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Orthonormal basis/Gram-Schmidt [Easy?]

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider the surface patch σ(t,θ) = (coshtcosθ, coshtsinθ, t) where t is an element of the set of real numbers and θ is an element from (-pi, pi).

    Show that σ defines a regular surface patch and find an orthonormal basis for the tangent space (TpS) at points of the form P = (cosht, 0, t)

    3. The attempt at a solution
    I have done the regular surface patch part. Now just wondering how I go about the orthonormal basis part.

    By differentiation:
    u1 = ∂σ/dt = (sinhtcosθ, sinhtsinθ, 1) and
    u2 = ∂σ/dθ = (-coshtsinθ, coshtcosθ, 1)

    Using Gram-Schmidt, I found that u1 and u2 are already orthogonal since the inner product is 0.

    I can then normalize u1 and u2:
    v1 = u1/|u1| = (sinhtcosθ, sinhtsinθ, 1)/cosht
    v2 = u2/|u2| = (-coshtsinθ, coshtcosθ, 1)/cosht

    Now I don't know where to go from here...
    I thought maybe I should check for a linear combination such that
    (cosht, 0, t) = Av1 + Bv2

    and then see if the constants A & B satisfy all 3 equations, but that didn't seem to work...
    Any other suggestions?
     
  2. jcsd
  3. Sep 16, 2009 #2

    Dick

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    You DEFINITELY don't want to try and solve P=Av1+Bv2 or anything like that. P is a point on the surface and v1 and v2 are in the tangent space. They don't really mix. I think you are basically done, except you've got a typo in the last component of u2 and v2, right? The points P=(cosh(t),0,t) are just the points on the surface where theta=0. You could put theta=0 into v1 and v2.
     
  4. Sep 16, 2009 #3
    thanks for the swift reply!
    yeah its a typo, last component of u2/v2 should be 0, just an error in copy & paste!

    I'm a bit confused with the second part you mentioned, putting theta = 0 into v1 and v2. How did you come to the conclusion that P are points where theta = 0?

    Anyhow, if I do this, I'll get...

    v1 = (sinht, 0, 1)
    v2 = (0, cosht, 0)

    Confused!
     
  5. Sep 16, 2009 #4

    Dick

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    If I put theta=0 into sigma(t,theta), I get P=(cosh(t),0,t). P is just a curve on the surface where theta=0. (sinh(t),0,1) and (0,cosh(t),0) are an orthonormal basis along that curve. Of course, you did more. You know an orthonormal basis everywhere.
     
  6. Sep 16, 2009 #5
    ah of course, what am I talking about!

    so just to make things crystal clear..
    σ(t,θ) = (coshtcosθ, coshtsinθ, t)
    σ(t,0) = (cosht, 0, t)

    Then I have an orthonormal basis
    v1 = u1/|u1| = (sinhtcosθ, sinhtsinθ, 1)/cosht
    v2 = u2/|u2| = (-coshtsinθ, coshtcosθ, 1)/cosht

    Now I plug θ=0 into my orthonal basis vectors and get..
    v1 = (sinht, 0, 1)
    v2 = (0, cosht, 0)

    The end.
    Correct?
     
  7. Sep 16, 2009 #6

    Dick

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    That's all I can think of to do.
     
  8. Sep 16, 2009 #7
    I was confused before, read the question a little different.
    But your suggestion makes alot more sense when I re-read the question.

    Thanks for the help!
     
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