2

I can't seem to understand how this cross product is computed, maybe I am missing something obvious, so any help would be appreciated.

We have $${\mathbf r}(s,\theta) =\gamma(s)+a({\mathbf n}(s)\cos\theta+{\mathbf b}(s)\sin\theta),$$ where ${\mathbf n},{\mathbf b}$ are the main vectors and $ka<1$ with $a>0$,

$$ r_{s}= (1-ka \cos\theta )t-\tau {\mathbf n}a \sin\theta +\tau {\mathbf b}a \cos\theta,$$ and $$r_{\theta}= -a {\mathbf n} \sin\theta + a {\mathbf b} \cos\theta .$$

So, the end result (from the solution manual) is $$r_{s}\times r_{ \theta}= -a(1-ka \cos\theta)({\mathbf n} \cos\theta +{\mathbf b} \sin\theta )$$ which I have no idea how he got there...

This task can be found from diff geometry book of Andrew Presley (4.2.7) Thanks.

Mark Fantini
  • 5,523
p0ffer
  • 231
  • Your notation is hard to read, please use \sin and make vectors bold \mathbf{b}. – Rene Schipperus May 31 '14 at 17:49
  • in the derivative of $r_s$ the first term is supposed to have vector $t$ at the end and I offer that this mysterious $t$ is nothing more than the tangent to the curve. Moreover, I believe this is the equations of a tube with radius $a$ attached to a given curve $\gamma$. Selling this question in terms of such would attract better answers faster :) – James S. Cook May 31 '14 at 18:36

1 Answers1

0

I have a theory. You're given a regular curve $\gamma$. With respect to this curve we can calculate the Frenet frame; $\vec{T},\vec{N}, \vec{B}$. The change in these along the curve can be expressed once more in terms of the Frenet frame in terms of the Frenet-Serret equations. Those equations bring in the curvature $k$ (?) and torsion $\tau$ to describe the evolution of the frame as it is attached along $\gamma$. Ok, the main point here is that $\vec{T},\vec{N}, \vec{B}$ for a right-handed frame of vectors as $$\vec{B} = \vec{T} \times \vec{N}$$ so $$\vec{T} = \vec{N} \times \vec{B}$$ and $$\vec{N} = \vec{B} \times \vec{T}.$$ Note, by construction $\vec{N} \cdot \vec{T}=0$. So, I think this will help you sort through the cross-products which you find mysterious. Also, there is a "$t$" which ought to be a $\vec{T}$ (my notation) in your post, it arises from $\gamma'(s)$ which, assuming $s$ is arclength, is precisely the unit-tangent $\vec{T}$.

James S. Cook
  • 16,755