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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 $b\to 0$. –  Henning Makholm Jul 6 '12 at 20:01
    
@HenningMakholm: Thank you! –  renne Jul 6 '12 at 20:29

1 Answer 1

No, the smallness of function $a$ does not given any information about the size of $\dot a$. The derivative can take arbitrarily large values, if $a$ oscillates rapidly. Taking the idea of Henning Makholm's example, let $$a(t) = \epsilon b t^2 \sin(t^{-10}) \tag1$$ where $\epsilon$ can be as small as you wish. The derivative of (1) exists at all $t$: it is $0$ when $t=0$, and $$\dot a(t) = 2\epsilon b t \sin(t^{-10}) - 10 \epsilon b t^{-9}\cos(t^{-10}),\quad t\ne 0 \tag2$$ The formula (2) shows that $\dot a$ is unbounded in any neghborhood of (2).

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