So I am familiar with the theory of Frenet frame and how it is used first to create a unit frame using Frenet formulae.

This problem extends on this idea as follows

"Instead of taking the Frenet frame along a UNIT SPEED curve $\alpha$ : [a,b]$\rightarrow$ $\mathbb{R}$$^{3}$, we can define a frame {T , U , V } by taking T (not unit) to be the tangent vector of $\alpha$ and let U be ANY UNIT vector field along along $\alpha$ such that T $\cdot$ U =0. the problem continues…

My question is I am asked to show that

T' = $\omega$$_3$ $\cdot$ U - $\omega$$_2$V and U' = -$\omega$$_3$ $\cdot$ T + $\omega$$_1$V

where $\omega$ are coefficient functions.

So I am not to sure how to start the problem . I have read the derivation of the frenet frame for Unit vector tangent direction vectors, and then the derivation of the principal normal vector etc. This question asks nearly the same, except that T is not unit.

Just need help getting started..

Thank you in advance.


You want to compare this with the proof we did in lecture (and that's in the textbook) for actually getting to the Frenet formulae.

Specifically, you can write any vector in the orthonormal {T, U, V} frame as just some aT + bU + cV. Try doing that for something like T', ie: write T' = aT + bU + cV.

But then you can also write $$T' = ( T' \cdot T)T + ( T' \cdot U)U + ( T' \cdot V)V $$ One of these terms is zero. By inspection with what you have to prove, you can see what the other two have to be (just coefficients, with indices as given). Repeat the process for U' and V'. You'll need some other differentiation to make the signs work.

Hope this helps,


  • $\begingroup$ thanks I just solved it, the idea that was weird was understanding what a vector field was as opposed to the principal normal. even though that idea doesn't really help much in the calculation. The other thing that took me awhile was seeing how to manipulate an equation of a spanning set. Thank you so much! $\endgroup$ Jan 31 '14 at 5:42
  • $\begingroup$ WHat is the main difference with a vector field of normal vectors perpendicular to T vector, and the idea of a principal normal at a specific point?? as opposed to a field of points? $\endgroup$ Jan 31 '14 at 5:47
  • $\begingroup$ Essentially the meaning of this equation that you had written is as follows: The T' term vanishes which means that the vectors U and V span the space that T' belongs to. so since we know that T'dot U is 0 this is perpendicular, so this is a measure of curvature normal to the plane. But there is another way of measuring curvature that is along the projection of V' which is a different form of curvature, it is a tangent plane. this plane is the key idea of have a vector field with many U(t) perp to it. Why? because the curvature with V dot T' is I guess the plane of which they're tangent to $\endgroup$ Jan 31 '14 at 6:21

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