0
$\begingroup$

The radius of curvature of a given path is $$r = \frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}.$$

This is kind of a formula I've been seeing for two years without knowing why it is so. Any help is appreciated.

  • 1
    Have you tried drawing a diagram - if you were given the task of determining the radius of curvature at a point given the equation, how would you proceed? If you knew the coordinates of three points on a circle, how would you calculate the radius? Now put these points close together... – 2017-02-17
  • 0
    I can't figure it out. :( – 2017-02-17
  • 0
    -1. No research effort. This is a question about mathematics not physics. – 2017-02-17
  • 0
    "Radius of curvature equation derivation" search finds http://www.solitaryroad.com/c361.html - see if that gets you going – 2017-02-17
  • 0
    This is a results from differential geometry and not terribly hard to derive. Differential geometry is very cool. Gauss figured out a lot of it, without sharing it much, like usual. There are plenty of books on the subject. One that is not ancient is by Kreyszig, and very good, and should be on the used market. Not the paperback. This one "Introduction to Differential Geometry and Riemannian Geometry". Really lovely stuff. – 2017-02-17
  • 2
    Would [math.se] be a better home for this question? – 2017-02-17
  • 0
    Well, all the books I've read about it are Physics books. That's why I posted it here. – 2017-02-17
  • 0
    This formula looks familiar to the formula of the length of normal in mathematics...If a circle's centre lies on the origin..Then I think you would be able to get this answer....Have you tried doing this..? – 2017-02-17
  • 0
    I did try. Nothing worked. I'm checking on the link provided by #Floris – 2017-02-17

0 Answers 0