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 $\frac{v^2}{a_\perp}$
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$\begingroup$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}
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ $\begingroup$I derived something, and it may be exactly what you were (or maybe still are) looking for. I show every step, so the math is a bit stretched (4 pages), but I feel it is better to be simpler than shorter. It's based on the definition you stated in your question.
An alternative derivation of radius of curvature (2D functions). How valid is it?
$\endgroup$ $\begingroup$The radius of curvature of a curve at a point is the radius of the circle that best approximates the curve at that point.
So first, let us find the differential equation representing the family of circles with a particular radius $r_0$. The equation of a circle is$$(x-h)^2 + (y-k)^2 = r_0^2$$Differentiating two times and finding the expressions for h and k in terms of $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ gives
$$y-k = -\left[\cfrac{1+ \left(\frac{dy}{dx} \right)^{2}}{\frac{d^2y }{dx^2}}\right]$$$$x-h = \left(\frac{dy}{dx}\right)\left[\cfrac{1+\left(\frac{dy}{dx}\right)^{2}}{\left(\frac{d^2y}{dx^2}\right)}\right]$$
and$$\left[\cfrac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}\right] = r_0$$
Let $f(x)$ be a general function a general point on its graph would be $(t, f(t))$
Now, consider all the circles passing through $(t,f(t))$.
If we want to approximate the function in the neighbourhood of that point with a circle,
it would be good if as many of the derivatives as possible of the function at that point equal the derivatives of the circle at the same point.
[Similar to how the polynomial approximation (i.e. Taylor series) in general increases in accuracy with greater number of terms]
But we see that if we fix the values:$\frac{dy}{dx} = f'(t)$ and$\frac{d^2y}{dx^2} = f''(t)$, it fixes both the centre of the circle as well as the radius.
Hence we cannot expect any further derivatives (i.e. $\frac{d^3y}{dx^3}$ etc) to be equal for the function and the circle, and this is the best possible approximation of the curve as a circle, at that point. And the radius of this circle would be the radius of curvature of the graph at that point.
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