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In the Frenet Formula, I can see why the derivative of the tangent vector is a function of the curvature times the normal vector, and the derivative of the binormal vector is a negative function of the torsion times the normal vector, (where B=TxN). But I don't quite understand why the derivative of the normal vector is function of BOTH the curvature and torsion.

Apparently, there is a "matrix symmetry" proof of this theorem using these vectors and their derivatives. But I'm looking for a calculus-based proof (if it exists). Can you express the normal as some product of the binormal and tangent vectors and then use a product rule? Or can you do some kind of convolution, inversion-type manipulation.

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If $v$ and $w$ are orthogonal unit vectors in $\mathbb{R}^3$ and $x=v\times w$ is their cross product, then you also have $v=w\times x$ and $w=x\times v$. In this case $B=T\times N$, so $N'=(B\times T)'=B'\times T+ B\times T'=-\tau N\times T +\kappa B\times N=-\tau(-B)+\kappa (-T)=-\kappa T+\tau B$.

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Observe that $B, T, N$ are orthonormal. So in particular $B\cdot N = 0 = T \cdot N$, which implies that

$$ 0 = \frac{d}{ds} (B\cdot N) = \dot{B} \cdot N + B \cdot \dot{N} $$

You already know what $\dot{B}$ is. Similarly we take the derivative of $T\cdot N$. Lastly, since $N\cdot\dot{N} = \frac12 \frac{d}{ds} (N\cdot N) = 0$, we have that

$$ \dot{N} = (\dot{N}\cdot N) N + (\dot{N}\cdot B) B + (\dot{N}\cdot T) T = 0N - (\dot{B}\cdot N) B - (\dot{T}\cdot N) T $$

where you can plug in the values from the expressions you already have.

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  • $\begingroup$ Alternatively, using that $B = T \times N$ is a choice of orientation on the basis vectors $B,T,N$, we see that the cross product relations between the three requires $N = B\times T$. Then you can compute $\dot{N}$ using the product rule as you mentioned. $\endgroup$ – Willie Wong Jun 25 '11 at 18:51

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