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Let $M$ be a surface parametrized by $X(u,v)=(u,v,uv(1-u))$. Find the principal curvature $k(v)$ for all unit vectors $v\in T_{(u,0,0)}M$ (the tangent plane to $M$ at $(u,0,0)$).

I find that the shape operator wrt basis $X_u$ and $X_v$ is

$A=\frac{1-2u}{(1+(u^2-u)^2)^{3/2}}\left(\begin{matrix} 0 & 1+(u^2-u)^2\\ 1 & 0 \end{matrix}\right)$ (same as the given solution).

Then for all unit vector $w$ on the tangent plane, $w=\cos\theta X_u+\sin\theta/\sqrt{1+(u^2-u)^2}X_v$. Then normal curvature $k(w)=A(w)\cdot w=w^TA^Tw=\frac{(1-2u)(2+(u^2-u)^2)}{(1+(u^2-u)^2)^{3/2}}\sin2\theta$.

But the given answer for the principal curvatures are $\pm\frac{1-2u}{1+(u^2-u)^2}$. Please tell me what's wrong with my solution. Thanks.

JSCB
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    Your operator $A$ does not depend on $v$, is that right? Also, the values $\pm\frac{1-2u}{1+(u^2-u)^2}$ are not vectors. The principal curvatures are the eigenvalues of the shape operator, so it should be easy to check with WolframAlpha. – Yuri Vyatkin Jun 01 '14 at 03:51
  • That's a typo, i mean principal curvature. – JSCB Jun 01 '14 at 04:28
  • I just checked, my answer is wrong, but I still don't know why... – JSCB Jun 01 '14 at 04:33
  • You can post your computation as an answer, and we can check it step by step. Or, add to your question the derivation of the shape operator. Otherwise, we have to do it for you, and I personally am a little bit busy at the moment. – Yuri Vyatkin Jun 01 '14 at 04:37
  • I believe the that of matrix multiplication is correct (checked by WA), I suspect the problem is how I let $w$ or the way of finding $k(w)$ using $w^TA^Tw$. – JSCB Jun 01 '14 at 04:47

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