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 $|k|\leq 1$.
Principal Curvatures of a surface
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geometry
differential-geometry
curvature
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0The'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. – 2011-05-24
1 Answers
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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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0It'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 :) – 2011-05-25