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?

Question

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.

1. The direction we would go if the car stopped traversing the curve and began traveling in a perfectly straight line.
2. 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.

Answer 1

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.