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Suppose that a 3-D surface has the property that $|k_1|\leq 1$ and $|k_2|\leq 1$ everywhere, where $k_1$ and $k_2$ are the principal curvatures. Prove or disprove that the curvature $k$ of a curve on that surface also satisfies $|k|\leq 1$.

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Please do not write questions in the imperative, it is not polite. If this is homework, please tag it as so, and if it is not, please include motivation for the question. How did you come upon this question? Finally, let us know what you have already tried. All of these things will help you in the end. –  Glen Wheeler Mar 17 '11 at 12:16
    
Do you mean normal curvature of the curve? –  lhf May 24 '11 at 17:59
    
The's a formula relating $k$, $k_1$ and $k_2$, and it's an equality. Find that formula and you'll have your answer. From the way you write your question I'm assuming it's a homework question so there's really no need for any more hints than the above. –  Ryan Budney May 24 '11 at 19:16
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up vote 2 down vote accepted

I don't think so. Take, for example, a plane, which has principal curvatures zero. Pick a small circle, with radius $R$ smaller than $1$ in that plane. The circle has curvature $1/R>1$.

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I was thinking that the OP might be speaking of non-flat surfaces (by which I mean that we exclude those points on a surface which make the matrix of the II fundamental form identically zero) –  Andy May 24 '11 at 18:03
    
Even if the surfaces are not flat, I don't think the result in the question is valid. The principal curvatures are the minimum and maximum curvatures of curves obtained by normal sections of the surface. But if you consider an arbitrary curve contained in the given surface, you could locally make the curvature as big as possible (I think). –  Beni Bogosel May 24 '11 at 19:42
    
It's probably right. I wasn't thinking of the problem, I was just making a remark (in our differential geometry class we never dealt with flat points). Even now I don't have time to think - it's almost 6am :) –  Andy May 25 '11 at 3:45
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