Differential Geometry: Unit Normal Field

[badgers14]

Let M be the surface defined by z=x²+3xy-5y². Find a unit normal vector field U defined on a neighborhood of p on M.

First, I reparameterized the equation for the surface to get x(u,v)=(u,v,u²+3xy-5y²). Next I found two tangent vectors x_u(u,v)=(1,0,2u+3v) and x_v=(0,1,3u-10v). The next step is where I'm unsure. In the text it gives an equation for the unit normal function, U=(x_u X x_v)/||x_u X x_v||. When I use this equation, I come up with
U=(-2u-3v,10v-3u,1)/(sqrt(13u²-54uv+109v²+1)

This just seemed messy to me, not sure if I'm missing something or if that is actually what the answer should look like. Any verification/help would be appreciated

 

[Dick]

Sure, it's a little messy. I think you are doing everything basically correctly, except you mean x=u and z=v and in z(u,v) you forgot to replace some of the x's and y's with u's and v's. And in the final result I don't agree with your coefficient of uv in the denominator.

 

[badgers14]

Thank you, and here is a follow up question that I'm having a little difficulty with.

Compute the covariant derivatives $$\nabla$$_u1U and $$\nabla$$_u2U at p=(0,0,0). Where U is the unit normal vector computed in part (a) and with tangent vectors u₁=(1,0,0) and u₂=(0,1,0).

First, I computed U at (0,0,0) which was the upward unit normal (0,0,1). From here I'm not sure what the next step is.

 

[Dick]

Thank you, and here is a follow up question that I'm having a little difficulty with.

Compute the covariant derivatives $$\nabla$$_u1U and $$\nabla$$_u2U at p=(0,0,0). Where U is the unit normal vector computed in part (a) and with tangent vectors u₁=(1,0,0) and u₂=(0,1,0).

First, I computed U at (0,0,0) which was the upward unit normal (0,0,1). From here I'm not sure what the next step is.

I'm not sure either. What's your definition of covariant derivative? It looks to me like they must want the covariant derivative in three dimensional space with the usual metric, in which case the Christoffel symbols are zero, and it's pretty much the usual derivative.

 

[Bozen]

I'm not sure either. What's your definition of covariant derivative? It looks to me like they must want the covariant derivative in three dimensional space with the usual metric, in which case the Christoffel symbols are zero, and it's pretty much the usual derivative.

I found this via Google while looking for something else, but it seems by the opening poster's name and the question that we are in the same class at the same university.

We are using Elementary Differential Geometry by Barrett O'Neill and our definition of covariant derivative is:

Let W be a vector field on R³, and let v be a tangent vector field on R³ at a point p. Then the covariant derivative of W with respect to v is the tangent vector $$\nabla$$_vW = W(p + tv)^' at the point p.

@badgers14, I have not done the computations for these problems yet, but I will make another post and let you know what I come up with.