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Curves & line integrals

In this part of the course we work on the following skills:

  • Work with parametric paths
  • Evaluate and work with scalar line integrals
  • Evaluate and work with vector line integrals
  • Work with potentials and conservative vector fields

See also the graded exercises and additional exercises associated to this part of the course.

Curves have played a part in earlier parts of the course and now we turn our attention to precisely what we mean by this notion. Up until now we relied more on an intuition, an idea of some type of 1D subset of higher dimensional space. We will also define how we can integrate scalar and vector fields along these curves. These types of integrals have a natural and important physical relevance. We will then study some of the properties of these integrals. To start let's recall a random selection of curves we have already seen:

  • Circle: x2+y2=4
  • Semi-circle: x2+y2=4,   x0
  • Ellipse: 14x2+19y2=4
  • Line: y=5x+2
  • Line (in 3D): x+2y+3z=0,   x=4y
  • Parabola (in 3D): y=x2,   z=x

In the above list the curves are written in a way where we are describing a set of points using certain constraint or constraints. In some cases in implicit form, in some cases in explicit form. For example, for the circle we formally mean the set {(x,y):x2+y2=4}. We have the idea that the curves should be sets which are single connected pieces and we vaguely have an idea that we need curves that are sufficiently smooth. To proceed we need a precise definition of the 1D objects we can work with. As part of the definition we force a structure which really allows us to work with these objects in a useful way.

Curves, paths & line integrals

Let α:[a,b]Rn be continuous. For convenience, in components we write α(t)=(α1(t),,αn(t)). We say that α(t) is differentiable if each component αk(t) is differentiable on [a,b] and αk(t) is continuous

Definition

We say that α(t) is piecewise differentiable if [a,b]=[a,c1][c1,c2][cl,b] and α(t) is differentiable on each of these intervals.

Definition

If α:[a,b]Rn is continuous and piecewise differentiable then we call it a path.

Note that different functions can trace out the same curve in different ways. Also note that a path has an inherent direction. We say that this is a parametric representation of a given curve. We already saw examples of paths in spiral and circular motion. A few examples of paths are as follows.

  • α(t)=(t,t), t[0,1]
  • α(t)=(cost,sint), t[0,2π]
  • α(t)=(cost,sint), t[π2,π2]
  • α(t)=(cost,sint), t[0,2π]
  • α(t)=(t,t,t), t[0,1]
  • α(t)=(cost,sint,t), t[10,10]

Observe how some of these paths represent the same curve, perhaps traversed in a different direction.

Let α(t) be a (piecewise differentiable) path on [a,b] and let f:RnRn be a continuous vector field. Recall that we consider α(t) and f(x) as n-vectors. I.e., in the case n=2, then

α(t)=(α1(t)α2(t)),f(x)=(f1(x)f2(x)).

Definition (line integral of vector field)

Let α(t) be a (piecewise differentiable) path on [a,b] and let f:RnRn be a continuous vector field. The line integral of the vector field f along the path α is defined as

fdα=abf(α(t))α(t) dt.

Sometimes the same integral is written as Cfdα to emphasize that the integral is along the curve C. Alternatively the integral is sometimes written as f1 dα1++fn dαn or f1 dx1++fn dxn. Each of these different notations are in common usage in different contexts but the underlying quantity is always the same.

Example

Consider the vector field f(x,y)=(y,x3+y) and the path α(t)=(t2,t3) for t(0,1). Evaluate fdα.

Solution

We start by calculating

α(t)=(2t3t2),f(α(t))=(t32t6+t3).

This means that f(α(t))α(t)=2t52+3t8+3t5 and so

fdα=01(2t52+3t8+3t5) dt=5942.

Now we consider the question of defining the line integral for scalar fields. Such a line integral allows us also to define the length of a curve in a meaningful way. Again let α(t), t[a,b] be a path in Rn and let f:RnR be a continuous scalar field.

Definition (line integral of scalar field)

Let α(t), t[a,b] be a (piecewise differentiable) path in Rn and let f:RnR be a continuous scalar field. The line integral of the scalar field f along the path α is defined as

f dα=abf(α(t)) α(t) dt.

Subsequently we will primarily work with the line integral of a vector field. However the analogous results hold also for this integral and the proofs are essentially the same. Namely it is linear and also respects how a path can be decomposed or joined with other paths which changing the value of the integral. Moreover, the value of the integral along a given path is independent of the choice of parametrization of the curve. In this case, even if the curve is parametrized in the opposite direction then the integral takes the same value. Consequently it makes sense to define the length of the curve as the line integral of the unit scalar field, i.e., the length of a curve parametrized by the path α is abα(t) dt.

Basic properties of the line integral

Having defined the line integral, the next step is to clarify its behaviour, in particular the following key properties.

Theorem

Linearity: Suppose f, g are vector fields and α(t) is a path. For any c,dR, then

(cf+dg)dα=cfdα+dgdα.

Joining / splitting paths: Suppose f is a vector field and that

α(t)={α1(t)t[a,c]α2(t)t[c,b]

is a path. Then

fdα=fdα1+fdα2.

Alternatively, if we write C, C1, C2 for the corresponding curves, then

Cfdα=C1fdα+C2fdα.

As already mentioned, for a given curve there are many different choices of parametrization. For example, consider the curve C={(x,y):x2+y2=1,y0}. This is a semi-circle and two possible parametrizations are α(t)=(t,1t2), t[1,1] and β(t)=(cost,sint), t[0,π]. These are just two possibilities among many possible choices. For a given curve, to what extent does the line integral depend on the choice of parametrization?

Definition (equivalent paths)

We say that two paths α(t) and β(t) are equivalent if there exists a differentiable function u:[c,d][a,b] such that α(u(t))=β(t).

Furthermore, we say that α(t) and β(t) are

  • in the same direction if u(c)=a and u(d)=b,

  • in the opposite direction if u(c)=b and u(d)=a.

With this terminology we can precisely describe the dependence of the integral on the choice of parametrization.

Theorem

Let f be a continuous vector field and let α, β be equivalent paths. Then

fdα={fdβif the paths in the same direction,fdβif the paths in the opposite direction.

Proof

Suppose that the paths are continuously differentiable path, decomposing if required. Since α(u(t))=β(t) the chain rule implies that β(t)=α(u(t)) u(t). In particular

fdβ=cdf(β(t))β(t) dt=cdf(α(u(t)))α(u(t)) u(t) dt.

Changing variables, adding a minus sign if path is opposite direction because we need to swap the limits of integration, completes the proof.

Gradients & work

Let h(x,y) be a scalar field in R2 and recall that the gradient h(x,y) is a vector field. Let α(t), t[0,1] be a path. Now let g(t)=h(α(t)), consider the derivative g(t)=h(α(t))α(t) and evaluate the line integral

hdα=01h(α(t))α(t) dt=01g(t) dt=g(1)g(0)=h(α(1))h(α(0)).

This equality has the following intuitive interpretation if we suppose for a moment that h denotes altitude. In this case the line integral is the sum of all the infinitesimal altitude changes and equals the total change in altitude.

As a first example of work in physics let's consider gravity. The gravitational field on earth is f(x,y,z)=(00mg). If we move a particle from a=(a1,a2,a3) to b=(b1,b2,b3) along the path α(t), t[0,1] then the work done is defined as fdα. We calculate that

fdα=01f(α(t))α(t) dt=01mg α3(t) dt=mg [α3(t)]01=mg(b3a3).

This coincides we what we know, work done depends only on the change in height.

As a second example of work in physics let's consider a particle moving in a force field. Let f be the force field and let x(t) be the position at time t of a particle moving in the field. Let v(t)=x(t) be the velocity at time t of the particle and define kinetic energy as m2v(t)2. According to Newton's law f(x(t))=mx(t)=mv(t) and so the work done is

fdx=01f(x(t))v(t) dt=01mv(t)v(t) dt=01ddt(m2v(t)2)=(m2v(1)2m2v(0)2)

In this case we see, as expected, the work done on the particle moving in the force field is equal to the change in kinetic energy.

The second fundamental theorem

Recall that, if φ:RR is differentiable then abφ(t) dt=φ(b)φ(a). This is called the second fundamental theorem of calculus and is one of the ways in which we see that differentiation and integration are opposites. The analog for line integrals is the following.

Theorem (second fundamental theorem for line integrals)

Suppose that φ is a continuously differentiable scalar field on SRn and suppose that α(t), t[a,b] is a path in S. Let a=α(a), b=α(b). Then

φdα=φ(b)φ(a).

Proof

Suppose that α(t) is differentiable. By the chain rule ddtφ(α(t))=φ(α(t))α(t). Consequently

φdα=01φ(α(t))α(t) dt=01ddtφ(α(t)) dt.

By the 2nd fundamental theorem in R we know that 01ddtφ(α(t)) dt=φ(α(b))φ(α(a)).

Our earth has mass M with centre at (0,0,0). Suppose that there is a small particle close to earth which has mass m. The force field of gravitation and potential energy are, respectively,

f(x)=GmMx3x,φ(x)=GmMx.

We can calculate φ(x) and see that it is equal to f(x).

The first fundamental theorem

First we need to consider a basic topological property of sets. In particular we want to avoid the possibility of the set being several disconnected pieces, in other words we want to guarantee that we can get from one point to another in the set in a way without every leaving the set (see figure).

Definition

The set SRn is said to be connected if, for every pair of points a,bS, there exists a path α(t),t[a,b] such that