This is problem 34.4 in Munkres' Analysis on Manifolds.
Let $\mathcal{C} = \{ \ (x,y,z) \in \mathbb{R}^{3}\mid x^{2} + y^{2} = 1 \text{ and } 0 \leq z \leq 1 \ \}$.
Orient $\mathcal{C}$ by declaring that the coordinate patch $\alpha : (0,1)^{2} \to \mathcal{C}$ given by
$\alpha(u,v) = \left( \ \cos(2\pi u), \ \sin(2\pi u), \ v\ \right)$
to belong to the orientation.
Describe the unit normal field corresponding to this orientation of $\mathcal{C}$. Describe the unit tangent field corresponding to the induced orientation of $\partial \mathcal{C}$.
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My Attempt:
Pick $(u,v) \in (0,1)^2$. Then a basis for the tangent space $T_{\alpha(u,v)}\mathcal{C}$ at the point $\alpha(u,v)$ is given by:
$\left\{\ \frac{\partial \alpha}{\partial u} ,\ \frac{\partial \alpha}{\partial v} \ \right\} = \left\{\ \left[ \begin{matrix} - 2 \pi \sin(2\pi u) \\ 2 \pi \cos(2\pi u) \\ 0 \\ \end{matrix} \right] ,\ \left[ \begin{matrix} 0 \\ 0 \\ 1 \\ \end{matrix} \right] \ \right\}$
So am I correct in saying that the unit normal must be the unit vector perpendicular to both of these vectors? I get:
$$\left[ \begin{matrix} \cos(2 \pi u) \\ \sin(2 \pi u) \\ 0 \\ \end{matrix} \right]$$
Is this correct? This has length one and is perpendicular to both of the above. So does this means that the unit normal field is $N : \mathcal{C} \to \bigsqcup_{p \in \mathcal{C}} T_{p}\mathcal{C}$ is as follows?
$$ N(p) = \left( p ;\ \left[ \begin{matrix} \cos(2 \pi u) \\ \sin(2 \pi u) \\ 0 \\ \end{matrix} \right] \ \right)$$
in the above, $(u,v) = \alpha^{-1}(p)$.
Furthermore, how do I get the unit tangent field corresponding to the induced orientation of $\partial \mathcal{C}$?