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If $|a(t)|\ll b$ is it alright to take $a\left({a\cdot \dot{a} \over b^2}\right)$ as $0$?

Would the following argument make sense?

I know that we can take $\left({a\cdot a \over b^2}\right)$ as $0$ and $0={d\over dt}\left({a\cdot a \over b^2}\right)=2\left({a\cdot \dot{a} \over b^2}\right)$.

So since $a$ is bounded by finite number (namely $b$) therefore $a\left({a\cdot \dot{a} \over b^2}\right)$ can be taken to be $0$.

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    No -- just because $a$ is small doesn't mean that $\dot a$ has to be. Just consider, for example, $a =b^2\sin(b^{-10})$ for $b\to 0$.2012-07-06
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    @HenningMakholm: Thank you!2012-07-06

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