4

I don't know how to judge whether it's a developable surface or not.I think firstly I should find a way to write the surface as a form of a ruled surface,that is,write the surface something like $$r(u,v)=a(u)+vl(u)$$,but setting $x=u,y=v,z=k^3/uv$ I can't get the desired form.Does anyone know how to do it.Thank you

Edit1:I find there is a fact that a surface is a developable surface iff there is a length-preserving mapping to a plane locally.So the surface above has a first fundamental form of $$(1+k^6/u^4v^2)(du)^2+(2k^6/u^3v^3)dudv+(1+k^6/u^2v^4)(dv)^2$$,hence letting $u'=k^3/uv,v'=u+v$,we can change the first fundamental form into something like $$(du')^2+(dv')^2$$ which is the first fundamental form of a plane,so I think it's a developable surface.The problem seems solved.

Edit2:When I read afterward context of the book I find it's very easy to do it by judging whether the Gauss curvature is zero or not.Now another question is that if I change the wrong transformation above into a complex one,using$$u'=k^3/uv+0i,v'=u+vi$$ we can get $|du'|^2+|dv'|^2$,but I'm not sure that whether such complex transformation belongs to a length-preserving mapping or not?

math
  • 533
  • Can you find a single line on this surface? – Ted Shifrin Sep 17 '21 at 16:26
  • How did you calculate $(du')^2+(dv')^2$?!!!! – Ted Shifrin Sep 18 '21 at 22:18
  • @TedShifrin I made a little mistake before.Now as for $ax^2+bxy+cy^2$ we can write it as something like $a(x+dy)^2+ey^2$,where $x$ and $y$ is $du$ and $dv$ respectively,so we can let $u'=\sqrt{a} (u+dv)$ and $v'=\sqrt{e} v$.But this seems not right,because if the method is right,then for any surface there will be a length-preserving mapping to a plane since any surface is like $ax^2+bxy+cy^2$,this is obviously not the case.So I'm stuck again – math Sep 19 '21 at 11:03
  • 1
    Go back to my first question. – Ted Shifrin Sep 19 '21 at 14:30

0 Answers0