Professor @TedShifrin, on the question of differential geometry, I can not seem to help that person out. What am I doing wrong?
I am assuming that all derivatives of $\alpha$ are with respect to $t$. Now there is an identity we can use:
$$|a\times b|^{2} = |a|^{2}|b|^{2}-(a\cdot b)^{2}.$$
Therefore,
$$\begin{align}
|\alpha'\times \alpha''|^2 &= |\alpha'|^{2}|\alpha''|^{2}-(\alpha'\cdot \alpha'')^{2}\\
& = |\alpha'|^{2}|\alpha''|^{2},\\
\end{align}
$$
since $\alpha'$ and $\alpha''$ are orthogonal. Therefore we have that