Radius of Curvature

I was asked to show that the expression is constant in a circle :

$\dfrac{\left[1+\left(\dfrac{\operatorname d \!y}{\operatorname d \!x}\right)^2\right]^{\frac 3 2}}{\dfrac{\operatorname d \!^2y}{\operatorname d \!x^2}}$

I found it to be equal to Radius.

Curiosity led me to search more about it on internet. That says that it is radius of curvature of a curve. But I can't find a proof of it.

Please give a proof which I can understand (as simple as possible). Please use a geometric approach. I see that the numerator is cube of $dl$ (differential arc length)

I see that we are lacking a definition of radius of curvature : I want to use the most obvious definition(to me) : Distance of point from centre of curvature at that point where the centre is defined as intersection of two infinitesimally close normals.

Or we could use the physics like $\frac{v^2}{a_\perp}$

Solution 1:

For a curve, $ds = \sqrt{dx^2+dy^2}$.

Also, $\tan \theta=\frac{dy}{dx}$.

So, we have

\begin{eqnarray} \frac{ds}{d\theta}&=&\frac{ds}{dx}.\frac{dx}{d\theta}=\frac{\frac{ds}{dx}}{\frac{d\theta}{dx}}\\ &=&\frac{\sqrt{1+\left(\frac{dy}{dx}\right)^2}}{\frac{d\left(\tan^{-1}\frac{dy}{dx}\right)}{dx}} \\&=&\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}} \end{eqnarray}

Solution 2:

It's easier to start with curvature. Given a curve $$\gamma:\quad t\mapsto {\bf z}(t)=\bigl(x(t),y(t)\bigr)\qquad(a<t<b)$$ the vector $\dot{\bf z}(t)=\bigl(\dot x(t),\dot y(t)\bigr)$ points for each $t$ in the direction of the forward tangent vector at ${\bf z}(t)$. Curvature is about the speed by which this tangent vector turns. As this is a purely geometric concept time $t$ should not enter into the definition. This means that we have to measure this speed with respect to arc length $s$.

The polar angle of the tangent vector is given by $\theta(t)=\arg\bigl(\dot {\bf z}(t)\bigr)$. It follows by the chain rule that $$\dot\theta(t)=\nabla\arg\bigl(\dot{\bf z}(t)\bigr)\cdot\ddot{\bf z}(t)\ .$$ As $\nabla\arg(x,y)=\left({-y\over x^2+y^2},{x\over x^2+y^2}\right)$ we obtain $$\dot\theta(t)=\left({-\dot y\over \dot x^2+\dot y^2},{\dot x\over \dot x^2+\dot y^2}\right)\cdot\bigl(\ddot x(t),\ddot y(t)\bigr)={\dot x \ddot y-\dot y\ddot x\over \dot x^2+\dot y^2}\ .$$ Now we need ${d\theta\over ds}$ instead of ${d\theta\over dt}$. Since $\dot s=\sqrt{ \dot x^2+\dot y^2}$ applying the chain rule again gives $$\kappa={d\theta\over ds}={\dot\theta\over\dot s}={\dot x \ddot y-\dot y\ddot x\over \left(\dot x^2+\dot y^2\right)^{3/2}}\ .\tag{1}$$ Apply formula $(1)$ to the case of a graph $x\mapsto\bigl(x, y(x)\bigr)$ (here $x$ is the parameter) and take the reciprocal to obtain the radius of curvature.