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Math Help - shape operator

  1. #1
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    shape operator

    Let M be a regular surface with shape operator [Tex]\begin{bmatrix}2 & 3\\3 & 6\end{bmatrix}[\Tex]. In each case find the tangent vector v or explain why it doesn't exist.

    (i) a unit vector v such that K(v)=-3 where K is normal curvature function
    (ii)unit vector v such that K(v)=10
    (iii) a vector v (possibly non unit) such tha K(v)=12

    for (i) and (ii) I found a quartic equation and found for (i) there is a solution but in (ii) the solutions were complex. My problem is this was quite hard work, plus for (iii) I would not have the condition that the sum of the squares of the components of v is 1, thus I would have one equation in 2 unknowns. So what to do?

    Ps. Why is the code not working?
    Last edited by Plato13; December 15th 2012 at 12:14 PM.
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  2. #2
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    Re: shape operator

    \begin{bmatrix}-2 & 3\\3 & 6\end{bmatrix} is the matrix. I should also say that the normal curvatue function is K(u)=u.S(u) where S is the shape operator and . is the dot product.
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  3. #3
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    Re: shape operator

    question SOLVED
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