Formula for Curvature

Question

How do I derive the formula for curvature through differentiation?

$$ k(t)= \dfrac{(x'y''-x''y')}{(x'^2+y'^2)^\frac32} $$

I know that $k(t)=$ the modulus of $ T'(s) $ where $T$ is the unit tangent vector

Answer 1

From the formula where primes denote differentiation w.r.t single independent variable x we can make sharing by new independent variable say $t$. This is more convenient when sticking still to the osculating plane, keeping away from torsion effects.

$$ \dfrac{d \, \phi}{ds} =\dfrac{d(\tan^{-1}y^{'})} {ds} = \dfrac{y''}{(1+y^{'2})^\frac32}$$

plug in

$$ y^{'}= \dfrac{\dot y}{\dot x} ;\quad ds= dx \sec \phi $$

with a different independent variable say $t$ into the middle term above

$$ \dfrac{1}{1+\dfrac{\dot y^2}{\dot x^2}}\cdot \dfrac { \dot x \ddot y-\ddot x \dot y}{\dot x^2}\cdot \dfrac{1}{\dot x\sqrt{ {1+\dfrac{\dot y^2}{\dot x^2}}}} $$

and simplify to get the new parametric formula