# Given a specific curve $\alpha(s)$, Prove that $\kappa(s) \geq 1/R$, where $\kappa(s)$ is the curvature of $\alpha$ at $s$.

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?

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, $$R |\alpha''(s_0)| = |\alpha(s_0)|\, |\alpha''(s_0)| \geq - \alpha(s_0) \cdot \alpha''(s_0) \geq 1.$$

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