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There are two questions to this problem:

1) Prove that the curve $\alpha(s)$ is a straight line.

2) Does the conclusion hold if $\alpha$ is not regular, ie. $\alpha'(s)=0$ for some $s\in I$.

I proved the first part: Since the lines pass through a fixed point $p$ we can write the condition: $$\alpha(s) +\lambda(s)t(s)=d$$ Taking the derivative on both sides we get: $$\alpha'(s)+\lambda'(s)t(s)+\lambda(s)t'(s)=0$$ $$(1+\lambda'(s))t(s)+\lambda(s)k(s)n(s)=0$$ Since the vectors $t$ and $n$ are perpendicular we have $1+\lambda'(s)=0$ and $\lambda(s)k(s)=0$. Here it follows that $\lambda'(s)=-1$ and $k(s)=0 \implies t'(s)=0$. So $\lambda(s)=-s+c$ and $t(s)=t_0$. Hence we can rewrite the condition: $$\alpha(s)=\underbrace{a_0}_{d-a}+st_0$$ which is a line.

What I am having trouble with is the 2nd question. My guess is that it should not be true but I don't see where the condition that the curve $\alpha(s)$ is regular has been used in the proof above. Clearly if $t=0$ at some point we can have a problem with the results that $\lambda'(s)=-1$ and $k(s)=0$ but these should be isolated points and not interfere with the general idea.

user53970
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  • Are we talking about plane or space ($\Bbb R^n$) curves? – Alex M. Dec 22 '15 at 14:48
  • It is a space curve. But why does it matter? – user53970 Dec 22 '15 at 14:50
  • I've encountered properties that were only valid for plane curves, and I as wondering whether this is one such instance, nothing more. Apparently not. – Alex M. Dec 22 '15 at 14:54
  • Please don't write part of a question exclusively in the title; that's quite confusing. Anyway, I assume that $\alpha(s)$ is the curve mentioned in the title, but what are $t(s), \lambda(s), k(s)$, and $n(s)$? – A.P. Dec 22 '15 at 15:11
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    @A.P. the literature that I use about differential geometry is "do Carmo". There $t(s)$ is the vector tangent to the curve, $n(s)$ is the vector normal to the curve, $k(s)$ is the curvature and $\lambda(s)$ is just the scalar that makes the condition true. From what I have seen in mathstackexchange this is a standard notation. – user53970 Dec 22 '15 at 21:32

1 Answers1

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Think about pieces-wise linear curves like the graph of $|x|$. all tangents intersect at the origin. Any curve whose tangent line intersect at a point will have a trace which is a union of line segments with a common end point.

Mars
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