Proving this curve has speed $r$ and torsion $-r^2$

Question

Let $ \beta : I \subset \mathbb{R} \rightarrow \mathbb{E}^3 : s \mapsto \beta(s)$ be an arc length parameterized curve defined on the sphere with radius $r$. Define the curve $\alpha$ as follows: $$ \alpha (t) = \int_a^t \beta(s) \times \beta' (s) ds. $$ Prove that $\alpha$ has speed $r$ and torsion $-r^2$.

Attempt: By definition the speed is $v = || \alpha' (t) ||$. Now we have that $\alpha' (t) = \beta(t) \times \beta' (t)$. But I'm not sure what to do with this crossproduct. Also, we have the torsion $\tau$ as $$ \tau= \frac{ \alpha' \times \alpha'' \cdot \alpha'''}{ || \alpha' \times \alpha'' ||^2 } $$ I computed \begin{align*} \alpha'' &= \beta' (t) \times \beta' (t) + \beta(t) + \beta'' (t) \\ &= \beta(t) \times T'(t) = \kappa v \beta(t) \times N(t) \end{align*} where $\kappa$ is the curvature and $N(t)$ is the unit normal. But I'm not really sure how to continue and how to work out these crossproducts.

Answer 1

From the fact that $\beta$ is on a sphere we know $\|\beta\|$ = r, or $\beta \cdot \beta = r^2$. Differentiating over $s$ you get that $\beta$ and $\beta'$ are orthogonal. The norm of a cross product is the product of norms of factors times the $\sin(angle)$. Therefore $\|\alpha'\| = \|\beta\| \|\beta'\| \sin \frac\pi2 = r\cdot 1\cdot 1 = r$

Answer 2

Here's a big hint: Try reparametrizing $\alpha$ with respect to arc length. Since you know that the speed of alpha is $r$, you can reparametrize setting $t=rs$. From there it's just a matter of computation.