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The above figure illustrates how the points $Q_0$, $Q_1$, ..., $Q_n$ (with position vectors $\vec R_0$, $\vec R_1$, ..., $\vec R_n$, respectively) partition the curve and generate the inscribed polygonal path, whose length we calculate by summing the lengths of the sides $\left|\Delta\vec R_k\right|=\left|\vec R_k-\vec R_{k-1}\right|$. The length of the curve is then taken to be the limit of these sums, as the partitions are refined in such a manner that the largest length $\left|\Delta\vec R_k\right|$ goes to zero.

Moreover, this limit can be evaluated if the curve is parametrized by $\vec R=\vec R(t)$, $a\leqslant t\leqslant b$. Recall that the interval $[a,b]$ is partitioned $a=t_0<t_1<t_2<\cdots<t_n=b$ to correspond with the points $\vec R_k=\vec R(t_k)$, and the approximation

$$\Delta\vec R_k\approx\frac{d\vec R}{dt}\Delta t_k$$

is used to argue that

$$\int_C\left|d\vec R\right|\equiv\lim\sum_{k=1}^n\left|\Delta \vec R_k\right|=\lim\sum_{k=1}^n\left|\frac{d\vec R}{dt}\Delta t_k\right|=\int_a^b\left|\frac{d\vec R}{dt}\right|dt$$

as the $\Delta t_k$ go to zero.

How is the approximation

$$\Delta\vec R_k\approx\frac{d\vec R}{dt}\Delta t_k$$

obtained? I am having a hard time understanding the geometrical meaning behind its RHS.

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$\frac{\Delta\vec R_k}{\Delta t_k}\approx\frac{d\vec R}{dt}|_{t=t_k}$ for a sufficiently fine partition. –  Andrew Aug 5 '12 at 21:57
    
@Andrew, I suppose this is only doable because the $a=t_0<t_1<\cdots<t_n=b$ correspond with the $\vec R_k=\vec R(t_k)$? –  Josué Molina Aug 5 '12 at 22:21
    
Yes, that's right. –  Andrew Aug 5 '12 at 22:30
    
If you want a purely intuitive justification, just imagine a particle flowing along the vector $\frac{d\vec{R_{k}}}{dt}$ for a duration $\Delta t_{k}$ at the point $\vec{R_{k}}$. You can convince yourself that you "land" reasonable close to the next $\vec{R_{k+1}}$, or in other words, the approximation above holds. You can then prove analytically that the approximation converges as $\Delta t_{k}\to0$ if the curve itself is sufficiently smooth. –  Taylor Martin Nov 10 '12 at 21:24
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