Proving Regularity of M(s,t) and Deriving the Unit Normal Vector Field N(s,t)

[Poirot1]

let y:R^3->R^2 unit speed curve of nonvanishing curvature and let its scalar curvature,torsion and Frenet frame be denoted κ, τ and [u,n,b] respectively. You may aassume the frenet formulae are true.

Let M(s,t)=y(s) +tb(s).

1)Show that M is regular surface.

Partial derivatives are y'(s)+tb'(s) and b(s). I have to show the cross product is zero.
I get that it's equal to b x (u-τn). I think this is probably non zero but the next question is 2)show that the unit normal vector field is

N(s,t)= $-\frac{n+tτu}{(1+t^{2}τ^{2})^{0.5}}$, which I cannot derive.

 

[jvicens]

To show that M is a regular surface, we need to show that the partial derivatives ∂M/∂s and ∂M/∂t are linearly independent at every point on M.

∂M/∂s = y'(s) + tb'(s) and ∂M/∂t = b(s)

To show linear independence, we need to show that the cross product of these two vectors is non-zero.

∂M/∂s x ∂M/∂t = (y'(s) + tb'(s)) x b(s)

Using the Frenet formulas, we can rewrite b'(s) as -κn(s) - τu(s).

∂M/∂s x ∂M/∂t = (y'(s) - κb(s) - tτn(s)) x b(s)

Using the properties of cross product, we get:

∂M/∂s x ∂M/∂t = y'(s) x b(s) - κ(b(s) x b(s)) - tτ(n(s) x b(s))

Since b(s) is a unit vector, b(s) x b(s) = 0. Also, n(s) x b(s) = u(s) by the Frenet formulas.

Therefore, ∂M/∂s x ∂M/∂t = y'(s) x b(s) - tτu(s)

Since y'(s) and b(s) are orthogonal unit vectors, their cross product is also a unit vector perpendicular to both of them. Therefore, y'(s) x b(s) ≠ 0.

Hence, we have shown that ∂M/∂s x ∂M/∂t ≠ 0, which means that ∂M/∂s and ∂M/∂t are linearly independent and M is a regular surface.

To show that the unit normal vector field is N(s,t)= $-\frac{n+tτu}{(1+t^{2}τ^{2})^{0.5}}$, we need to show that it is perpendicular to the tangent plane of M at every point.

The tangent plane of M at a point (s,t) is spanned by the vectors ∂M/∂s and ∂M/