I am trying to understand the curvature of a plane curve. The Curvature is given by: $$\kappa = \frac{i+f'(x)j}{(1+f'(x)^2)^{\frac{1}{2}}} \times \frac{f''(x)j}{(1+f'(x)^2)^{\frac{1}{2}}} \div \sqrt{1+f'(x)^2} =\frac{|f''(x)|}{(1+f'(x)^2)^{\frac{3}{2}}} $$
and also by: $$\kappa = \frac{d\theta}{ds}$$
I can see that $\sqrt{1+f'(x)^2}$ is an arc length $ds$ and equivalent to $|r'(t)|$.
$\frac{i+f'(x)j}{(1+f'(x)^2)^{\frac{1}{2}}}$ is equivalent to T, $\frac{r'}{|r'|}$ and $\frac{f''(x)}{(1+f'(x)^2)^{\frac{1}{2}}}$ is equivalent to $\frac{r''(t)}{|r'(t)|}$ in 3D.
But where in this equation is $d\theta$ equivalent to?
In the plane we have the argument function ${\rm arg}:\>\dot{\mathbb R}^2\to{\mathbb R}/(2\pi)$, also known as polar angle. When $x>0$ then ${\rm arg}(x,y)=\arctan{y\over x}$. This ${\rm arg}$ function has a well defined gradient $$\nabla{\rm arg}(x,y)=\left({-y\over x^2+y^2}, {x\over x^2+y^2}\right)\ .\tag{1}$$
The $\theta$ in the above formulas is the argument of the tangent vector $\bigl(1,f'(x)\bigr)$, resp. $\bigl(x'(t),y'(t)\bigr)$: $$\theta(t)={\rm arg}\bigl(x'(t),y'(t)\bigr)\ .$$ The formula $$\kappa:={d\theta\over ds}={\theta'(t)\over s'(t)}$$ 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 $${\theta'(t)\over s'(t)}={1\over s'(t)}\>\nabla{\rm arg}\bigl(x'(t),y'(t)\bigr)\cdot(x''(t),y''(t)\bigr)={x'(t)y''(t)-x''(t)y'(t)\over s'^3(t)}\ .$$