The change of the angle of the gradient as moving along the curve

Question

I'm given a curve $g = 0$ in 2D specified by g(x,y) = f(x) - y. The normal to the curve is the gradient of $g$ - $(f', -1)$. Now I want express the change in the angle $\theta$ of the normal as I move along the curve. However, I'm not sure how to express this quantity. (For instance if $f$ is linear, this means that $d\theta = 0$ as the normal does not change direction$.

Answer 1

Might be useful, $$\theta = \arctan{\Big(-\frac{1}{f^{'}}\Big)},$$ therefore $$d\theta = \partial_x\Big[\arctan{\Big(-\frac{1}{f^{'}}\Big)}\Big]dx=\frac{f^{''}}{(f^{'})^2+1}dx$$

Answer 2

Convert the presentation of the curve into a parametric representation: $$\gamma:\qquad t\mapsto\left\{\eqalign{x(t)&:=t \cr y(t)&:=f(t)\cr}\right.\tag{1}$$ Now the tangent vector at the point $\bigl(t,f(t)\bigr)$ is given by ${\bf t}=\bigr(1,f'(t)\bigr)$, and the normal vector by turning ${\bf t}$ by $90^\circ$ counterclockwise, hence ${\bf n}=\bigl(-f'(t),1\bigr)$.

Since $x'(t)\ne0$ for all $t$ the representation $(1)$ is regular; therefore both ${\bf t}$ and ${\bf n}$ are $\ne{\bf 0}$ for all $t$.