Definition of curve length by integrals

Question

Let $\gamma:[0,1]\to \mathbb R^2$ be differentiable, injective, hence a nice curve. The length of the curve is defined to be $$ L(\gamma) = \int_0^1 |\dot\gamma(t)| dt. $$ This formula is often motivated by approximating the curve by a polygonal curve, taking length of straight pieces, and passing to the limit.

However, in $\mathbb R^2$ and $\mathbb R^3$ there is no need to define the length of the curve. We can just measure it by a measuring tape (normed such that the length of $[0,1]$ is $1$).

My question is: How can we prove that the curve length by integration gives the same length if we measure length by a measuring tape.

Edit: As Brahadeesh argues in his answer (and others have mentioned this as well), it is unclear what is the meaning of 'measuring by tape' mathematically. So let me expand the question in the following way:

Is it possible to define a length of a curve just by (elementary?) geometric considerations without using differential and integral calculus?

Answer 1

Lay out the measuring tape. Now slightly alter the shape of the measuring tape, so that it stays near the curve, with slope near the slope of the curve, but is piecewise linear. In the limit as the alteration is made smaller, the piecewise linear curve approaches the original curve uniformly, and its derivative (defined almost everywhere) approaches in Lebesgue $L^1$ sense, and so the integral giving the length converges.

Answer 2

Do you agree with the following assumptions?

• Instead of a single long measuring tape we can use many short ones placed on the curve end to end. We simply add up the length that we read off and get the total curve length.
• Approximating the length of a sufficiently short tape (on a sufficiently well-behaved curve) by the distance of its end points is a valid estimation.
• Our approximation gets better for shorter measuring tapes and the error goes to zero with the length of the tape going to zero.

In order to prove something about reality, we have to create a model. And in order to create a model we have to make assumptions. Above I try to fix some minimal assumptions which suffice to justify the mentioned integral formula. Even if you disagree with one of the assumptions, this would be helpful information in order to answer your question even more satisfying.

It follows the justification: Your curve can be parametrized by a function $\phi:[a,b]\to\Bbb R^3$. A partition of the curve are points $\phi(t_1),\phi(t_2),...,\phi(t_n)$ on the curve with $t_1=a$ and $t_n=b$ which mark start/end points of the short measuring tapes. One can prove that for any sequence of such partitions with

$$\max_i \|\phi(t_{i+1})-\phi(t_i)\|\to 0$$

for $n\to\infty$ the real number sequence

$$\ell_n:=\sum_{i=1}^{n-1} \|\phi(t_{i+1})-\phi(t_i)\|$$

is a Cauchy sequence (again, we assume the curve is sufficiently well-behaved). As a Cauchy sequence in $\Bbb R$ it does converge to some value $L\in\Bbb R$. By the nature of convergence and if you accept above assumptions, by increasing the accuracy of our approximation, we get closer and closer to measuring the value $L$ for the length approximation. So it is forced on us by logical reasoning that $L$ equals the length of the curve. Now, independent of this motivation, one can prove that $L$ has this exact integral representation from your question just by considering the introduced formalisms and limits.

Answer 3

It is an act of faith to admit that our physical world matches Euclidean geometry. This cannot be proven, but centuries of experimentation lead us to believe in this model, at least between the limits of quantum mechanics and general relativity.

This said, the length of a straight tape is given by Pythagoras' theorem and its deformations to match a curve are described by length-preserving transformations, leading to the above integral.

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Note that the artifacts we are talking about are idealized objects with zero stiffness, and in a way we are doing experiments of thoughts. In the true world, we should have to take into account the physics of continuous media and account for elasticity effects inside the tape, probably leading to tiny corrections of the formulas where curvature appears.

Answer 4

How can we measure the length of a curve in $\mathbb{R}^2$ by using a measuring tape?

Well, as mentioned in a comment above:

Take the tape, fit/attach it to the curve, read off the length. Like it works irl.

But there is some subtlety here.

First of all, it is not possible to measure the lengths of curves just as we do for "linear" objects in real life. For mathematical lines have no width or volume whereas real objects do, and real objects are discrete at the atomic level whereas mathematical lines have no holes in them.

But, even keeping this objection aside, the rest of the procedure still has some subtlety behind it. Given a "mathematical" tape measure or an idealized tape measure (one that has no width yadda yadda), surely we can fit/attach it to the curve and thereby read off its length?

Well, for that we would need a mathematical tape measure. Do we have one?

Yes! The real line comes to our rescue. It is a "tape measure" which is naturally normalized so that the length of $[0,1]$ is $1$, and which does not suffer from any of the drawbacks of a physical tape measure. So, the real line can be used to measure the lengths of curves.

Great, but how do we fit/attach the real line to a given curve, so that we can read off its length?

Here it gets a bit tricky.

When we fit the real line to a curve, we are defining a function $f$ from a subset of $\mathbb{R}$ to the curve, which is "distance-preserving". Here, distance-preserving has a different meaning: it does not mean that $d(\mathbf{x},\mathbf{y})$ is the Euclidean distance between the points $\mathbf{x}$ and $\mathbf{y}$ on the curve when they are viewed as points in $\mathbb{R}^2$. Instead, the distance between these points is the length of the curve between these points. So, we are viewing the curve as a metric space with distance between two points defined as the length on the curve between these points.

But this is utterly hopeless because it is circular. The mapping $f$ is characterised by being distance-preserving, but the distance between two points is defined to be the length of the curve between those points. Which is defined by $f$. Circularity.

And this is not even getting to the fact that we are subtly assuming that the curve is simple (no self-intersections). Even more disastrously, what is the subset of $\mathbb{R}$ on which $f$ is defined? It can be taken to be $[0,l]$ where $l$ is the length of the curve. So $f$ is defined by the length of the curve and defines the length of the curve at the same time. More circularity.

Maybe if we instead fitted the curve to the tape? Even that doesn't work because to define a "distance-preserving" function $f$, we need to know what the distance between two points on the curve is, and that distance is given by $f$ itself. Circularity remains.

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A little reflection on the above discussion should make it clear that the problem with measuring the length of a curve using a measuring tape (by fitting it to the curve and reading off its length) is that we are defining the length of a curve in $\mathbb{R}^2$ as follows:

The length of a curve is whatever its length is.

This is correct! But utterly useless.

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How do we fix this state of affairs? Consider, for example, how we measure the area below the graph of a function $f : [0,1] \to \mathbb{R}$. We could say:

Well, to measure the area we just rearrange the shape while preserving its "mass" so that it now looks like a $1 \times l$ rectangle. We can then read off the area from its length $l$.

Again, this is correct! But how is it helpful?

Observe, on the other hand, how useful the idea of the Riemann sum and Riemann integral is compared to the above!

Whatever be the area, it is surely greater than the sum of areas of the rectangles in a lower Riemann sum and lesser than the sum of areas of the rectangles in an upper Riemann sum. In particular, whatever be the area, it lies between $\,\sup L(f,P)\,$ and $\,\inf U(f,P)$. If these two values are equal, then it makes most sense to say that the area is this common value. If these two values are not the same, then it makes most sense to say that the area is not defined for that particular function $f$.

Not only have we found a definition amenable to computation, we have managed to clearly specify when the area is even defined in the first place, something which did not come out of our earlier definition of area. It is in fact so robust that we can even compute the area under some functions that are discontinuous at many many points, functions which render the idea of "shape" of the area redundant and thus the first definition as well.

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This is why we use the same intuition in defining the length of a curve as well.

Whatever be the length of the curve, it is surely not lesser than the sum of lengths of line segments that connect finitely many points along the curve from start to end. If the curve is described by the parametrization $\gamma$, then some simple geometric arguments, followed by taking the supremum of the set of values computed for different choices of line segments gives the expression $$\int_0^1 |\dot{\gamma}(t)| dt$$ for the supremum. Then, it makes most sense to say that this integral is the length of the curve when this integral exists. When the integral does not exist, we say that the length of the curve is not defined.