# Proof of the Frenet-Serret formulae

by CAF123
Tags: formulae, frenetserret, proof
 PF Gold P: 2,307 1. The problem statement, all variables and given/known data 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.
Mentor
P: 18,293
 Quote by CAF123 1. The problem statement, all variables and given/known data 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.
PF Gold
P: 2,307
 Quote by micromass 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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