If we look at a curve drawn on a flat plane, what is the angle between the tangent line and where we would end up if we traveled in a perfect circle? Suppose that $f$ is a differentiable function from $\mathbb{R}$ to $\mathbb{R}$.
For example, maybe $\forall x \in \mathbb{R}$, $f(x) = x^{7} + 8 x^{5} + 301 x^{2} + 99$
At any point on curve $f$ we have two approximations of interest:

*

*an approximation of $f$ by a perfect circle with the same curvature.

*an approximation of $f$ by the line tangent to $f$ at point $(x, f(x))$.

Suppose that you form a triangle whose three vertices are:

*

*$v_{1} = (x, f(x))$

*$v_{2}$ is the point we would arrive at after traveling $1$ unit of distance along the circle which has the same rate of curvature as function $f$ at the point $(x, f(x))$

*$v_{3}$ is the unique point on line tangent to curve $f$ at point $(x, f(x))$ such that $1 = \begin{vmatrix} v_{3} - v_{1} \end{vmatrix}$ and the x-coordinate of $v_{3}$ is greater than the x-coordinate of $v_{1}$
Let $\theta$ be the interior angle at the triangle at vertex  $v_{1} = (x, f(x))$ .
What formula in terms of function $f$ and real-number $x$ will give us angle $\theta$?
This is intended to be a measure of curvature.
Informally, if you were driving a car, I am trying measure the angle between two directions vectors where the vectors lay in the same plane.

*

*The direction we would go if the car stopped traversing the curve and began traveling in a perfectly straight line.

*The direction the car would go if the car's steering wheel was locked in place (frozen in place) and the car began traveling in perfect circles.

 A: Suppose the perfect-circle approximation has radius $r$, and the center of that circle is at $v_4$. Then in order for arc $v_1 v_2$ to have length $1$, the angle $\angle v_1 v_4 v_2$ must measure $\frac1r$ radians.
The other two angles of the isosceles triangle $\triangle v_1 v_4 v_2$ must be $\frac\pi2 - \frac1{2r}$. In particular, the measure of $\angle v_4 v_1 v_2$ is $\frac\pi2 - \frac1{2r}$. Also, we know that $\angle v_4 v_1 v_3 = \frac\pi2$: the radius of the circle is perpendicular to the tangent line to the circle, and the tangent line to the circle at $v_1$ corresponds to the tangent line to the curve. Therefore $\theta = \angle v_2 v_1 v_3 = \angle v_4 v_1 v_3 - \angle v_4 v_1 v_2 = \frac1{2r}$. That's your measure of curvature!
Up to a constant, this is equivalent to the traditional definition of curvature which is $\kappa = \frac1r$. Wikipedia gives a number of formulas for $\kappa$; in the case of a twice-differentiable function $f \colon \mathbb R \to \mathbb R$, we get $\kappa = \frac{|f''(x)|}{(1 + f'(x)^2)^{3/2}}$. Then of course $\theta$ is half of this.
I will also note that the definition of $\theta$ gets into trouble a little if $r$ is too small: specifically, if $r \le \frac1{2\pi}$. Then, traveling $1$ unit of distance around the circle will make multiple entire loops around the circle, and give an angle $\theta$ that depends on how many loops you've made.
