Show that a point lies on the line of striction

[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!

 

[mighty2000]

To show that any point on M that satisfies xu×xv=0 must lie on the line of striction, we can use the fact that length(β')=1 and length(δ)=1, and the given reparametrization y(u,w)=ϒ(u)+wδ(u).

First, we can rewrite xu×xv as (β'(u) +vδ'(u))(δ(u))=ϒ'⋅δ'. Since we know that length(β')=1 and length(δ)=1, we can substitute these values into the equation to get (β'(u) +vδ'(u))(δ(u))=1⋅1=1.

Next, we can rewrite ϒ'⋅δ' as (β'(u)+r(u)δ'(u))⋅δ'(u)=β'(u)⋅δ'(u)+r(u)δ'(u)⋅δ'(u). Since δ'≠0, we can divide both sides by δ' to get β'(u)+r(u)δ'(u)=β'(u)+r(u)=1.

Now, we can solve for r(u) to get r(u)=1-β'(u). Substituting this into our reparametrization, we get y(u,w)=β(u)+(v-1+β'(u))δ(u). This shows that any point on M that satisfies xu×xv=0 must lie on the line of striction, as the reparametrization is dependent on the value of v.