0
$\begingroup$

This question already has an answer here:

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

$\endgroup$

marked as duplicate by Robert Z, GNUSupporter 8964民主女神 地下教會, Raskolnikov, Nosrati, Did Jan 27 '18 at 13:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Is it possible without using the dot product? $\endgroup$ – user519957 Jan 27 '18 at 10:14
  • $\begingroup$ A second answer has been posted at the site of the first posting of this question: math.stackexchange.com/questions/275248/… $\endgroup$ – Christian Blatter Jan 27 '18 at 16:35
0
$\begingroup$

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

$\endgroup$