Arc Length and the Fundamental Theorem of Calculus

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In this section, we will interpret the plane curvature of a curve in a new way. For that, we will introduce the notion of arc length. This will bring us to the idea of integration, an art of approximation, by means of which, as we have said, Galileo was able to understand the principle of uniform acceleration. Integration is a powerful technique that was used haltingly and with great difficulty since the time of Archimedes. With the invention of Calculus however, Newton made it available to everyone. He did this by cultivating a new way of thinking about measurement. We will explore that in this section, as we develop the "Fundamental Theorem of Calculus," but first, let us describe the problem.

Arc length Parameterization

Recall that if we have a plane curve where are smooth (infinitely differentiable) functions of their parameter t, then a different parameterization of , say

would yield the same points along the curve.

As we saw, each smooth curve defines another curve and

We call the velocity curve. As in the previous section, we will assume that

That is, the velocity never vanishes. Represent the velocity curve

Now given a parameterization of the curve by the parameter t as above, there is a unique strictly increasing function up to the addition of a constant,

with the property that the new curve with parameter s

has the following property

We will call this parameterized curve the arc length parameterization. It will require a short excursion into the Integral Calculus to justify the existence and uniqueness of this parameterization, and we will do that below, because it will be good preparation for Chapter 5. Meanwhile, let us assume that we have such a parameterized curve and see what that entails.

We saw that for the curve where so , the curvature is independent of the parameterization so that if we write , then at we have the same velocity frame . And if we write the acceleration

in the form

and if

then

Now for the arc-length parameterization, . Therefore the curvature at is simply

Curvature in arc length parameterization: angular acceleration

It is the absolute value of the angular component of acceleration . There is more. The radial component of acceleration in the velocity frame is 0.

To see this, observe that since the speed is constant 1. (Compare with the general case:

Therefore writing as above we see that

Question 1: Show that if then is a multiple of , the second vector of the velocity frame at .

End of Question

This shows that the radial component of acceleration:

Curvature as measure of curve turning

Therefore the curvature is simply the absolute value of the acceleration for the arc length parameterization, and the acceleration itself is purely angular. It means that the velocity vector only turns as you traverse the curve. It does not grow or shrink. The curvature measures the rate of turning. We will make that idea more precise when we introduce the Gauss map in the next section. For now, recall the osculating circles at each point of the curve. This "rate of turning" is simply the reciprocal of the radius of the osculating circle.

Now we cannot illustrate this idea (except in the simplest cases) in the microworld for one reason. Given a curve it is not easy to say what the arc length parameterization is! We know that under light restrictions on that it exists, as we shall see, but it may not be possible to express the function in terms of the elementary ones (such as polynomials, trig functions, exponentials, and so on).

So now we will see why such a function may exist at all, and what conditions must be imposed on the curve. First of all we assumed that and that are smooth (infinitely differentiable) functions of their parameter t. And we assumed that . Next, it is convenient to restrict the curve to a closed interval, its domain of definition that we will call the interval . So our curve is now a mapping

and we seek an increasing smooth function t defined on another closed interval such that

Now one condition that must be satisfied immediately for to be parameterized by arc length is that

Therefore, since t is increasing and has positive derivative,

where .

We will show in a moment that there is a function

Such a function is called an antiderivative for . Since the derivative of will be strictly positive, will be one-to-one.

Question 2: Use the Mean Value Theorem and the existence of one-to-one above to show that if are two functions mapping that satisfy the condition above, and = a, then for all

End of Question

Now the heart of our problem is to establish the existence of a function ,

This will be called the arc-length function for our original curve . It will be strictly increasing, and map

for some non-negative number , which will be the length of the curve. This function will have a differentiable inverse where

so that for all

Question 3: Show that (assuming differentiability of ) that if we define then

is indeed parameterized by arc length, that is, show that (using the chain rule)

so that

End of Question

Partition of an interval

We will give an intuitive sketch here how we will construct the function for smooth . The actual construction will follow the discussion of Riemann integration below. First, we introduce the idea of a partition of the interval for . To fix ideas, we will let for now, and explain how to extend the construction for general .

Definition: A partition of size n, , for the interval is a sequence of n+1 numbers

The mesh of the partition is the maximum of the length of subintervals so defined:

End of Definition

Polygonal approximation of a curve

Each partition allows us to make a polygonal approximation to the curve in the following way. Consider the sequence of points

In the exploration, we chose the curve

These points lie in the curve, and we may connect each to the next by a straight line segment as in the picture:

We chose a partition of size 9 and determined 9 line segments. The length of arc of all but the final segment is p. These segments "approximate" the curve on the interval . We may calculate the length of the polygonal path so determined. The length will be:

Now since we have assumed that is smooth (infinitely differentiable) on it follows (from a fairly deep theorem in set topology, and we will explain why it does below) that the numbers are all bounded above by some positive number. In fact, there is a smallest number such that

for all partitions . It is also obvious from the triangle inequality that if in the sense that all points of the first partition belong to the second, then

Arc length and rectifiability

These facts taken together will allow us to define the length of the arc on to be the upper bound of the lengths of polygonal segments: . Now we apply the same reasoning for each and define the length of the arc on in the same way to be .

The function obviously satisfies . It is also increasing, in the sense that if then . This brings us to the problem: how do we establish the crucial property:

We take that question up now.

Integration: A new Art of Approximation

Galileo asked: "How far does an object fall over an interval of time T if it falls freely?" He answered this question, as we said, by measuring that distance using (what we would call) integration. We shall explain how in a moment once we establish a working understanding, for our arc length problem, of the Fundamental Theorem of Calculus.

Integration strategy

The strategy of integration is to measure a quantity in two steps.

First, we imagine that there is a parameter or variable that varies from one fixed number to another, and determines as it increases, the growing amounts of the quantity to be measured. This simply means that the quantity to be measured, call it Q, is a function of the parameter, call it t. We are implying that Q is an increasing function of the parameter, but it needn't be. It will simplify the discussion to think of it that way for now. So Q depends on t and it is an increasing function of t. We imagine that at some starting value of the parameter: to we know the value and then that we would like to determine the value of Q at a "later" value of t: which we will call .

It is important to note that the parameter need not be time (even though we have denoted it t). And again, it is not necessary that Q increase with t. That assumption just makes it easier to visualize what is going on. Speaking of simplifications, we will often assume that and that in our examples in this lecture. An example to keep in mind throughout this discussion is, of course, the arc length function .

So Q "starts" at parameter value a at "nothing" or 0 and at some later parameter value, it is "something". We want to measure how much it is. This introduction of a parameter is a very important step. Let us illustrate with a few examples.

Example 1: Galileo's freely falling body.

The "Quantity" Q is the distance from the height at which it was dropped. The parameter that Galileo used (as we shall see) is the time. At the starting time, the distance fallen is 0, or no distance at all. At later times, the total distance fallen is some positive amount. Thus, Q(t) is the amount of distance fallen through time t.

Example 2: The arc length function .

A smooth curve . Assume for now (we justify it shortly) that for each the numbers are all bounded above by some positive number. In fact, there is a least upper bound such that for all partitions . We define . The question is, how do we show that the derivative is the speed, ? This is the inverse of Example 1, but it is a problem of integration, as we will see.

We could go on, but these examples illustrate the need to introduce a parameter in order to do the measurement (although we have not said how the measurement is to be done yet!). This parameter is often called the "variable of integration" in this strategy.

Having established that our quantity to be measured is determined in this way by a parameter, special to the measurement problem, we take the second critical step. In this lecture, we are interested in determining the value .
Let us now assume that . We will show (once we define ) that , and further that increases with t. For simplicity, let us call simply, T. So we want to measure .

Now we imagine that if we partitioned (divided) the parameter interval with a finite increasing sequence of n+1 parameter values

we would have a sequence of values for Q

This is obvious, and it tells us nothing new. But now look at the sum

This is equal to

For simplicity, let us call

and

for this partition.

The idea is that we use our partition to decompose the amount Q(T) into parts. We may finally write, for this partition of the parameter interval that

Now, we still have said nothing new, but we have subtly changed our point of view. We are now in possession of a means to write the quantity we want to measure as a sum of smaller quantities. In a limiting process, to be described below, we imagine, along with Newton, that there are an "infinite" number of "infinitely small" quantities that somehow give a finite sum! Adding together these "infinitely small" quantities is what we shall call integration. We can usually visualize these "smaller" quantities. For example

• For the freely falling body, is a small increment in x, the distance from the height from which the object was dropped, that results from falling through a small interval of time.
• For the arc length, is the length of arc on the interval .
• For Q the area swept out in a planetary orbit, is the area of a small wedge of the orbit, with vertex at the Sun, as the planet moves through a short interval of time.

The basic idea behind measurement by integration is this.

Suppose that we can find a continuous function

called a simple approximating function, such that, for each partition,

of the parameter interval, with all small enough, we may choose a sequence of parameter values:

such that for each

then obviously

We often write

The trick is to find a single continuous approximating function f such that for any T^¢ with 0 < T^¢ < T, and for any partition of

with all small enough, there is a sequence with

that gives the equality

Integral as Antiderivative

Question 4: Show that if such a continuous function f is found for quantity Q, then

for all . Thus, we say that Q is an antiderivative for f.

End of Question

It is not as difficult as it may seem to find such a simple approximating function. These functions are usually obtained by approximating the DQi. Here, as we said, we encounter another "Art of Approximation."

Simple approximating function

Definition 1: Given a quantity Q as above, a continuous function defined on a parameter interval , with parameter t. Suppose that Then a continuous function

is a simple approximating function for Q if it is continuous on all of and it satisfies the condition that if is small enough, and then there is an such that .

End of Definition

General approximating function

This simple approximating function will not do for arc length, however. We will, in fact, consider a more general condition for approximating functions.

Definition 2: Given a quantity Q as above, a continuous function defined on a parameter interval , with parameter t. Suppose that Then a continuous function

will be called a general approximating function of order k for Q if there is a continuous function

with

and such that if is small enough, and , then there are such that

.

End of Definition

The need for general approximating functions will be clear also when we consider in Harmony of the Spheres, the rate of change of area in planetary orbits, as stated in Kepler's second Law. We will observe that the constant function of t

yields an approximation to the area swept out over time interval from . We will see, however, that this f is a general approximating function of order 2 for the area and not a simple approximating function (order 1). The difference between simple and general approximating functions is a technical one, though, and our main theorem will apply to them both.

Now, in the case of a simple approximating function, to measure we look for a continuous approximating function such that whenever is small enough, there is an such that

Next suppose we have found such an approximating function f associated with Q. If we knew what the values of s

were for some partition we would be finished, because we could simply evaluate

But of course, in general we do not know what the values of s are, even for one partition of [0,T]. That is the reason that integration is a powerful tool.

Riemann Sums

Definition 3: Now suppose next that we have found an approximating function f associated with Q. Then for a partition

we choose any points

whatsoever satisfying for 1 £ i £ n

then the sum (for those points)

is called a Riemann Sum associated with the partition and the approximating function f.

In general, these Riemann Sums approximate but do not equal exactly. As it happens, though, if Q has a continuous approximating function f, simple or general, then we may choose partitions with finer and finer mesh, that is, where the maximum distance between neighboring points gets smaller and smaller, and then these Riemann Sums will be seen to approach a limit as the meshes of the partitions approach 0. This is the consequence of a deep topological fact about continuous functions defined on closed intervals, that asserts that such functions are uniformly continuous.

The common limit that those Riemann Sums approach as

is written

and it is equal to the value of the quantity to be measured. This is because f is a general approximating function in the sense described earlier, and is a consequence of a theorem we state next. Newton did not think of this operation in terms of limits in the way that we do. That came later. He imagined that the really did go to 0 (or became infinitesimal).

End of Definition

The intuitive idea is that the entire quantity is the sum of these smaller quantities. It didn't bother Newton to imagine that the sum would be an 'infinite' sum of 'infinitesimal' amounts. Everything rests on the following Theorem (whose proof, as we said above, is beyond the scope of these lectures).

Integral as limit of Riemann sums

Theorem 1: Suppose that Q is a quantity to be measured, parameterized as above, on the interval and suppose in addition that

and there is a continuous function

with

and such that if is small enough, and , then there are such that

.

Then for any positive number e chosen in advance, there is a positive number with the property that whenever a partition

is chosen with mesh less than , that is, so that all

then if

and

are any two Riemann Sums (defined as above) for that partition and for approximating function f, then

End of Theorem

In particular, for any given partition, we may always choose Riemann sums R and S in such a way that

R < < S

Simply construct R in such a way that that on each interval of the partition f(s') is the minimum value attained by f on that interval. And construct S in such a way that that on each interval of the partition f(s") is the maximum value attained by f on that interval. This can be done because f is continuous. Now since there is an s in the interval such that and since f(s') < f(s) < f(s") we have the desired result. These special Riemann sums are called lower Riemann sums and upper Riemann sums respectively.

Now this means that if we have an approximating function f then the Riemann sums formed in any way we like approximate and, what is more, as the meshes of the partitions become finer and finer, the approximation becomes better and better. This is the sense in which we measure . We represent it as the limit of Riemann Sums.

Fundamental Theorem of Calculus

But of course, if that were the end of the story, things would be pretty dismal, because those limits are not easy to evaluate. Fortunately, there is more. Recall that you showed that for f a simple approximating function for Q

It is easy to see that this is also so for general approximating functions. This means that the function is an antiderivative of . Suppose that you wanted to evaluate

One way, certainly, would be to take the limit of the Riemann Sums. But suppose that you were able, algebraically, to produce a function such that

This would be different from for which, presumably, you have no algebraic formula. (If you did, you wouldn't be trying to measure it by integration). You find this function by looking at and "guessing" a function whose derivative is f.

Now and are antiderivatives of .

Question 5: Show that there is a constant C such that for all . Then show that

End of Question

It is as easy as that. The trick now, of course, is to discover a general approximating function f that really does give Riemann Sum approximations to the quantity you want to measure. That requires intelligence and craft. But once you've done that, you needn't estimate even a single Riemann sum if you can antidifferentiate f. The quantity is

for that antiderivative !

Consider Galileo's problem: the distance that a body falls from rest from time 0. Galileo observed that for a uniformly accelerating body,

That is, it is the average of initial and final velocities (or the velocity at the midpoint) multiplied by the time. In any case, we know from the Mean Value Theorem that for some

Therefore a simple approximating function for is for Galileo's constant g. Therefore we seek an antiderivative for . An obvious choice is

Thus

This was a difficult (but brilliant) calculation for Galileo. For us, it is child's play. But not all integrations are child's play as you will now see.

Arc length function defined

Let us return to the arc length function . Recall that we claimed that the lengths of polygonal paths: are all bounded above by some positive number. (In fact this implies by a basic property of the real numbers , that there is a least upper bound such that for all partitions .)

If we write then we see that for any choice

By the Mean Value Theorem, there are two points in the interval such that

Defined by

and let

Since F is continuous on , it is uniformly continuous (that "deep" topological fact again). That is, for any there is a such that, whenever for in the interval , then

Now choose a partition with mesh . Then for each interval of the partition

for in the interval . And

Then it follows from uniform continuity and the triangle inequality that

Now the terms

And so we know that for any that we choose, then when the mesh of the partition is less than some we can guarantee first that

and second, that letting

and so, by the triangle inequality again

This shows that the lengths of polygonal paths approach the integral of the speed. It guarantees that these polygonal lengths are bounded above, and that the least upper bound is:

From the last statement, and continuity of the speed, we may conclude that,

This shows that the arc length parameterization of is given by

Exploration: Arc length and integration

The exploration for this page combines integration and arc length measurements in a visual way. First, you define a curve on the interval as in the previous exercise. Here, we define a circle of radius 6:

Use the Graph/Clear buttons to draw your curve.

Now comes the interesting part. Select a Step Size to determine the length of arc the system will measure along your curve as it creates a series of points for a polygonal approximation. The arc length from each point to the next will be your step size. We chose 2p for this example, so the circle of radius 6 will be divided into 6 equal arcs of length 2p.

When you press the Calculate Length Button,

the system draws the polygonal path that passes through these points and hence "approximates" the arc length. If the path does not close, it adds a final segment to close it up. It reports the length of the entire arc: here it is 12p and the system reports 37.6991117 -- accurate to one part in 10,000,000. And it reports the length of the polygon path. Here it reports 36.0398506 (It should be 36, of course).

And of course, it draws the picture (here a regular hexagon).:

You will see, whatever your curve, that the polygonal path lengths are shorter than the arc length, but they approximate it more and more closely as you reduce the step size. For example, for step size p/4 the polygonal length is 37.6759920 and the picture is:

An interesting thing to do is experiment with other shapes, such as ellipses with nonzero eccentricity:

We chose the ellipse

which is certainly a "projection" of the original one. and we continue to use the step size 2p. But look what happened:

The system closed the polygon, but it is fairly clear that the polygonal segments have different lengths. We do not in any case get a projected hexagon. As we said above, arc length is a metric property. It requires an absolute notion of length. The projection does not preserve length. It is not an isometry. And so we should not expect the projection to preserve arc length.

Now some isometries are rotations, translations, and inversions across a line in general. Try them and you will see that the shape of the polygon -- as well as that of the curve -- is preserved.