# Principal Curvatures of a surface

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

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 $1/R>1$.