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I've always felt that the notion of 'area under the curve' is more precise than 'length of a curve', because of the Riemann integral and the reasoning used there:

First, we assume the quantity we call area $A$ between the graph of a function and x-axis eixsts. Next, we show that it must necessarily be the limit of upper and lower Riemann sums, if their limits (as we take finer partitions) are equal.

It would be less obvious if we only considered lower Riemann sums, because the fact that it converges doesn't necessarily mean it converges to the quantity we call area $A$ under the graph.

But by using both lower and upper sums, we know that the sum of areas of rectangles of lower sums will always be less or equal to $A$ (because we basically defined it to be like that), similarly upper sums will be greater or equal to $A$, so $A$ is 'trapped' between them.

In the case of arc length it's less certain that $L=\int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }$ is the length of the graph of $f(x)$ on $[a,b]$, assuming $f'$ is continuous on $[a,b]$. Even if it converges, how can we prove that it converges to what we could call the length of a curve? It's like defining area using lower sums only - fine, lower sums might converge to a certain value, but it doesn't mean the area exists (upper sums can have a different limit).

You will probably say we cannot prove it, because it's a definition. However, mathematicians have chosen to define it in this, not some other way for some reason (the could have, right?), thus it is considered more 'correct'. So the question - why should I treat it as the 'right' definition?

user4205580
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Actually that integral is not really the definition. The definition of the length of the graph of $f$ for $a\le x\le b$ is the sup of the sums $$\sum_{j=1}^n\left((x_j-x_{j-1})^2+(f(x_j)-f(x_{j-1})^2\right)^{1/2}$$over all choices $a+x_0<\dots<x_n=b$. That sum is just the length of a polygonal path joining points of the graph of $f$, so it makes sense as the definition of the length.

Now if we assume that $f'$ is continuous it's possible to prove that the length is equal to that integral.

EDIT: Regarding the question of what makes this definition "correct": Of course that's meaningless in a mathematical sense, a definition is a definition. It's not meaningless to ask why it makes sense as a definition of arc length.

One way to look at it is this: What is the "length" of a curve anyway? It's the net distance you travel if you walk along the curve. But your pedometer doesn't have a magic arc-lenght measurer - all it can do is measure the distance between successive steps and add. That's what that sum does. Except of course oops, measuring the distance between this step and the next underestimates the arc length because it ignores wiggles on a scale smaller than the step size. Hence the "sup".

David C. Ullrich
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  • Oh, sorry, right. But then, what makes this definition correct? – user4205580 Sep 30 '15 at 15:04
  • A definition is correct by definition. This definition gives the answer we want for polygons. And it makes sense, at least to most of us: You consider the least upper bound of the length of those polygonal interpolants... – David C. Ullrich Sep 30 '15 at 15:09
  • And I'm also not sure if the supremum definition given above is complete. I also feel this definition is okay, but well, you can and should always have doubts, right? Questioning what is considered 'true' leads to improvements. – user4205580 Sep 30 '15 at 16:39
  • @user4205580 For one thing, if we assume that the line is the shortest path between two points, then we know that the length of a curve $S$ is bigger than the length of any polygonal approximation to $S$. (You're breaking $S$ into pieces, underestimating the lengths of the pieces with straight lines, and then adding up the underestimations.) So, we know that the length of $S$ is at least (that is, $\ge$) the supremum of the lengths of the polygonal approximations. We know all this even before coming up with a formal definition. – Akiva Weinberger Sep 30 '15 at 16:39
  • @Akiva Weinberger columbus Yeah, so what? As you said, we know that $S$ is at least the supremum of lengts of the polygonal approximation. The question is if we can eventually 'reach' $S$ with polygonal approximations or not. – user4205580 Sep 30 '15 at 16:42
  • @user4205580 A big part of the problem is that we have a lower bound (the supremum thing) but no upper bound. With area we had an upper bound. I'm in the middle of writing an answer that tries to address this. – Akiva Weinberger Sep 30 '15 at 16:44
  • @AkivaWeinbergercolumbus How do you plan on getting either bound before giving a definition of the arc length? – David C. Ullrich Sep 30 '15 at 21:49
  • The current definition approximates the true arc length $S$ 'from below', that is the arc length is at least the supremum given in your answer. Personally I feel it would be nice if we could prove that approximating the arc length from above (taking infimum) gives the same number as approximation 'from below'. It would eliminate any objections one could have regarding the current definition. And that would be exactly the same approach as in Riemann's integral (upper and lower sums). I cannot come up with an example of such a function whose upper and lower approximations would differ, but... – user4205580 Oct 01 '15 at 16:55
  • it doesn't mean they do not exist. – user4205580 Oct 01 '15 at 16:55
  • @user4205580 What are you taking the infimum of? – Akiva Weinberger Oct 01 '15 at 16:59
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    @user4205580 There are no objections to the current definition, except in this thread. And you haven't given any reason for your objection, other than that you don't believe it and it would be nice if we had another one. – David C. Ullrich Oct 01 '15 at 17:00
  • Ullrich, have you ever heard of an axiomatic definition? Instead of defining $\operatorname{Length}(S)$ for every curve $S$, we give a list of properties that the function $\operatorname{Length}$ should have. The list of properties is complete if it uniquely determines the function $\operatorname{Length}$. Area has a nice, axiomatic definition (barring pathologies such as nonmeasurable sets). We want one for length, too. – Akiva Weinberger Oct 01 '15 at 17:03
  • @AkivaWeinbergercolumbus I'm taking the infimum of the approximations from above (based on your reasoning). You mentioned we only have lower bound, but no upper bound. So all we know is the true length is equal to or greater than the number produced by the definition of arc length. It would be nice if we could take the infimum of approximations with curves longer than the true curve in respective intervals (just like straight lines we are using right now are known to be the shortest path between two points). – user4205580 Oct 01 '15 at 17:06
  • I know, I can't take a longer curve if I don't have a definition of length. – user4205580 Oct 01 '15 at 17:10
  • @AkivaWeinbergercolumbus "Ullrich, have you ever heard of an axiomatic definition?" You really feel it's important to be insulting? An axiomatic definition like that would be great. I've never seen one. – David C. Ullrich Oct 01 '15 at 17:11
  • Sorry, I wasn't trying to sound insulting. :( I was just trying to clarify the problem. – Akiva Weinberger Oct 01 '15 at 17:12
  • @AkivaWeinbergercolumbus Fine. It is true that asking for an axiomatic definition of that sort gives a much more coherent statement of "the problem" than we've seen her previously. But it could be that there is no such definition - I've never seen one, not that that proves anything, but I doubt that anyone's ever given one. – David C. Ullrich Oct 01 '15 at 17:17
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    @AkivaWeinbergercolumbus Ok, axioms: Finite additivity, the length of a line segment is what it is, a straight line is the shortest curve joining two points. (So far we can show the length is greater than or equal to the standard definition.) And if $C_n\to C$ uniformly and the length of $C_n$ is no larger than $M$ for every $n$ then the length of $C$ is no larger than $M$. That last seems quite reasonable to me - we know examples of curves approaching shorter curves but no examples in the other direction. Those axioms imply that arc length is equal to what the standard definition says. – David C. Ullrich Oct 01 '15 at 17:28
  • What are $C_n$, $C$ and $M$? You say 'we know examples of curves approaching shorter curves but no examples in the other direction' - isn't it like the other way round? The more line segments we take, the better the approximation? And the sum of lengts of line segments is always less or equal to the curve length. – user4205580 Oct 01 '15 at 18:27
  • $C_n$ is a sequence of curves. $C$ is a curve. $M$ is a real number. I was a bit imprecise with "shorter" and "longer". There are plenty of examples out there of curves $C_n$ approaching a curve $C$ where, say, the length of $C_n$ is $2$ but the length of $C$ is $\sqrt 2$. (Staircase approachinng the diagonal of the unit square.) There are no examples of $C_n$ approaching $C$ where the length of $C_n$ is less than or equal to $M$ for every $n$ but the length of $C$ is greater than $M$. – David C. Ullrich Oct 01 '15 at 18:31
  • Going to work now. Just so you won't assume I'm dumbfounded when I don't have an immediate reply to your next comment... – David C. Ullrich Oct 01 '15 at 18:33
  • There are no examples of $C_n$ approaching $C$ where the length of $C_n$ is less than or equal to $M$ for every $n$ but the length of $C$ is greater than $M$ - and I could restate my question this way. It's was all about whether such $C_n$s exist or not (but you made it precise and short). The fact we don't know such examples means nothing really. But if it's your axiom, fine. – user4205580 Oct 01 '15 at 18:56
  • @DavidC.Ullrich and it's impossible to prove or disprove this statement (the one in italics in my last comment) without the definition of arc length... – user4205580 Oct 02 '15 at 17:47
  • You usually try to prove axioms? – David C. Ullrich Oct 02 '15 at 18:36
  • No. I usually try to find out if it's necessary for a statement to be an axiom or if it can be deduced from other axioms. – user4205580 Oct 02 '15 at 20:06
  • By the logic in your post, we should define the integral to be the lower Riemann sum and forget about the upper sum. I don’t think that’s correct reasoning…. – Vivaan Daga Mar 28 '23 at 14:20
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Rather than write a new answer, I'll just link to an answer I wrote a while back that touches on this issue:

False proof: $\pi=4$, but why?

It's surface area rather than perimeter, but it's basically the same thing. The relevant point is the last paragraph: If everything's convex, we can get an upper bound and use the squeeze theorem, like with Riemann sums. (To get a lower bound, we can use polygonal approximations, or just use convexity again.) We can then find the perimeters of almost all curves, by cutting them into convex pieces and finding the lengths of those. Eventually you can prove that this is equivalent to the usual definition.

(Exception: Cantor's staircase isn't convex anywhere, so this won't work. I'm not quite sure what to do with this. It has length $2$ on the unit interval.)

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Akiva Weinberger
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  • With regard to this exception, I don't think there has to be a way to 'do something with it'. People defined continuous function trying to express the idea of drawing a graph of a function without lifting the pencil. But can you actually draw Weierstrass function? Or can you measure the area under its graph? Yes, because it's integrable. But its graph 'never stops changing'... – user4205580 Sep 30 '15 at 17:15
  • True. I'm just saying that we have to use @DavidCUllrich's supremum definition for it, rather than trying to go around it like we can do for convex shapes. – Akiva Weinberger Sep 30 '15 at 17:20
  • Maybe we should look at it this way: arc length has been defined so that it works for a narrow set of functions with 'nicely looking graph', and it's possible to prove the correctness of this definition for them using the method you provided. But it's not our fault there are 'weird' functions outside the set of 'nicely looking functions' for which this definition also generates a number (as you say, Cantor's staircase has length $2$, by definition of arc length). – user4205580 Sep 30 '15 at 19:09
  • @user4205580 Yeah. And we're reasonably sure that, at least for those graphs, it's the "right" definition (whatever that means). And since it agrees with the supremum definition for those graphs, that means that we're pretty sure that the supremum definition is the "right" generalization to all curves. – Akiva Weinberger Sep 30 '15 at 19:11
  • On the other hand, maybe we should restrict ourselves to defining a curve length if both upper and lower bounds exist and have equal limits, i.e. for those nicely looking functions. The number produced by definition, as you said, only tells us the number the 'true arc length' is greater than. But it doesn't mean it's equal to it. – user4205580 Sep 30 '15 at 19:16
  • I still believe that the Cantor's staircase's "true" length is $2$. Partly because I'm pretty sure that a curve that only goes up and to the right, from $(0,0)$ to $(1,1)$, can't be more than $2$ units long. (That's true for "nice" curves, anyway.) – Akiva Weinberger Sep 30 '15 at 19:21
  • Fine, of course it might be the case, but it doesn't solve the more general problem, right? – user4205580 Sep 30 '15 at 19:26
  • @AkivaWeinbergercolumbus Actually no, defining surface area and arc length are not more or less the same; surface area is much harder. I think the story is about Gauss sitting in class, professor says something about surface area, he comes up with a counterexample. If you pick a bunch of points on a curve, spaced closely together, the su of the lengths of the line segments joining the points approximates the length, while if you triangulate a surface similarly, so the corners of the triangles lie on the surface, the sum of the areas of the triangles need not approximate the area, – David C. Ullrich Sep 30 '15 at 21:46
  • @AkivaWeinbergercolumbus But anyway, say we have a convex curve and interior and exterior convex polygons. How do you know that the length of the exterior polygon is greater than the length of the curve? – David C. Ullrich Sep 30 '15 at 21:48
  • @user4205580 What problem? – David C. Ullrich Sep 30 '15 at 21:48
  • @DavidC.Ullrich We can't prove it without a formal definition, clearly. (We'd have to add it as an axiom.) With the definition as "the supremum of polygonal approximations" it can be proven, though. – Akiva Weinberger Sep 30 '15 at 22:24
  • @AkivaWeinbergercolumbus Certainly. But that definition is unacceptable here for some reason I haven't quite got yet... – David C. Ullrich Sep 30 '15 at 22:29
  • @AkivaWeinbergercolumbus Although my question was actually "how do you prove it", not "can it be proved". I never thought about this much. Thinking about it, the only approach I can think of definitely does not work: Say $O$ is a point interior to both curves. Imagine two rays emanating from $O$, making a very small angle. It seemed perhaps one could show by some calculusishh argument that the part of the interior curve in the angle between the two rays was shorter than the corresponding part of the exterior curve. But that's not true. Hence I really have no idea how to prove it... – David C. Ullrich Sep 30 '15 at 23:15
  • @AkivaWeinbergercolumbus Example: The interior curve is the square with vertices $(\pm 1,0)$ and $(0,\pm 1)$. The exterior curve is the square with vertices $(\pm 1,\pm 1)$. The point $O$ is $(0,0)$. Consider the part of each curve very close to the point $(1,0)$, trapped between those two rays. That bit of the interior curve is longer than that bit of the exterior curve. Not a counterexample to the result, of course, but a counterexample to the only plan of attack I can think of. – David C. Ullrich Sep 30 '15 at 23:18
  • @AkivaWeinbergercolumbus I think I got it. Say $C_1\subset C_2$ are convex subsets of the plane. Say $x_1,\dots,x_n$ are some points running along the boundary of $C_1$ "in order". At each point $x_j$ there is an angle formed by that point, the "previous" point and the "next". Convexity shows the exterior angle at $x_j$ is larger than $\pi$. Project "outward", "away from $C_2$", along the ray bisecting that exterior angle until you get to a point of the boundary of $C_2$. Call that point $y_j$. Then the sum of the distances for the $y_j$ is $\ge$ the sum for the $x_j$. – David C. Ullrich Oct 01 '15 at 00:33
  • @DavidC.Ullrich what problem? Quoting the comment below: ' A big part of the problem is that we have a lower bound [of the curve length] (the supremum thing) but no upper bound. With area we had an upper bound [too].' – user4205580 Oct 01 '15 at 16:25
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    @user4205580 And how is that a problem? You seem to want a definition of arc length such that we can prove the definition gives the True Arc Length. But trying to prove that without a definition is meaningless. It does make sense to ask for a definition that people agree gives the "right" answer, that is, in accord with their intuitive notion of arc length. The standard definition has this property - I don't know anyone but you who has any problem with it, – David C. Ullrich Oct 01 '15 at 16:28
  • @DavidC.Ullrich 'Others don't have this problem, so this problem doesn't exist.' Nice logic. – user4205580 Oct 01 '15 at 16:37
  • @user4205580 It would be ridiculous logic if we applied it to a mathematical question. The current "question" is not a mathematical question, it has to do with whether a certain definition is consistent with people's intuitions. The only way to determine whether that definition is consistent with people's intuitions is to ask them. The answer is that yes it is. – David C. Ullrich Oct 01 '15 at 16:40
  • @user4205580 See edit to my answer. And also tell me whether you have any examples in mind where the standard definition gives an answer that you feel is wrong. – David C. Ullrich Oct 01 '15 at 16:46