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Midterm Review

  1. Sep 30, 2014 #1
    1. The problem statement, all variables and given/known data

    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).

    2. Relevant equations

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

    3. 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?
     
    Last edited: Sep 30, 2014
  2. jcsd
  3. Sep 30, 2014 #2

    HallsofIvy

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    If u= 0 then [tex]\sigma_u\times \sigma_v= 0[/tex].
     
  4. Sep 30, 2014 #3
    Yes. I'm a bit confused. It says to show that the surface is regular.
     
  5. Oct 1, 2014 #4

    Ray Vickson

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    The original description excluded the boundaries ##u =0##, ##u = \pi##, ##v = 0##, ##v = 2 \pi##.
     
  6. Oct 1, 2014 #5
    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.
     
  7. Oct 1, 2014 #6

    Ray Vickson

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    Who cares? You just want to know if it = 0 or not.
     
  8. Oct 1, 2014 #7
    (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).
     
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