I have to prove that for any curve, the radius of curvature, $\rho$ , the following relation is true where $x = f(t)$, $y = \phi(t)$, i.e., the parametric form is to be used
$$\frac{1}{\rho^2} = \frac{\bigg\{\bigg(\frac{d^2x}{dt^2}\bigg)^2+\bigg(\frac{d^2y}{dt^2}\bigg)^2-\bigg(\frac{d^2s}{dt^2}\bigg)^2\bigg\}}{\bigg(\frac{ds}{dt}\bigg)^4}$$
Now to proceed with the proof, I'm trying to use the relation, $$\frac{ds}{dt} = \sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}$$
But, if I square the above relation and differentiate it with respect to $t$, I end up with $$\frac{ds}{dt}\frac{d^2s}{dt^2}= \frac{dx}{dt}\frac{d^2x}{dt^2}+ \frac{dy}{dt}\frac{d^2y}{dt^2}$$
Then squaring it, because I want to get the $\bigg(\frac{d^2s}{dt^2}\bigg)^2$ term, as is given in the RHS of the given relation, I end up with something much more complicated which doesnt seem to reduce to anything useful. How do I approach this proof?
If $x=x(t)$ and $y=y(t)$, we set :
$\frac{dx}{dt}=\dot{x}$, $\frac{dx}{ds}=x’$ and $\frac{dt}{ds}=t’=\frac{1}{\dot{s}}$
$\frac{dy}{dt}=\dot{y}$ , $\frac{dy}{ds}=y’$,
then the first Frenet formula says:
$\frac{1}{\rho}*\vec{n}=\frac{d\vec{\tau}}{ds}$
where
$\vec{\tau}$ is the tangent in generic point of the curve line and $\vec{n}$ is the normal line.
In the Cartesian plane, the tangent equation is
$\vec{\tau}=x’\vec{i}+y’\vec{j}$,
the first derivative is
$\frac{d\vec{\tau}}{ds}= x’’\vec{i}+y’’\vec{j}$.
The normal line equation is
$\vec{n}=\frac{-\dot{y}}{\dot{s}}\vec{i}+\frac{+\dot{x}}{\dot{s}}\vec{j}$.
Squaring the first Frenet formula,we have:
$\frac{1}{\rho^{2}}=x’’^{2}+y’’^{2}$
because
$\vec{n}.\vec{n}=1$
$\big(\frac{\vec{d\tau}}{ds}\big)^{2}= \big(x’’\vec{i}+y’’\vec{j}\big)^{2}=x’’^{2}+y’’^{2}$.
That said, we express the formula as a function of the variable t.
$x’=\frac{\dot{x}}{\dot{s}}$,
$x’’=\frac{\dot{s}\ddot{x}t’-\dot{x}\ddot{s}t’}{\dot{s}^{2}}$,
$x’’=\frac{\dot{s}\ddot{x}-\dot{x}\ddot{s}}{\dot{s}^{3}}$,
(because $t’=\frac{dt}{ds}=\frac{1}{\dot{s}}$),
$y’’=\frac{\dot{s}\ddot{y}t’-\dot{y}\ddot{s}t’}{\dot{s}^{2}}$,
$y’’=\frac{\dot{s}\ddot{y}-\dot{y}\ddot{s}}{\dot{s}^{3}}$;
and more
$x’’^{2}=\frac{\ddot{x}^{2}\dot{s}^{2}+\dot{x}^{2}\ddot{s}^{2}-2\dot{x}\ddot{x}\dot{s}\ddot{s}}{\dot{s}^6}$,
$y’’^{2}=\frac{\ddot{y}^{2}\dot{s}^{2}+\dot{y}^{2}\ddot{s}^{2}-2\dot{y}\ddot{y}\dot{s}\ddot{s}}{\dot{s}^6}$,
and making their sum, we get
$x’’^{2}+y’’^{2}=\frac{\dot{s}^2\big(\ddot{x}^2+\ddot{y}^2\big)+\ddot{s}^2\big(\dot{x}^2+\dot{y}^2\big)-2\dot{s}\ddot{s}\big(\dot{x}\ddot{x}+\dot{y}\ddot{y}\big)}{\dot{s}^6}$;
making the derivative of the equation of the differential of the curve arc:
$\dot{x}^2+\dot{y}^2=\dot{s}^2$,
we have that
$\dot{x}\ddot{x}+\dot{y}\ddot{y}=+\dot{s}\ddot{s}$,
The sum of the squares of the second derivatives with respect to the variable $s$ is,
$x’’^{2}+y’’^{2}=\frac{\dot{s}^2\big(\ddot{x}^2+\ddot{y}^2\big)+\ddot{s}^2\big(\dot{x}^2+\dot{y}^2\big)-2\dot{s}^2\ddot{s}^2}{\dot{s}^6}$,
that is,
$x’’^{2}+y’’^{2}=\frac{\dot{s}^2\big(\ddot{x}^2+\ddot{y}^2\big)+\ddot{s}^2\dot{s}^2-2\dot{s}^2\ddot{s}^2}{\dot{s}^6}$,
$\frac{1}{\rho^{2}}= \frac{\ddot{x}^2+\ddot{y}^2-\ddot{s}^2}{\dot{s}^4}$.