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Let $ \alpha :\left[ {a,b} \right] \to R^3 $ be a differentiable curve, and let $ a= t_0 < .... < t_n = b $ a partition consider $ \eqalign{ & L\left( {\alpha ,P} \right) = \sum\limits_{i = 1}^n {\left| {\alpha \left( {t_i } \right) - \alpha \left( {t_{i - 1} } \right)} \right|} \cr & P = \max (\left( {t_i - t_{i - 1} } \right),i = 1..n) \cr} $ Prove that given any $ \varepsilon > 0 $ there exist a $ \delta > 0 $ such that if $ \left| P \right| < \delta $ then $ \left| \int_a^b \left| \frac{d}{dt}\alpha(t)\right|\,dt - L(\alpha ,P)\right| < \varepsilon $

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    ...and the outer is kinda useless.2011-08-09

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A few hints, which I can expand if need be. I assume that your curve is $C^1$.

When you see an equation like the last one, in which the derivative is related to some finite sum of differences, your mind should jump to the mean value theorem. Use this to find, for each $i = 1, \ldots, n$ and $j = 1, 2, 3$, an $s_{ij} \in [t_i, t_{i + 1}]$ such that (\alpha_j)'(s_{ij}) = \frac{\alpha_j(t_i) - \alpha_j(t_{i - 1})}{t_i - t_{i - 1}}.

Another observation is that because \alpha' is a continuous function on the compact set $[a, b]$, it is even uniformly continuous. For a given $\varepsilon$, it follows that if in the above $P$ is small enough then for $i, j$ and any $t \in [t_i, t_{i - 1}]$ we have |(\alpha_j)'(t) - (\alpha_j)'(s_{ij})| < \varepsilon. So |\alpha'(t) - ((\alpha_j)'(s_{ij}))_{j = 1}^3| < \sqrt{3}\varepsilon.

Putting this together involves some careful manipulations using the triangle inequality and some annoying square roots. Good luck!