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Proof of the Frenet-Serret formulae

by CAF123
Tags: formulae, frenetserret, proof
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CAF123
#1
Sep22-12, 08:23 AM
PF Gold
P: 2,307
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.
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micromass
#2
Sep22-12, 10:08 AM
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P: 18,293
Quote Quote by CAF123 View Post
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?
It's weird that you need to use the chain rule. But first, what is the definition of [itex]\kappa[/itex]??

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?
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.
CAF123
#3
Sep22-12, 12:36 PM
PF Gold
P: 2,307
Quote Quote by micromass View Post
It's weird that you need to use the chain rule. But first, what is the definition of [itex]\kappa[/itex]??
[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?


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.
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.


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