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Textbook by: Lawrence Baggett. E-mail the author

Integration with Respect to Arc Length

Module by: Lawrence Baggett. E-mail the author

Summary: We introduce next what would appear to be the best parameterization of a piecewise smooth curve, i.e., a parameterization by arc length. We will then use this parameterization to define the integral of a function whose domain is the curve.

We introduce next what would appear to be the best parameterization of a piecewise smooth curve, i.e., a parameterization by arc length. We will then use this parameterization to define the integral of a function whose domain is the curve.

Theorem 1

Let CC be a piecewise smooth curve of finite length LL joining two distinct points z1z1 to z2.z2. Then there exists a parameterization γ:[0,L]Cγ:[0,L]C for which the arc length of the curve joining γ(t)γ(t) to γ(u)γ(u) is equal to |u-t||u-t| for all t<u[0,L].t<u[0,L].

Proof

Let φ:[a,b]Cφ:[a,b]C be a parameterization of C.C. Define a function F:[a,b][0,L]F:[a,b][0,L] by

F ( t ) = a t | φ ' ( s ) | d s . F ( t ) = a t | φ ' ( s ) | d s .
(1)

In other words, F(t)F(t) is the length of the portion of CC that joins the points z1=φ(a)z1=φ(a) and φ(t).φ(t). By the Fundamental Theorem of Calculus, we know that the function FF is continuous on the entire interval [a,b][a,b] and is continuously differentiable on every subinterval (ti-1,ti)(ti-1,ti) of the partition PP determined by the piecewise smooth parameterization φ.φ. Moreover, F'(t)=|φ'(t)|>0F'(t)=|φ'(t)|>0 for all t(ti-1,ti),t(ti-1,ti), implying that FF is strictly increasing on these subintervals. Therefore, if we write si=F(ti),si=F(ti), then the sisi's form a partition of the interval [0,L],[0,L], and the function F:(ti-1,ti)(si-1,si)F:(ti-1,ti)(si-1,si) is invertible, and its inverse F-1F-1 is continuously differentiable. It follows then that γ=φF-1:[0,L]Cγ=φF-1:[0,L]C is a parameterization of C.C. The arc length between the points γ(t)γ(t) and γ(u)γ(u) is the arc length between φ(F-1(t))φ(F-1(t)) and φ(F-1(u)),φ(F-1(u)), and this is given by the formula

F - 1 ( t ) F - 1 ( u ) | φ ' ( s ) | d s = a F - 1 ( u ) | φ ' ( s ) | d s - a F - 1 ( t ) | φ ' ( s ) | d s = F ( F - 1 ( u ) ) - F ( F - 1 ( t ) ) = u - t , F - 1 ( t ) F - 1 ( u ) | φ ' ( s ) | d s = a F - 1 ( u ) | φ ' ( s ) | d s - a F - 1 ( t ) | φ ' ( s ) | d s = F ( F - 1 ( u ) ) - F ( F - 1 ( t ) ) = u - t ,
(2)

which completes the proof.

Corollary 1

If γγ is the parameterization by arc length of the preceding theorem, then, for all t(si-1,si),t(si-1,si), we have |γ'(s)|=1.|γ'(s)|=1.

Proof

We just compute

| γ ' ( s ) | = | ( φ F - 1 ) ' ( s ) | = | φ ' ( F - 1 ( s ) ) ( F - 1 ) ' ( s ) | = | φ ' ( F - 1 ( s ) | | 1 F ' ( F - 1 ( s ) ) | = | φ ' ( f - 1 ( s ) ) | 1 | φ ' ( f - 1 ( s ) ) | = 1 , | γ ' ( s ) | = | ( φ F - 1 ) ' ( s ) | = | φ ' ( F - 1 ( s ) ) ( F - 1 ) ' ( s ) | = | φ ' ( F - 1 ( s ) | | 1 F ' ( F - 1 ( s ) ) | = | φ ' ( f - 1 ( s ) ) | 1 | φ ' ( f - 1 ( s ) ) | = 1 ,
(3)

as desired.

We are now ready to make the first of our three definitions of integral over a curve. This first one is pretty easy.

Suppose CC is a piecewise smooth curve joining z1z1 to z2z2 of finite length L,L, parameterized by arc length. Recall that this means that there is a 1-1 function γγ from the interval [0,L][0,L] onto CC that satisfies the condidition that the arc length betweenthe two points γ(t)γ(t) and γ(s)γ(s) is exactly the distance between the points tt and s.s. We can just identify the curve CC with the interval [0,L],[0,L], and relative distances will correspond perfectly. A partition of the curve CC will correspond naturally to a partition of the interval [0,L].[0,L]. A step function on the dcurve will correspond in an obvious way to a step function on the interval [0,L],[0,L], and the formula for the integral of a step function on the curve is analogous to what it is on the interval. Here are the formal definitions:

Definition 1:

Let CC be a piecewise smooth curve of finite length LL joining distinct points, and let γ:[0,L]Cγ:[0,L]C be a parameterization of CC by arc length. By a partition of CC we mean a set {z0,z1,...,zn}{z0,z1,...,zn} of points on CC such that zj=γ(tj)zj=γ(tj) for all j,j, where the points {t0<t1<...<tn}{t0<t1<...<tn} form a partition of the interval [0,L].[0,L]. The portions of the curve between the points zj-1zj-1 and zj,zj, i.e., the set γ(tj-1,tj),γ(tj-1,tj), are called the elements of the partition.

A step fucntion on CC is a real-valued function hh on CC for which there exists a partition {z0,z1,...,zn}{z0,z1,...,zn} of CC such that h(z)h(z) is a constant ajaj on the portion of the curve between zj-1zj-1 and zj.zj.

Before defining the integral of a step function on a curve, we need to establish the usual consistency result, encountered in the previous cases of integration on intervals and integration over geometric sets, the proof of which this time we put in an exercise.

Exercise 1

Suppose hh is a function on a piecewise smooth curve of finite length L,L, and assume that there exist two partitions {z0,z1,...,zn}{z0,z1,...,zn} and {w0,w1,...,wm}{w0,w1,...,wm} of CC such that h(z)h(z) is a constant akak on the portion of the curve between zk-1zk-1 and zk,zk, and h(z)h(z) is a constant bjbj on the portion of the curve between wj-1wj-1 and wj.wj. Show that

k = 1 n a k L ( z k - 1 , z k ) = j = 1 m b j L ( w j - 1 , w j ) . k = 1 n a k L ( z k - 1 , z k ) = j = 1 m b j L ( w j - 1 , w j ) .
(4)

HINT: Make use of the fact that hγhγ is a step function on the interval [0,L].[0,L].

Now we can make the definition of the integral of a step function on a curve.

Definition 2:

Let hh be a step function on a piecewise smooth curve CC of finite length L.L. The integral, with respect to arc length of hh over CC is denoted by Ch(s)ds,Ch(s)ds, and is defined by

C h ( s ) d s = j = 1 n a j L ( z j - 1 , z j ) , C h ( s ) d s = j = 1 n a j L ( z j - 1 , z j ) ,
(5)

where {z0,z1,...,zn}{z0,z1,...,zn} is a partition of CC for which h(z)h(z) is the constant ajaj on the portion of CC between zj-1zj-1 and zj.zj.

Of course, integrable functions on CC with respect to arc length will be defined to be functions that are uniform limits of step functions. Again, there is the consistency issue in the definition of the integral of an integrable function.

Exercise 2

1. Suppose {hn}{hn} is a sequence of step functions on a piecewise smooth curve CC of finite length, and assume that the sequence {hn}{hn} converges uniformly to a function f.f. Prove that the sequence {Chn(s)ds}{Chn(s)ds} is a convergent sequence of real numbers.
2. Suppose {hn}{hn} and {kn}{kn} are two sequences of step functions on a piecewise smooth curve CC of finite length l,l, and that both sequences converge uniformly to the same function f.f. Prove that
limChn(s)ds=limCkn(s)ds.limChn(s)ds=limCkn(s)ds.
(6)
Definition 3:

Let CC be a piecewise smooth curve of finite length L.L. A function ff with domain CC is called integrable with respect to arc length on CC if it is the uniform limit of step functions on C.C.

The integral with respect to arc length of an integrable function ff on CC is again denoted by Cf(s)ds,Cf(s)ds, and is defined by

C f ( s ) d s = lim C h n ( s ) d s , C f ( s ) d s = lim C h n ( s ) d s ,
(7)

where {hn}{hn} is a sequence of step functions that converges uniformly to ff on C.C.

In a sense, we are simply identifying the curve CC with the interval [0,L][0,L] by means of the 1-1 parameterizing function γ.γ. The next theorem makes this quite plain.

Theorem 2

Let CC be a piecewise smooth curve of finite length L,L, and let γγ be a parameterization of CC by arc length. If ff is an integrable function on C,C, then

C f ( s ) d s = 0 L f ( γ ( t ) ) d t . C f ( s ) d s = 0 L f ( γ ( t ) ) d t .
(8)

Proof

First, if hh is a step function on C,C, let {zj}{zj} be a partition of CC for which h(z)h(z) is a constant ajaj on the portion of the curve between zj-1zj-1 and zj.zj. Let {tj}{tj} be the partition of [0,L][0,L] for which zj=γ(tj)zj=γ(tj) for every j.j. Note that hγhγ is a step function on [0,L],[0,L], and that hγ(t)=ajhγ(t)=aj for all t(tj-1,tj).t(tj-1,tj). Then,

C h ( s ) d s = j = 1 N a j L ( z j - 1 , z j ) = j = 1 n a j L ( γ ( t j - 1 ) , γ ( t j ) ) = j = 1 n a j ( t j - t j - 1 ) = 0 L h γ ( t ) d t , C h ( s ) d s = j = 1 N a j L ( z j - 1 , z j ) = j = 1 n a j L ( γ ( t j - 1 ) , γ ( t j ) ) = j = 1 n a j ( t j - t j - 1 ) = 0 L h γ ( t ) d t ,
(9)

which proves the theorem for step functions.

Finally, if f=limhnf=limhn is an integrable function on C,C, then the sequence {hnγ}{hnγ} converges uniformly to fγfγ on [0,L],[0,L], and so

C f ( s ) d s = lim C h n ( s ) d s = lim 0 L h n ( γ ( t ) ) d t = 0 l f ( γ ( t ) ) d t , C f ( s ) d s = lim C h n ( s ) d s = lim 0 L h n ( γ ( t ) ) d t = 0 l f ( γ ( t ) ) d t ,
(10)

where the final equality follows from (Reference). Hence, Theorem 2 is proved.

Although the basic definitions of integrable and integral, with respect to arc length, are made in terms of the particular parameterization γγ of the curve, for computational purposes we need to know how to evaluate these integrals using different parameterizations. Here is the result:

Theorem 3

Let CC be a piecewise smooth curve of finite length L,L, and let φ:[a,b]Cφ:[a,b]C be a parameterization of C.C. If ff is an integrable function on C.C. Then

C f ( s ) d s = a b f ( φ ( t ) ) | φ ' ( t ) | d t . C f ( s ) d s = a b f ( φ ( t ) ) | φ ' ( t ) | d t .
(11)

Proof

Write γ:[0,L]Cγ:[0,L]C for a parameterization of CC by arc length. As in the proof to Theorem 2, we write g:[a,b][0,L]g:[a,b][0,L] for γ-1φ.γ-1φ. Just as in that proof, we know that gg is a piecewise smooth function on the interval [a,b].[a,b]. Hence, recalling that |γ'(t)|=1|γ'(t)|=1 and g'(t)>0g'(t)>0 for all but a finite number of points, the following calculation is justified:

C f ( s ) d s = 0 L f ( γ ( t ) ) d t = 0 L f ( γ ( t ) ) | γ ' ( t ) | d t = a b f ( γ ( g ( u ) ) ) | γ ' ( g ( u ) ) | g ' ( u ) d u = a b f ( γ ( g ( u ) ) ) | γ ' ( g ( u ) ) | | g ' ( u ) | d u = a b f ( φ ( u ) ) | γ ' ( g ( u ) ) g ' ( u ) | d u = a b f ( φ ( u ) ) | ( ' g a m m a g ) ' ( u ) | d u = a b f ( φ ( u ) ) | φ ' ( u ) | d u , C f ( s ) d s = 0 L f ( γ ( t ) ) d t = 0 L f ( γ ( t ) ) | γ ' ( t ) | d t = a b f ( γ ( g ( u ) ) ) | γ ' ( g ( u ) ) | g ' ( u ) d u = a b f ( γ ( g ( u ) ) ) | γ ' ( g ( u ) ) | | g ' ( u ) | d u = a b f ( φ ( u ) ) | γ ' ( g ( u ) ) g ' ( u ) | d u = a b f ( φ ( u ) ) | ( ' g a m m a g ) ' ( u ) | d u = a b f ( φ ( u ) ) | φ ' ( u ) | d u ,
(12)

as desired.

Exercise 3

Let CC be the straight line joining the points (0,1)(0,1) and (1,2).(1,2).

1. Find the arc length parameterization γ:[0,2]C.γ:[0,2]C.
2. Let ff be the function on this curve given by f(x,y)=x2y.f(x,y)=x2y. Compute Cf(s)ds.Cf(s)ds.
3. Let ff be the function on this curve that is defined by f(x,y)f(x,y) is the distance from (x,y)(x,y) to the point (0,3).(0,3). Compute cf(s)ds.cf(s)ds.

The final theorem of this section sums up the properties of integrals with respect to arc length. There are no surprises here.

Theorem 4

Let CC be a piecewise smooth curve of finite length L,L, and write I(C)I(C) for the set of all functions that are integrable with respect to arc length on C.C. Then:

1.  I(C)I(C) is a vector space ovr the real numbers, and
C(af(s)+bg(s))ds=aCf(s)ds+bCg(s)dsC(af(s)+bg(s))ds=aCf(s)ds+bCg(s)ds
(13)
for all f,gI(C)f,gI(C) and all a,bR.a,bR.
2.  (Positivity) If f(z)0f(z)0 for all zC,zC, then Cf(s)ds0.Cf(s)ds0.
3. If fI(C),fI(C), then so is |f|,|f|, and |Cf(s)ds|C|f(s)|ds.|Cf(s)ds|C|f(s)|ds.
4. If ff is the uniform limit of functions fn,fn, each of which is in I(C),I(C), then fI(C)fI(C) and Cf(s)ds=limCfn(s)ds.Cf(s)ds=limCfn(s)ds.
5. Let {un}{un} be a sequence of functions in I(C),I(C), and suppose that for each nn there is a number mn,mn, for which |un(z)|mn|un(z)|mn for all zC,zC, and such that the infinite series mnmn converges. Then the infinite series unun converges uniformly to an integrable function, and Cun(s)ds=Cun(s)ds.Cun(s)ds=Cun(s)ds.

Exercise 4

1. Prove the preceding theorem. Everything is easy if we compose all functions on CC with the parameterization γ,γ, obtaining functions on [0,L],[0,L], and then use (Reference).
2. Suppose CC is a piecewise smooth curve of finite length joining z1z1 and z2.z2. Show that the integral with respect to arc length of a function ff over CC is the same whether we think of CC as being a curve from z1z1 to z2z2 or, the other way around, a curve from z2z2 to z1.z1.

REMARK Because of the result in part (b) of the preceding exercise, we speak of “integrating over CC” when we are integrating with respect to arc length. We do not speak of “integrating from z1z1 to z2,z2,” since the direction doesn't matter. This is in marked contrast to the next two kinds of integrals over curves that we will discuss.

here is one final bit of notation. Often, the curves of interest to us are graphs of real-valued functions. If g:[a,b]Rg:[a,b]R is a piecewise smooth function, then its graph CC is a piecewise smooth curve, and we write graph(g)f(s)dsgraph(g)f(s)ds for the integral with respect to arc length of ff over C=graph(g).C=graph(g).

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