# Arc length definition

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?

## 2 Answers

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.)

• 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

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".

• 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