- #1
Wolves Consortium
Homework Statement
Given M: x(u,v)=β(u)+vδ(u) is a ruled surface with length(β')=1 and length(δ)=1, also supposing δ'≠0. M may be parametrized by y(u,w)=ϒ(u)+wδ(u), where ϒ'⋅δ'=0 (this reparametrization is called the line of striction). Show that any point on M that satisfies xu×xv=0 must lie on the line of striction.
Homework Equations
- The problem asks for you to state that the reparametrization exists, which is fine if you allow ϒ(u)=β(u)+r(u)δ(u) for some function r(u) and allow w=v-r(u)
- For the problem described, xu×xv= (β'(u) +vδ'(u))(δ(u)), and since ϒ'⋅δ'=0, I have: (β'(u) +vδ'(u))(δ(u))=ϒ'⋅δ'
The Attempt at a Solution
From here, I am stuck. I have tried algebraic manipulation to show that xu×xv is equivalent to y(u,w) if xu×xv=0, but I can not find any way to make this happen succinctly. Am I even going about this correctly? I know that such a point indicates that xu×xv is linearly dependent, but I don't understand what this signifies to the problem. Any help I could receive would be greatly appreciated. Thanks!