Proof of the Frenet-Serret formulae

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SUMMARY

The discussion focuses on the proof of the Frenet-Serret formulas, specifically demonstrating that \(\frac{dT}{ds} = \kappa N\) and \(\frac{dB}{ds} = -\tau N\). Participants confirm the correctness of using the chain rule for deriving the tangent vector and discuss the implications of the scalar \(\kappa\), which represents the curvature of a curve. The necessity of proving that \(N \cdot \frac{dN}{ds} = 0\) is emphasized, with suggestions to utilize the derivative of the dot product \(N \cdot N\) to establish this fact.

PREREQUISITES
  • Understanding of vector calculus, particularly tangent, normal, and binormal vectors.
  • Familiarity with the concepts of curvature (\(\kappa\)) and torsion (\(\tau\)).
  • Knowledge of the chain rule in calculus.
  • Ability to compute derivatives of vector functions.
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  • Study the derivation of the Frenet-Serret formulas in detail.
  • Learn about the geometric interpretation of curvature and torsion in 3D space.
  • Explore applications of the Frenet-Serret formulas in physics and engineering.
  • Investigate the implications of vector derivatives in the context of differential geometry.
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Homework Statement


Consider the unit tangent vector T unit normal vector N and binormal vector B parametrized in terms of arc length s.
1) Show that \frac{dT}{ds} = \kappa\,N

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

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

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

Many thanks.
 
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CAF123 said:

Homework Statement


Consider the unit tangent vector T unit normal vector N and binormal vector B parametrized in terms of arc length s.
1) Show that \frac{dT}{ds} = \kappa\,N

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

It's weird that you need to use the chain rule. But first, what is the definition of \kappa??

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

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

Yes, it is true that N\cdot \frac{dN}{ds}=0, but it needs to be proven. To prove this, consider the function N\cdot N and take derivatives.
 
micromass said:
It's weird that you need to use the chain rule. But first, what is the definition of \kappa??
\kappa 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 N\cdot \frac{dN}{ds}=0, but it needs to be proven. To prove this, consider the function N\cdot N and take derivatives.
Thanks, so an argument could go;\frac{d}{ds} (N\cdot N) = 2(N\cdot \frac{dN}{ds}). 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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