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There is given curve parameterized by arc length $ \alpha (s), s \in I $, and there is the surface $ r(s,v)=\alpha (s) + vT(s) , s\in I, v \in \mathbb{R}$, where $T(s)$ is the tangent vector field.

I have to determine if the surface is regular, or to determine if $ || r_{s} \times r_{v} || $ is not equal to zero.

I tried this:

$ r_{s} = \alpha ' (s)+vT'(s), r_{v}=T(s)$.

$ || r_{s} \times r_{v} || = ||r_{s}|| || r_{v}|| \sin \angle(r_{s},r_{v}) = ||\alpha ' (s)+vT'(s)||||T(s)||\sin \angle(r_{s},r_{v}) $

Since the curve is parameterized by arc length, there are $|| \alpha '(s) = ||T(s)|| = 1$: $ || r_{s} \times r_{v} ||=(1+0)\sin \angle(r_{s},r_{v})=\sin\angle(r_{s},r_{v})$

Is the angle between the coordinate lines always equal to $\frac{\pi}{2}$, or if not, how do I determine the angle? Also, please tell me if I did something wrong up there.

Thanks!

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Note that $\alpha'(s)=T(s)$ and that $T(s)$ is orthogonal to $T'(s)$, so the scalar product between $r_s$ and $r_v$ is just $1$. If $\theta$ is the angle between the vectors $r_s,r_v$, then it follows that $$\cos\theta=\frac{r_s\cdot r_v}{\|r_s\|\cdot\|r_v\|}=\frac{1}{\sqrt{1+v^2}}$$ Here I used the fact that $T(s)$ is orthogonal to $T'(s)$ so the norm of $r_s$ is calculated by Pythagoras' theorem. It follows that $$\sin^2\theta=\frac{v^2}{1+v^2}$$ and this is zero only if $v=0$.

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