)bigr)$:
$$theta(t)=rm argbigl(x'(t),y'(t)bigr) .$$ The formula
$$kappa:=dthetaover ds=theta'(t)over s'(t)$$
then is saying that the curvature $kappa$ is the angular velocity (with respect to arc length) with which the tangent vector rotates. I think this is intuitively very satisfying. Using $(1)$ and the chain rule one obtains
$$theta'(t)over s'(t)=1over s'(t)>nablarm argbigl(x'(t),y'(t)bigr)cdot(x''(t),y''(t)bigr)=x'(t)y''(t)-x''(t)y'(t)over s'^3(t) .$$