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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 $1/R$. –  Alexei Averchenko Jun 29 '11 at 2:30

2 Answers 2

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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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thanks for the hints! –  Jr. Jun 29 '11 at 1:31

$\alpha(t) \equiv (x(t), y(t), z(t))$ 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. –  Jesse Madnick Jun 29 '11 at 1:30

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