Proof of the Frenet-Serret formulae

In summary: Since N is a unit vector the dot product of N and the derivative of N is zero, hence the result.In summary, the conversation discusses the unit tangent vector, unit normal vector, and binormal vector parametrized in terms of arc length s. The first part shows that the derivative of the tangent vector with respect to arc length is equal to the curvature times the normal vector. The second part uses this result to show the existence of a scalar -\tau such that the derivative of the binormal vector with respect to arc length is equal to -\tau times the normal vector. The conversation also discusses the definition of \kappa as the rate of change of the tangent vector with respect to arc length and proves that the dot product
  • #1
CAF123
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Homework Statement


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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  • #2
CAF123 said:

Homework Statement


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.
 
  • #3
micromass said:
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.
 

FAQ: Proof of the Frenet-Serret formulae

What is the Frenet-Serret formula?

The Frenet-Serret formula is a mathematical tool used to describe the curvature and torsion of a curve in three-dimensional space. It is commonly used in the study of differential geometry and has applications in physics, engineering, and computer graphics.

Why is the Frenet-Serret formula important?

The Frenet-Serret formula allows us to calculate the curvature and torsion of a curve at any given point, which is essential in understanding the behavior and properties of curves in three-dimensional space. It also helps us analyze and model the motion of objects in space, such as the trajectory of a rocket or the movement of a roller coaster.

How does the Frenet-Serret formula work?

The formula is based on three vectors known as the tangent, normal, and binormal vectors, which are perpendicular to each other and describe the orientation of the curve at a given point. By taking derivatives of these vectors, we can determine the curvature and torsion of the curve using the Frenet-Serret equations.

What are some real-world applications of the Frenet-Serret formula?

The Frenet-Serret formula has many practical applications, including analyzing the motion of objects in space, designing roller coasters and other amusement park rides, and creating computer graphics for video games and movies. It is also used in fields such as robotics, animation, and computer-aided design.

Are there any limitations to the Frenet-Serret formula?

While the Frenet-Serret formula is a powerful tool, it does have some limitations. It is only applicable to smooth curves, meaning that there can be no sudden changes in direction or sharp corners. It also assumes that the curve is non-self-intersecting, meaning that it does not intersect itself. Additionally, the formula does not take into account external forces acting on the curve, such as friction or gravity.

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