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I'd like some hint for this problem:

Show that every unit-speed curve $ \alpha : I \subset \mathbb{R} \longrightarrow \mathbb{R}^3$, whose image is in the sphere of radius $R$ , has curvature $\displaystyle k_{\alpha}(s)\geq \frac{1}{R}$ .

Thanks.

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    Intuitively, the least curved lines on a sphere are its geodesics, which are arcs of great circles. It is easy to see that those have curvature exactly $1/R$.2011-06-29

3 Answers 3

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Regard $\alpha\colon I \to \mathbb{S}^2_R$ as a map from the interval $I$ to the sphere of radius $R$. Let $\langle \cdot, \cdot \rangle$ denote the dot product. Here are two hints:

(1) Consider the quantity \frac{d}{ds}\langle \alpha(s), \alpha'(s)\rangle. What do you know about the quantity \langle \alpha(s), \alpha'(s)\rangle based on the geometry of the sphere?

(2) Since you're trying to prove an inequality, perhaps you know of some inequality involving the inner product $\langle v, w\rangle$ of two vectors.

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$\alpha(t) \equiv (x(t), y(t), z(t))$ verifies the identity $x^2 + y^2 + z^2 = R^2$ deriving twice the previous identity we deduce $\left\vert x\ddot x + y\ddot y + z\ddot z \right\vert = 1$ and applying the Cauchy–Schwarz inequality to the left side we obtain $k_\alpha R = \Vert (\ddot x, \ddot y, \ddot z) \Vert R \ge 1$

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    Ah, you just beat me to it.2011-06-29