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[Solution] Instead of taking the Frenet frame along a (unit speed) curve α:[a, b] \rightarrow R^3, we can define a frame T, U, V by taking T to be

Question: 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 $T$ to be the tangent vector of $\alpha$ as usual and letting $U$ be any unit vector field along $\alpha$ with $T \cdot U=0$. That is, $U:[a, b] \rightarrow \mathbb{R}^{3}$ associates a unit vector to each $t \in[a, b]$ which is perpendicular to $T(t)$. Define $V=T \times U$. Show that the natural equations (i.e., "Frenet Formulas") for this frame are

\[\begin{aligned} T^{\prime} &=& \omega_{3} U &-\omega_{2} V \\ U^{\prime} &=-\omega_{3} T & &+\omega_{1} V \\ V^{\prime} &=\omega_{2} T &-\omega_{1} U & \end{aligned}\]

where $\omega_{1}, \omega_{2}$ and $\omega_{3}$ are coefficient functions. Furthermore, show that the Darboux vector $w$ satisfying $T^{\prime}=\omega \times T, U^{\prime}=\omega \times U$ and $V^{\prime}=\omega \times V$ is given by $\omega=\omega_{1} T+\omega_{2} U+$ $\operatorname{V}_{\text {. }}$

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[Solution] Instead of taking the Frenet frame along a (unit speed) curve α:[a, b] \rightarrow R^3, we can define a frame T, U, V by taking T to be

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