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Consider a curve $\beta:I\subseteq R \rightarrow \mathbb{E}^3:s\mapsto\beta(s)$ parametrized by its arc length defined on a sphere with radius r. We define the curve $\alpha$ as: $$\alpha (t)=\int_a^t \beta (s)\times\beta'(s)ds$$ Prove that $\alpha$ has velocity $r$ and torsion $-r^2$.

I don't know how to begin? Normally, I can find the velocity very easily by calculating $\parallel \alpha' \parallel$. But I don't know how to proceed when dealing with an integral and a vector product in the parametrization of the curve?

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    Consider $\beta(t)=(\cos(t),\sin(t),0)$. Then $\alpha(t)=(0,0,t-a)$ has zero torsion. – 2017-01-15
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    So now I need to re-parametrize $\beta$ such that $\parallel \alpha ' \parallel=r$? This would be $\beta(t)=(r cos(s/R),r sin(s/r),0)$ such that $\alpha =(0,0,rt-ra)$ with velocity r? – 2017-01-15
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    Torsion is zero anyway. – 2017-01-15
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    Can you be a little more concrete? – 2017-01-15
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    In my example $\beta$ is a unit speed curve in $S^2$, that is, $r=1$. Then we have obviously that $\|\alpha'\|=r$ but the torsion of $\alpha$ is not $-1/r^2$. – 2017-01-15
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    So this is my new attempt: $\parallel \beta \parallel=r$ or $\beta \cdot \beta' = r^2$. After differentiating over s, you get that $\beta$ and $\beta'$ are orthogonal thus $\parallel \alpha' \parallel =\parallel \beta \parallel \parallel \beta' \parallel sin \frac{\pi}{2}=r \cdot 1 \cdot 1=r$ – 2017-01-15

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Hint: let $\beta$ be a Frenet curve with non-vanishing torsion $\tau$ and let $r$ be a positive number. Then show:

If $\beta$ is contained in the sphere with center $p_0$ with radius $r$ (i.e., $\langle \beta-p_0,\beta-p_0\rangle=r^2$), then $$\beta=p_0-\frac{1}{\kappa}N-\left(\frac{1}{\kappa}\right)'\cdot\frac{1}{\tau}\cdot B,$$ where $N$ and $B$ denote the normal and binormal vector, resp., and hence $$ \left(\frac{1}{\kappa}\right)^2+ \left(\left(\frac{1}{\kappa}\right)'\right)^2\cdot\left(\frac{1}{\tau}\right)^2=r^2.$$