We are describing a curve with a moving frame. Figure 1 says about the definition of curve
Figure 1
Specifications and Proprieties of the curve
We can make a rotation 3D Matrix $R(s)_{3 \times 3}$ by arranging the orthogonal directors $d_3(s),d_2(s),d_1(s)$ in column order. Means place $d_3(s)$ as column $1$,$d_2(s)$ as column $2$,$d_1(s)$ as column $3$
We can make a vector called Darboux vector at each point
$\Omega(s)=d_1(s)\kappa_1(s)+ d_2(s)\kappa_2(s)+d_3(s)\kappa_3(s) \tag1$ where $\kappa(s)=\begin{pmatrix}\kappa_3(s)\\ \kappa_2(s)\\\kappa_1(s)\end{pmatrix}$ as mentioned in Figure 1
$d_i'(s)=\Omega(s) \times d_i(s),i=1,2,3 \tag 2$
From Equation (2) we can derive the following relationship related to derivative of R(s) as following $R'(s)=[\Omega(s)]_{\times}R(s)\tag 3$ . It means if $\Omega(s)= \left(\begin{array}{c}x (s)\\ y(s) \\ z(s) \end{array}\right) ,[\Omega(s) ]_{\times}=\left( \begin{array}{ccc} 0 & -z(s) & y(s) \\ z(s) & 0 & -x(s) \\ -y(s) & x(s) & 0 \\ \end{array} \right)$
Question
- Assume you are only given the $\Omega(s)$ how can you calculate Rotation 3D matrix R(s)? Is it true that $e^{[\Omega(s)]_{\times}}=R(s)$? If not what could be that relationship