Regularity of Surface M and Matrix at Point P

In summary, the surface M parametrized by σ(u,v) = (sin u cos v, sin u sin v, 3cos u), 0 < u < π, 0 < v < 2π, is regular and the matrix Sp with respect to the basis {σu(P), σv(P)} in TP(M) can be found by evaluating the partial derivatives at the point P = (\sqrt{2}/2, \sqrt{2}/2, 0).
  • #1
Shackleford
1,656
2

Homework Statement



Consider the surface M parametrized by σ(u, v) = (sin u cos v, sin u sin v, 3 cos u), 0 < u < π, 0 < v < 2π.

(ii) Show that the surface is regular.

(iii) Let P = ([itex]\sqrt{2}/2, \sqrt{2}/2, 0)[/itex] in M. Find the matrix Sp with respect to the basis {σu(P), σv(P)} in TP(M).

Homework Equations



Surface is regular if σu x σv does not equal zero at any point in the domain.

The Attempt at a Solution



σu = (cos v cos u, sin v cos u, -3 sin u)

σv = (-sin u sin v, sin u cos v, 0)

σu x σv = (3 cos v sin2u, 3 sin2u sin v, sin u cos u)

||σu x σv|| = sqrt{9 sin4u + sin2u cos2u}

Of course, it looks like the norm of the vector product should simplify further.

I'm also unsure about the given basis for the matrix of the shape operator. By direction calculation, the matrix contains expressions of the first and second fundamental forms. Do I evaluate the partials at the point P and use those in the inner product calculations?
 
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  • #2
If u= 0 then [tex]\sigma_u\times \sigma_v= 0[/tex].
 
  • #3
HallsofIvy said:
If u= 0 then [tex]\sigma_u\times \sigma_v= 0[/tex].

Yes. I'm a bit confused. It says to show that the surface is regular.
 
  • #4
Shackleford said:
Yes. I'm a bit confused. It says to show that the surface is regular.

The original description excluded the boundaries ##u =0##, ##u = \pi##, ##v = 0##, ##v = 2 \pi##.
 
  • #5
Ray Vickson said:
The original description excluded the boundaries ##u =0##, ##u = \pi##, ##v = 0##, ##v = 2 \pi##.

Okay. It doesn't look like any combination of u and v in the domain yields (0, 0, 0), so I can state that it is regular. However, it seems that the norm argument should simplify a bit more.
 
  • #6
Shackleford said:
Okay. It doesn't look like any combination of u and v in the domain yields (0, 0, 0), so I can state that it is regular. However, it seems that the norm argument should simplify a bit more.

Who cares? You just want to know if it = 0 or not.
 
  • #7
Ray Vickson said:
Who cares? You just want to know if it = 0 or not.

(iii) Let P = [itex](\sqrt{2}/2, \sqrt{2}/2,0)[/itex] in M. Find the matrix Sp with respect to the basis {σu(P), σv(P)} in TP(M).
 

Related to Regularity of Surface M and Matrix at Point P

1. What is the concept of regularity of surface M at point P?

The regularity of a surface M at a point P refers to the smoothness and continuity of the surface at that specific point. It is a measure of how well-behaved the surface is at that point, with regards to its derivatives and curvature.

2. How is the regularity of surface M at point P determined?

The regularity of a surface M at point P can be determined by looking at the first and second derivatives of the surface at that point. If the derivatives exist and are continuous, the surface is considered to be regular at that point.

3. What does it mean for a surface to be regular at all points?

If a surface M is regular at all points, it means that the surface is smooth and continuous everywhere. This indicates that the surface is well-behaved and does not have any sudden changes or singularities.

4. Can a surface be regular at some points but not others?

Yes, a surface can be regular at some points but not others. For example, a surface may be regular at a point where the derivatives are continuous, but may not be regular at a point where the derivatives do not exist or are discontinuous.

5. What is the relationship between regularity of surface M and matrix at point P?

The regularity of a surface M at point P is closely related to the matrix at that point. The matrix at a point on a surface is used to calculate the first and second derivatives, which in turn determine the regularity of the surface at that point.

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