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Why is curve length

$$L(C)=\int_a^b |r'(t)| dt$$

(where $r(t)$ is a parametrization of the curve.)

an integral (an area/volume)?

mavavilj
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  • Why does an integral have to represent an area/volume? – imranfat Oct 31 '17 at 17:23
  • @imranfat Then what does it represent? – mavavilj Oct 31 '17 at 17:24
  • @imranfat Perhaps it means that one wants the integral function of $r'(t)$ as an answer, rather than the derivative? – mavavilj Oct 31 '17 at 17:26
  • Integrals can represent a lot of things. But going back to basics, have you studied Riemann Sums? These sums can "simulate" a lot of applications. Riemann Sums are also used to derive the arc length. Another application of the integral is Centroid. And those are coordinates (one dimensional) – imranfat Oct 31 '17 at 17:27
  • If you agree an integral can be used to calculate a volume (three-dimensional) and an area (two-dimensional), then why not a length (one-dimensional). Remember an integral is not a volume, an area, or a length. It is merely a limit of Riemann sums. But it can be used to calculate many different kinds of things. – Leonard Blackburn Oct 31 '17 at 17:42
  • By the way, areas and volumes are scalars (as opposed to vectors) so the title of your question has a faulty premise. – Leonard Blackburn Oct 31 '17 at 17:43

2 Answers2

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An intuitive, completely unrigorous definition is that the integrand(function being integrated) is the derivative of arc length (or a tiny differential (shudders) bit of arc length). Taking the integral sums up these bits of arc lengths over whatever times you specify.

rubikscube09
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  • In another way, the integral of a speed $\cdot dt$ is space: sum of $ds(t)/dt \cdot dt=ds(t)$. An area would be $scalar \cdot ds_1 ds_2$ etc. – G Cab Oct 31 '17 at 17:47
  • Ah yeah of course, since the integration is with regards to $t$, which is the parameter. – mavavilj Oct 31 '17 at 18:13
  • Does this also mean that the integral in this doesn't have a volume interpretation on $x-y$ -plane, because the $t$ -axis does not belong to it? – mavavilj Oct 31 '17 at 18:14
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As has already been said, integration is just generalized summation. This summation can be used to calculate area or volume, but it can also be used to calculate other things. In this case, to sum up infinitesimal lengths to get the full curve length.

To see why we get a length here, it may be useful to consider dimensional analysis. The parameter $t$ is measured in some unit, perhaps time (though it could be anything). $r$ is a vector in space, so it's components are all measured by length. Therefore the derivative $$\frac {dr}{dt}$$ (and its norm) would have units of $\frac {\text{length}}{\text{time}}$. When integrated, we multiply by $dt$, which has units of time. So our result is $$\frac {\text{length}}{\text{time}}(\text{time}) = \text{length}$$

And so we should expect to get a length from it.

Contrast this with your standard area-under-the-curve integral, say of $y = f(x)$. Since $f(x)$ is a height, it has units of length. And since $x$ is a horizontal distance, it too is in units of length. Thus $f(x) dx$ has units of $\text{length}\times\text{length}$, which is area.

Paul Sinclair
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