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Suppose $\alpha$ is an arclength-parameterized curve with the property that $|\alpha(s)| \leq |\alpha(s_o)| = R,\forall s$ sufficiently close to $s_o$. Prove that $\kappa(s) \geq 1/R$, where $\kappa(s)$ is the curvature of $\alpha$ at $s$.

A hint is given to define a function $f(s)=|\alpha(s)|^2$ and to consider it's second derivative at $s_o$

$\frac{d}{ds} f(s)=\frac{d}{ds}|\alpha(s)|^2=2\alpha(s)*\alpha'(s)$

and thus by the product rule $f''(s)=2(\alpha'(s)^2+\alpha(s)\alpha''(s))$.

So, i'm suppose to see something by examining this function apparently, but I am not. Is there something obvious i'm missing?

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By assumption, the function $f(s)$ has a maximum at $s = s_0$, hence $f'(s_0) = 0$ and $f''(s_0) \leq 0$. Moreover, $|\alpha'(s)| = 1$ for every $s$ (since $\alpha$ is parametrized by arc-length). Since $$ f''(s) = 2 |\alpha'(s)|^2+ 2 \alpha(s)\cdot \alpha''(s) = 2( 1 + \alpha(s)\cdot \alpha''(s)), $$ the condition $f''(s_0) \leq 0$ gives $$ \alpha(s_0) \cdot \alpha''(s_0) \leq -1, $$ so that, by the Cauchy-Schwarz inequality, $$ R |\alpha''(s_0)| = |\alpha(s_0)|\, |\alpha''(s_0)| \geq - \alpha(s_0) \cdot \alpha''(s_0) \geq 1. $$

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  • $\begingroup$ and $|\alpha''(s_o)|=\kappa(s_o)$ because $\alpha(s)$ is arc-length parameterized? $\endgroup$
    – user624065
    Jan 21, 2019 at 16:29
  • $\begingroup$ Yes, that's right. $\endgroup$
    – Rigel
    Jan 21, 2019 at 16:32

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