Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 4 down vote favorite
2

I know that a ruled surface is a surface with parametrization $x(u,v)$ = $c(u)$ + $vf(u)$ where if $I$ $\subset$ $R$ is an interval, $c$: $I$ $\mapsto$ $R^3$ and $f$ : $I$ $\mapsto$ $R^3$ are smooth curves with $f$ $\neq$ $0$ on $I$. The curve $c$ is called the directrix and the $f(u)$ are called the rulings. If I'm not mistaken, is a ruled surface regular at a point $p$ = $x(u,v)$ provided that $f(u)$ $\bigwedge$ $c'(u)$ = $vf(u)$ $\bigwedge$ $f'(u)$ and $x_u$ $\bigwedge$ $x_v$ $\neq$ $0$? I actually think that only the second condition is really required, or are both of them required? Are the two conditions equivalent, and is there any other extra requirement? I'd really appreciate some input on this, thanks.

share|improve this question
I could be wrong, but since the rulings $f(u)$ are actually a vector, won't $f(u)$ and $f'(u)$ be parallel and therefore $vf(u) \bigwedge f'(u)=0$? What am I missing? – Adrian Petrescu Apr 9 '11 at 3:40
@Adrian notice that $v$ is the linear parameter, $f(u)$ need not be linear. The statement that "the $f(u)$ are called the rulings" seems misleading. – yasmar Apr 9 '11 at 20:04
The surface is regular at $p$ if $x_u \wedge x_v \neq 0$. If you explicitly compute this from your given equation, you see that $p$ is not regular exactly when $f(u) \wedge c'(u) = vf'(u) \wedge f(u)$, which is not quite the same as what you wrote as the first "requirement". – yasmar Apr 9 '11 at 20:10
@yasmar So then only the second statement is enough to say that the ruled surface is regular? What about my first "requirement?" – Libertron Apr 9 '11 at 20:41
The second statement is more or less the definition of regular, regardless of whether or not the surface is ruled. I expect you miss-copied the first statement from somewhere. – yasmar Apr 9 '11 at 22:11

1 Answer

up vote 2 down vote accepted

Let $x(u,v)=c(u)+vf(u)$ where $u\in I$. Note that by definition, $x(u,v)$ is a regular surface if $x_u\wedge x_v\neq 0$. Simple computation gives $x_u(u,v)=c'(u)+vf'(u)$ and $x_v(u,v)=f(u)$, which implies that $$x_u\wedge x_v=c'(u)\wedge f(u)+vf'(u)\wedge f(u).$$ Therefore, $x(u,v)$ is a regular surface if and only if $c'(u)\wedge f(u)+vf'(u)\wedge f(u)\neq 0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.