# shape operator

• Dec 15th 2012, 11:01 AM
Plato13
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?
• Dec 16th 2012, 10:27 AM
Plato13
Re: shape operator
$\displaystyle \begin{bmatrix}-2 & 3\\3 & 6\end{bmatrix}$ 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.
• Dec 16th 2012, 11:17 AM
Plato13
Re: shape operator
question SOLVED