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If $r: J \times \mathbb{R} \rightarrow\mathbb{R}^3$, $r(s,t)=c(s)+tv(s)$ is a parametrization of a surface which Gaussian curvature is nonzero, then for each $s_0$ there exist limits of normal vectors $n_+(s_0):=\lim_{t \rightarrow \infty} n(s_0, t)$, $n_-(s_0):=\lim_{t \rightarrow -\infty} n(s_0, t)$ and $n_+(s_0)=-n_-(s_0)$.

I've managed to find a formula for the unit normal vector:

$$n(s,t)=\frac{r_s \times r_t}{|r_s \times r_t|}=\frac{c'(s) \times v(s)+tv'(s) \times v(s)}{\sqrt{\det\mathbb{I}_{r(s,t)}}}$$.

I also know that the gaussian curvature of this surface is negative, because it's not zero and must be nonpositive (i've proved ellier that the gaussian curvature of ruled surfaces is nonpositive). So every point on this surface is hyperbolic, but I'm not sure if it can be useful.

Edit:

Explicity it will be:$$n(s,t)=\frac{c'(s) \times v(s)+tv'(s) \times v(s)}{\sqrt{c'(s) \cdot c'(s)+2tc'(s) \cdot v'(s)+t^2v'(s) \cdot v'(s)- (c'(s) \cdot v(s))^2}}=\frac{\frac{c'(s) \times v(s)}{t}+v'(s) \times v(s)}{\sqrt{\frac{c'(s) \cdot c'(s)}{t^2}+2\frac{c'(s) \cdot v'(s)}{t}+v'(s) \cdot v'(s)- \frac{(c'(s) \cdot v(s))^2}{t^2}}}$$ I've divided the numerator and the denominator through t.

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  • $\begingroup$ So write this out explicitly and find the limit. $\endgroup$ – Ted Shifrin Aug 23 '17 at 3:01
  • $\begingroup$ Explicity it will be$$n(s,t)=\frac{r_s \times r_t}{|r_s \times r_t|}=\frac{c'(s) \times v(s)+tv'(s) \times v(s)}{\sqrt{\det\mathbb{I}_{r(s,t)}}}=\frac{c'(s) \times v(s)+tv'(s) \times v(s)}{\sqrt{c'(s) \cdot c'(s)+2tc'(s) \cdot v'(s)+t^2v'(s) \cdot v'(s)- (c'(s) \cdot v(s))^2}}$$ $\endgroup$ – mrnobody Aug 23 '17 at 13:19
  • $\begingroup$ So what is $\lim\limits_{t\to\infty} \dfrac{A+tB}{\sqrt{at^2+bt+c}}$? (Here $A$ and $B$ are vectors and $a,b,c$ are scalars.) $\endgroup$ – Ted Shifrin Aug 23 '17 at 14:31
  • $\begingroup$ So it should be $\frac{B}{\sqrt{a}}$. But why the limits with $t \rightarrow \infty$ and $t \rightarrow -\infty$ are opposite? $\endgroup$ – mrnobody Aug 23 '17 at 14:48
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    $\begingroup$ You can certainly figure that out! $\endgroup$ – Ted Shifrin Aug 23 '17 at 16:46

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