1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proof of the Frenet-Serret formulae

  1. Sep 22, 2012 #1


    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Consider the unit tangent vector [itex] T[/itex] unit normal vector [itex] N [/itex] and binormal vector [itex] B [/itex] parametrized in terms of arc length s.
    1) Show that [tex] \frac{dT}{ds} = \kappa\,N[/tex]

    I think this part is fine for me. What I did was: [tex]N(t) = \frac{T'(t)}{|T'(t)|}[/tex] and said, by the chain rule, [itex] \frac{dT}{ds} \frac{ds}{dt}= T'(t) [/itex] which simplified to [tex] N(s) = \frac{|r'(t)|}{|T'(t)|} \frac{dT}{ds} => \frac{dT}{ds} = \kappa N(s) [/tex]
    Can somebody confirm this is correct?

    2) Use a) to show that there exists a scalar [itex] -\tau [/itex] such that [tex] \frac{dB}{ds} = -\tau\,N [/tex]

    I was given a hint to try to show that [itex] \frac{dB}{ds} . B = 0 [/itex]
    I took the derivative [tex]\frac{d}{ds} B = \frac{d}{ds}(T ×N) = T ×\frac{dN}{ds}[/tex]
    Therefore, [tex] (T × \frac{dN}{ds}) . B = (B ×T) . \frac{dN}{ds} = N . \frac{dN}{ds}. [/tex] Am I correct in assuming the above is equal to 0?

    Many thanks.
  2. jcsd
  3. Sep 22, 2012 #2
    It's weird that you need to use the chain rule. But first, what is the definition of [itex]\kappa[/itex]??

    Yes, it is true that [itex]N\cdot \frac{dN}{ds}=0[/itex], but it needs to be proven. To prove this, consider the function [itex]N\cdot N[/itex] and take derivatives.
  4. Sep 22, 2012 #3


    User Avatar
    Gold Member

    [itex] \kappa [/itex] is the rate of change of the tangent vector of a curve with respect to arc length, ie a measure of how much the tangent vector changes in magnitude and direction as a point moves along a curve. Why is it weird using chain rule - did you have another method in mind - is mine ok?

    Thanks, so an argument could go;[tex] \frac{d}{ds} (N\cdot N) = 2(N\cdot \frac{dN}{ds}).[/tex] We know the dot product is a scalar quantity, so the left hand side is zero (derivative of a constant) and so the right hand side is only true provided this term in brackets is zero.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook