Proving the curvature formula for an arbitrary planar curve using perpendicular bisectors

Question

Given an arbitrary (i.e. not necessarily arc-length parameterised) planar parametric curve $C(t) = \Big(x(t), y(t)\Big)$, I'm looking to prove the formula for its (signed) curvature

$$\kappa = \frac{x'y'' - x''y'}{(x'^2 + y'^2)^\frac32}.$$

I would like to do so using perpendicular bisectors, as this seems the most natural approach to me.

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My attempt starts as follows. Given three points on the curve $P = C(t-t_1)$, $Q = C(t)$ and $R = C(t+t_2)$, the intersection of the perpendicular bisectors of the edges connecting these points (see the illustration below) yields the centre of the circle going through $P$, $Q$ and $R$. In the limit, as both $t_1 \to 0$ and $t_2 \to 0$, the circle through the three points is known as the osculating circle with a radius $r = \frac{1}{\kappa}$.

The intersection of the perpendicular bisectors can be expressed as $S + \alpha u = T + \beta v$, with $S$ and $T$ the midpoints of the edges $PQ$ and $QR$, $\alpha$ and $\beta$ unknown scalar values, and $u$ and $v$ chosen to be the vectors $Q-P$ and $R-Q$ rotated $90$ degrees counter-clockwise. That is,

$$u = \Big( -(Q-P)_y, (Q-P)_x \Big) = \Big( y(t-t_1)-y(t), x(t)-x(t-t_1) \Big), \\ v = \Big( -(R-Q)_y, (R-Q)_x \Big) = \Big( y(t)-y(t+t_2), x(t+t_2) - x(t) \Big).$$

Re-expressing, we have $\alpha u - \beta v = T - S$, or

$$\left(\begin{array}{cc}y(t-t_1)-y(t) & y(t+t_2)-y(t) \\ x(t)-x(t-t_1) & x(t) - x(t+t_2)\end{array}\right) \left(\begin{array}{c}\alpha \\ \beta\end{array}\right) = \frac12 \left(\begin{array}{c}x(t+t_2) - x(t-t_1) \\ y(t+t_2) - y(t-t_1) \end{array}\right).$$

We can solve for $\alpha$ and $\beta$ by inverting this $2 \times 2$ matrix. Expressing the matrix entries symbolically, recall that

$$\left(\begin{array}{cc}u_x & -v_x \\ u_y & -v_y\end{array}\right)^{-1} = \frac{\left(\begin{array}{cc}-v_y & v_x \\ -u_y & u_x\end{array}\right)}{v_x u_y - u_x v_y}.$$

Taking the limit of $t_1 \to 0$ and $t_2 \to 0$, we should then get $r = \alpha \|u\| = \beta \|v\|$.

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Unfortunately, this doesn't appear to result in the desired curvature formula (or well, its reciprocal). It probably means different lengths for the vectors $u$ and $v$ should be used — dividing them by $t_1$ and $t_2$, respectively, yields something that in the limit looks like the components of the tangent vector $C'(t) = \Big( x'(t), y'(t) \Big)$. However, that does not resolve the second derivatives appearing in the formula (which I suppose come from $T - S$). How to proceed?

Answer 1

You have $Q = C(t)$ and $R = C(t + h) $

The center of the circle passing through these two points satisfies

$(S - Q) \cdot (S - Q) = (S - R) \cdot (S - R) $

so that

$ S \cdot S - 2 S \cdot Q + Q \cdot Q = S \cdot S - 2 S \cdot R + R \cdot R $

from which

$ 2 S \cdot (R - Q ) = R \cdot R - Q \cdot Q \hspace{15pt}(1) $

Using Taylor series approximation of $R$, we get

$R = C(t + dt) = C(t) + h C'(t) + \frac{1}{2} h^2 C''(t) $

Plugging this into $(1)$ and retaining only differentials of first and second order, we get

$ 2 S \cdot ( h C' + \frac{1}{2} h^2 C'' ) = 2 h C \cdot C' + h^2 (C \cdot C'' + C' \cdot C' ) $

From this, it follows that

$ S \cdot C' = C \cdot C' \hspace{15pt}(2)$

and

$ S \cdot C'' = C \cdot C'' + C' \cdot C'\hspace{15pt}(3) $

Equation $(2)$ can be written as

$ (S - C) \cdot C' = 0 \hspace{15pt}(4) $

So that $(S - C)$ is perpendicular to $C'$. It follows that

$ S = C + \alpha R(90^\circ) C' \hspace{15pt} (5) $

Explicitly writing the components of $C= (x, y)$, $C' = (x', y')$, and $C'' = ( x'', y'' ) $, then $(5)$ becomes

$ S = (x - \alpha y' , y + \alpha x' ) $

Plugging this into $(3)$

$ x'' (x - \alpha y') + y'' (y + \alpha x') = x x'' + y y'' + x'^2 + y'^2$

Cancelling equal terms on both sides of the equation and solving for $\alpha$

$\alpha = \dfrac{ x'^2 + y'^2 }{ x' y'' - y' x'' } $

Now the radius of the circle is $\| S - Q \| = \alpha \| C' \| = \alpha \sqrt{x'^2 + y'^2 } $

Hence

$ r = \dfrac{ (x'^2 + y'^2)^{\frac{3}{2}} } { x' y'' - y' x''} $

and the curvature is

$ \kappa = \dfrac{1}{r} =\dfrac { x' y'' - y' x''}{ (x'^2 + y'^2)^{\frac{3}{2}} } $

Answer 2

Using perpendicular bisectors we can define the osculating circle whose center $C$ is the center of curvature and whose radius is $\rho$.

The circle lying in the osculating plane at the curve at the point $P$, having radius $\rho$ is called the "osculating circle". We can show that: "the osculating circle is the limit figure of a circle passing through three points of the given curve, when they tend to coincide in only one. Let the center of this last circle be $C$, its radius $r$ and $P$, $P_{1}$, $P_{2}$ the three points that tend, in the passage to the limit, to coincide with $P$. Setting

$F(t)=(C-M)^{2}-r^{2}$,
which represents the equation of a circle of center $C$ and radius $r$ , passing through the generic point $M$.

Then the function $F(t)$ must be null when instead of $M$ we place $P$, $P_{1}$, $P_{2}$, since it is null when $M$ is a point of the circle of center $C$ and radius $r$.

By Rolle's theorem $F(t)$ is null at a point $P_{1}*$ between $P$ and $P_{1}$ and at a point $P_{2}*$ between $P_{1}$ and $P_{2}$;

and again for the same theorem $F''(t)$ is null at a point $P*$ between $P_{1}*$ and $P_{2}*$.

When $P_{1}$ and $P_{2}$ tend to $P$, $P_{1}*$, $P_{2}*$ and $P*$ also tend to $P$, then the center $C$ is defined by the equations:

$F(t)=0$, $F’(t)=0$, $F’’(t)$ ;

i.e.

$(C-P)^{2}=r^{2}$,

$(C-P)\dot{P}=0$,

$(C-P)\ddot{P}=\dot{P}^{2}$.

The first equation is the equation of the center circle $C$ and radius $r$, the second says that $C$ lies in the plane normal to the curve in $P$; therefore $C$ is on the main normal, because, having to be coplanar with $P$, $P_{1}$, $P_{2}$ at the limit, it must be on the osculating plane.

The third equation taking as a parameter the arc $s$, becomes:

$(C-P) \frac{1}{\rho} \vec{n}=1$,

i.e. remembering that $r$ is the module of $C-P$ e che $C-P$ ed $\vec{n}$ are parallel:

$r=\rho$.

At this point referring to Prove a relation for radius of curvature, using parametric coordinates

and in particular the formula:

$\frac{1}{\rho^{2}}= \frac{\ddot{x}^2+\ddot{y}^2-\ddot{s}^2}{\dot{s}^4}$

Placing in place of

$\ddot{s}=\frac{\dot{x}\ddot{x}+\dot{y}\ddot{y}}{\dot{s}}$,

and in place of

$\dot{s}^2=\dot{x}^2+\dot{y}^2$,

we get:

$\frac{1}{\rho}=\frac{\dot{x}\ddot{y}-\dot{y}\ddot{x}}{\Big(\dot{x}^2+\dot{y}^2\Big)^{\frac{3}{2}}}$.