# 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 $|k|\leq 1$.

• 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 $1/R>1$.