How can we simplify the limit of the angle at a vertex of a traingle as the triangle gets arbitrarily small? (the limit involves the law of cosines)

Question

Let $f$ be a mapping $\mathbb{R} \implies \mathbb{R}$

Let $\forall x, r \in \mathbb{R}$ let $T(x, r)$ be the triangle whose three vertices are the following points:

1. $(x-r, f(x-r))$
2. $(x+0, f(x+0))$
3. $(x+r, f(x+r))$

Let $\theta(x, r)$ be the the interior angle at point $(x, f(x))$ on triangle $T(x, r)$.

Let $\Theta(x)$ be the limit of $\theta(x, r)$ as $r$ approaches $0$.

Question: How can we simplify the following formula?

$\begin{align} \Theta(x) &= \lim \limits_{r \to 0} \theta(x, r) \\ &= \lim \limits_{r \to 0} \cos^{-1}\begin{pmatrix} \dfrac{-(r^{2}) + a^{2}(x,r) + b^{2}(x, r)}{2*a(x,r)*b(x,r)} \end{pmatrix} \\ \end{align}$

where

$\qquad \qquad a(x, r) = \sqrt{r^{2}+ \begin{pmatrix}f(x)-f(x-r)\end{pmatrix}^{2}} $
$\qquad \qquad b(x, r) = \sqrt{r^{2}+ \begin{pmatrix}f(x)-f(x+r)\end{pmatrix}^{2}}$

The law of cosines was used here.

A Remark

If $f$ is differentiable then $\forall x \in \mathbb{R}, Θ(x) = 180º$

However, I am interested in calculating the angle when $f$ represents a polygonal-chain

Answer 1

Define $A = \frac{f(x) - f(x - r)}{r}$ and $B = \frac{f(x + r) - f(x)}{r}$. If $f$ is differentiable, then $A$ and $B$ both converge to $f'(x)$ as $r \rightarrow 0$. But this generalization allows for piecewise functions that "turn the corner" at $(x, f(x))$. Anyhow, if we make the substitutions

$$f(x - r) = f(x) - rA$$ $$f(x + r) = f(x) + rB$$

in your definitions of functions $a$ and $b$, then these functions simplify to:

$$a(x, r) = r \sqrt{1 + A^2}$$ $$b(x, r) = r \sqrt{1 + B^2}$$

Plug these into your formula for $\cos(\theta)$.

$$\cos(\theta(x, r)) = \dfrac{-r^{2} + a^{2}(x,r) + b^{2}(x, r)}{2 a(x,r) b(x,r)}$$ $$= \dfrac{-r^{2} + r^2(1 + A^2) + r^2(1 + B^2)}{2 r \sqrt{1 + A^2} r \sqrt{1 + B^2}}$$ $$= \dfrac{r^2 + r^2A^2 + r^2B^2}{2 r^2 \sqrt{(1 + A^2)(1 + B^2)}}$$ $$= \dfrac{1 + A^2 + B^2}{2\sqrt{(1 + A^2)(1 + B^2)}}$$

If $f$ is differentiable at $x$, then $A = B = f'(x)$, and the above expression simplifies to $1 - \frac{1}{2(1+f'(x)^2)}$. Otherwise,

$$\Theta(x) = \cos^{-1}\left(\dfrac{1 + f'_-(x)^2 + f'_+(x)^2}{2\sqrt{(1 + f'_-(x)^2)(1 + f'_+(x)^2)}}\right)$$

Where $f'_-(x)$ and $f'_+(x)$ denote left and right derivatives.

Answer 2

The angle a line makes with the horizontal in the inverse tangent of the slope. The exterior angle between two line segments is the absolute difference of the angles the line segments make with the horizontal. The interior angle is $\pi$ minus the exterior angle. So,

$\pi - \left|\arctan f'_+(x)-\arctan f'_-(x) \right|$

Taking limits is a bit weird when you're dealing with line segments, though. You can just look at the slopes, or take rise over run over an interval that doesn't include any other vertices.