ES.182A Topic 36 Notes Jeremy Orloff

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1 ES.82A Topic 36 Notes Jerem Orloff 36 Vector fields and line integrals in the plane 36. Vector analsis We now will begin our stud of the part of 8.2 called vector analsis. This is the stud of vector fields (defined below) and various tpes of integrals involving them. As alwas, we ll be able to think of the integrals as sums (or more technicall, as limits of sums). We will learn several big, somewhat abstract theorems: Green s, Gauss and Stokes. As usual, we ll tr to convince ou that these are not as difficult as the seem at first. In fact, we will see that the are all, in essence, fanc versions of the fundamental theorem of calculus. These theorems will offer us magical was of of computing different tpes of integrals in the plane and in space, which allow us to avoid the hard work we ve alread encountered with double integrals. There are wide applications of these theorems to science and engineering. You will use them etensivel in Vector Fields (or vector valued functions) A vector field in the plane is an assignment of a vector to ever point (, ) in the plane. Algebraicall a vector field is a function of (, ) whose value is a vector, i.e. F(, ) is a vector field if F(, ) = M(, )i + N(, )j = M, N So algebraicall, a vector field is nothing more than two ordinar functions of two variables. Once we get to vector fields in space, we will see that man phsical things are most naturall modeled as vector fields. That is, even before we know an algebraic description, we ll have a phsical description that attaches a vector to each point in space. Eample 36.. point (, ). (a.) A constant field: F(, ) = gj, i.e. the vector is the same at each (a.2) (A quick 3D eample.) Near the surface of the earth there is a (roughl) constant gravitational field F(,, z) = gk. This is another wa of saing that a mass m placed in this field eperiences a force mgk. (b.) In a radial vector field the vector at each point, points along the radial line from the origin. The vector field F(, ) =, is radial (see figures below) because, at the point (, ), the vector, can be drawn as a vector from the origin to the point. This is clearl along the radial line from the origin. (b.2) If we scale, at each point it is still radial. For eample, (Here r is our usual polar r.) V(, ) = i +, 2 j = + 2 r 2.

2 36 VETOR FIELDS AND LINE INTEGRALS IN THE PLANE 2 This vector field ehibits another important feature for us: it is not defined at the origin because the denominator becomes zero there. We will sa that V has a singularit at the origin. (c) In a tangential vector field the vector at each point is tangent to the circle through the point, centered at the origin. For eample, we have the unit tangential field F =,. r We know this is tangential because it is perpendicular to the radial vector,. (See below for a visualization.) You should be able to show that each vector is a unit vector. (d) We alread know about the gradient of a function. Now we get to call it a gradient vector field. That is, suppose f(, ) is an ordinar function of two variables, then using several of our notations F(, ) = f(, ) = f f f (, ), (, ) =, f = f i + f j. For eample, if f(, ) = 2 then f = 2, 2 = 2 i + 2 j Visualization of vector fields This can be summarized as: draw little arrows in the plane. More specificall, for a field F, at each of a number of points (, ) draw the vector F(, ) Eample Sketch the vector fields, (a.), (b.2) and (c) from the previous eample. answer: These are given below. Note that in (b.2), the denominator r 2 increases as we move awa from the origin, so the vectors get smaller. We call this a shrinking radial field. You can imagine the field as representing the velocit of water pouring outward from a source at (,).) (a.) onstant vector field (b.2) Shrinking radial field It s amusing a mabe useful to combine (b.2) and (c) in one figure: (c) Unit tangential field

3 36 VETOR FIELDS AND LINE INTEGRALS IN THE PLANE 3 A radial field (blue) and the unit tangent field (orange) A word on scales for vector fields Suppose and have dimensions of length in meters and the vector field F has dimension of force in Newtons. What scale should we use when we visualize F as arrows in the -plane? For eample, suppose F(, ) = 2, 3. We need to plot F in the same direction as 2, 3, but because the dimensions are different we can choose a different scale for the length, i.e. we don t have to plot it as 2 units in the -direction and 3 units in the -direction, we could plot it as unit in and.5 units in. The confusion is even greater if and have different dimensions and if the coordinates of F have different dimensions. We ll pla the math card and not think too hard about this. In the eamples above, we did the simplest thing and used the same scales for F as for Line integrals: introduction Line integral (also called path integrals) are computed along a curve. We will start b learning a straightforward method of computing line integrals. This won t be motivated, but ou will see that the computations are not difficult. After that, we will motivate line integrals b computing the work done b a force. This will use our familiar idea of dividing the curve into little pieces and constructing an integral as a sum from a basic formula. Finall, we will go back and uncover some ke properties of line integrals and get more practice computing them Definition and computation of line integrals along a parametrized curve We need the following ingredients: A vector field F(, ) = M(, )i + N(, )j = M, N A parametrized curve : r(t) = (t)i + (t)j = (t), (t), with t running from a to b. Note: since r =,, we have dr = d, d.

4 36 VETOR FIELDS AND LINE INTEGRALS IN THE PLANE 4 Definition. The line integral of F along is defined as F dr = M, N d, d = M d + N d. omment: The notation F dr is common in phsics and M d + N d in thermodnamics. (Though everone uses both notations.) We ll see what these notations mean in practice with some eamples. Eample Let F(, ) = 2, 2 and let be the curve r(t) = t, t 2, with t running from to. ompute the line integral I = F dr. Do this first using the notation M d + N d. Then repeat the computation using the notation F dr. answer: First we draw the curve, which is the part of the parabola = 2 running from (, ) to (, ). (i) Using the notation M d + N d. We have r =,, so = t, = t 2. In this notation F = M, N, so M = 2 and N = 2. We put everthing in terms of t: d = dt d = 2t dt M = (t 2 )(t 2 ) = t 4 N = t 2t 2 Now we can put all of these in the integral. Since t runs from to, these are our limits. I = M d + N d = t 4 dt + (t 2t 2 )2t dt = t 4 + 2t 2 4t 3 dt = 2 5. (ii) Using the notation F dr. Again, we have to put everthing in terms of t: F = M, N = t 4, t 2t 2 dr dt dr =, 2t, so d r = dt =, 2t dt dt

5 36 VETOR FIELDS AND LINE INTEGRALS IN THE PLANE 5 Thus, F dr = t 4, t 2t 2, 2t dt = t 4 + (t 2t 2 )2t dt. So the integral becomes I = F dr = This is eactl the same integral as in method (i). t 4 + (t 2t 2 )2t dt Work done b a force along a curve Having seen that line integrals are not unpleasant to compute, we will now tr to motivate our interest in doing so. We will see that the work done b a force moving a bod along a path is naturall computed as a line integral. Similar to integrals we ve seen before, the work integral will be constructed b dividing the path into little pieces. The work on each piece will come from a basic formula and the total work will be the sum over all the pieces, i.e. an integral Basic formula: work done b a constant force along a small line We ll start with the simplest situation: a constant force F pushes a bod a distance s along a straight line. Our goal is to compute the work done b the force. The figure shows the force F which pushes the bod a distance s along a line in the direction of the unit vector T. The angle between the force F and the direction T is θ. F length= s θ T=unit vector vector= r = s T We know from phsics that the work done b the force on the bod is the component of the force in the direction of motion times the distance moved. That is, work = F cos(θ) s We want to phrase this in terms of vectors. Since T = we know F T = F cos(θ). Using this in the formula for work we have work = F T s. () Equation is important and we will see it again. For now, we want to make one more substitution. We ll call the vector s T = r. This is the displacement of the bod. (Note, it is essentiall the same as our formula ds dt T = dr.) Using this, Equation becomes dt work = F r. (2) This is the basic work formula that we ll use to compute work along an entire curve

6 36 VETOR FIELDS AND LINE INTEGRALS IN THE PLANE Work done b a variable force along an entire curve Now suppose a variable force F moves a bod along a curve. Our goal is to compute the total work done b the force. The figure shows the curve broken into 5 small pieces, the jth piece has displacement r j. If the pieces are small enough, then the force on the jth piece is approimatel constant. This is shown as F j. r 5 r 4 F 3 F 4 F F 2 r 3 F 5 r r 2 Also, if the pieces are small enough, then each segment is approimatel a straight line and the force is approimatel constant. So we can appl our basic formula for work and approimate the work done b the force moving the bod along the jth piece as W j F j r j. The total work is the sum of the work over each piece. total work = W j F j r j. Now, as usual, we let the pieces get infinitesimall small, so the sum becomes an integral and the approimation becomes eact. We get: total work = F dr. The subscript indicates that it is the curve that has been split into pieces. That is, the total work is computed as a line integral of the force over the curve! 36.6 Properties of line integrals In this section we will uncover some properties of line integrals b working some eamples. Eample First look back at the value found in Eample Now, use the same vector field as in Eample But, in this case, let be the straight line from (, ) to (, ), i.e. same endpoints, but different path. ompute the line integral F dr. answer: As alwas, start b sketching the curve:

7 36 VETOR FIELDS AND LINE INTEGRALS IN THE PLANE 7 We ll use the notation M d + N d. Parametrize the curve: = t, = t, with t from to. Put everthing in terms of t: Now we put this into the integral I = M d + N d = d = dt d = dt M = 2 = t 3 N = 2 = t t 3 dt t dt = t 3 t dt = 4. This is a different value from Eample 36.3, which leads to the important principle: Important principle for line integrals. Line integrals over two different paths with the same endpoints ma be different. Eample Again, look back at the value found in Eample Now, use the same vector field and curve as Eample 36.3 ecept use the following (different) parametrization of. = sin(t), = sin 2 (t); t π/2. ompute the line integral F dr. answer: We won t sketch the curve it is identical to the one in Eample Putting everthing in terms of t we have d = cos(t) dt We put these in the integral I = I = = = π/2 π/2 d = 2 sin(t) cos(t) dt M = 2 = sin 2 (t) sin 2 (t) = sin 4 (t) N = 2 = sin(t) 2 sin 2 (t) M d + N d and compute sin 4 (t) cos(t) dt + (sin(t) 2 sin 2 (t))2 sin(t) cos(t) dt ( sin 4 (t) + 2 sin 2 (t) 4 sin 3 (t) ) cos(t) dt (Let u = sin(t), du = cos(t) dt.) = 2 5. u 4 + 2u 2 4u 3 du

8 36 VETOR FIELDS AND LINE INTEGRALS IN THE PLANE 8 This is the same value we got in Eample 36.3! In fact, the u substitution led to eactl the same integral! This leads us to the important principle: Important principle for line integrals. The parametrization of the curve doesn t affect the value of line the integral over the curve. You should note that our work with work make this reasonable, since we developed the line integral abstractl, without an reference to a parametrization List of properties of line integrals. Independent of parametrization: The value of the line integral F dr is independent of the parametrization of. 2. Reversing direction on the curve changes the sign: If is a curve, then we write for the same curve traversed in the opposite direction. In this case F dr = F dr. (See the net eample.) Eample Let be the curve from Eample Sketch and and give a parametrization of. answer: follows the parabola = 2 from (,) to (,), so the curve covers the same section of the parabola, but goes from (,) to (,), i.e. we reversed the direction of the arrow. goes from (,) to (,) goes from (,) to (,) The curve can be parametrized as r(t) = t, t 2, with t running from to. The easiest wa to reverse this is to have t run from to With this parametrization the t limits on the integral are reversed, which, we know from 8., changes the sign of the integral. If ou insist on an increasing parameter, we can parametrize b r(u) = u, ( u) 2, with u runnning from to. 3. (Intrinsic formula) We can write the line integral as F dr = F T ds where T = unit tangent vector to and ds = differential of arclength.

9 36 VETOR FIELDS AND LINE INTEGRALS IN THE PLANE 9 Reason: We know from our work on parametrized curves that dr dt = Tds. So, dr = T ds. dt (A comparison of Equations and 2 above, essentiall shows the same thing.) 4. If is a closed curve we use the notation F dr = M d + N d. The little circle on the integral sign indicates the curve is closed, i.e. starts and ends at the same point Rectangular paths Eample Evaluate I = (,) to (,) shown below. d+( + 2) d where is the rectangular path from (, ) 2 answer: The path is given in two pieces labeled and 2. This means we will have to split the integral into two pieces, i.e. I = d + ( + 2) d = d + ( + 2) d + d + ( + 2) d. 2 We ll do the integration one piece at a time. First, d + ( + 2) d. Parametrize : We ll use as the parameter: =, =, with running from to Put everthing in terms of : =, =, d = d, d =, M =, N (skip, since d = ). Put this in the integral and compute: M d + N d = M d = Net, the integral over 2. d =. Parametrize 2 : Use parameter : =, =, runs from to. Put everthing in terms of : =, =, d =, d = d, M (skip, since d = ), N = + 2 = + 2

10 36 VETOR FIELDS AND LINE INTEGRALS IN THE PLANE Put this in the integral and compute M d + N d = N d = 2 2 Adding, the pieces we have I = 2 =. + 2 d = 2. Shorthand. Because d = on and d = on 2 we can write M d + N d = + 2 M d + N d. 2 Using the shorthand will save us some writing in the future. Eample Evaluate I = d + d where is the unit circle traversed in a counterclockwise (W) direction. answer: Parametrization: = cos(t), = sin(t), t 2π. So, d = sin(t) dt, d = cos(t) dt. We get I = 2π sin t( sin t) dt + cos t(cos t) dt = 2π dt = 2π Some super-duper, reall seriousl important eamples In these eamples we are going to integrate a tangential field around a closed loop. We will see in the net few topics that these are ke computations. In the following r is the usual polar distance r 2 = Eample Let F =, r 2 r, and let be the unit circle traversed in a counterclockwise (W) direction. ompute I = F 2 dr answer: Sketch and the vector field F.

11 36 VETOR FIELDS AND LINE INTEGRALS IN THE PLANE Parametrize : = cos(t), = sin(t), t 2π. Put everthing in terms of t: (Note, on the unit circle r =.) d = cos(t) dt, d = sin(t) dt, M = r 2 = sin(t), N = r 2 = cos(t). Put this in the integral and compute: I = 2π sin(t)( sin(t)) dt + cos(t)(cos(t)) dt = 2π Eample 36.. Redo the previous using the intrinsic formula: sin 2 (t) + cos 2 (t) dt = 2π F dr = dt = 2π. F T ds. answer: T = unit tangent to the unit circle. That is, on the unit circle T = i + j. So, on the unit circle, Thus, F T = /r 2, /r 2, = r 2 =. I = F T ds = ds = 2π. The value 2π comes because we know the arclenth of the unit circle. Lesson: It pas to think geometricall. Eample 36.. Let F be the same as the previous eamples. Let 2 be the unit circle centered on (2,) traversed counterclockwise. ompute I 2 = F dr. 2 answer: Parametrize 2 : = 2 + cos(t), = sin(t), t from to 2π. Put everthing in terms of t: (Note, r 2 is not constant.) d = sin(t) dt d = cos(t) dt r 2 = = (2 + cos(t)) 2 + sin 2 (t) = cos(t) M = r 2 = sin(t) cos(t) N = r 2 = 2 + cos(t) cos(t)

12 36 VETOR FIELDS AND LINE INTEGRALS IN THE PLANE 2 Put this in the integral: I 2 = M d + N d = 2 2π sin 2 (t) + 2 cos(t) + cos 2 (t) cos(t) O! We put this into Wolfram Alpha and found I 2 =. dt = 2π + 2 cos(t) cos(t) dt Note. We should suspect that the value of is no accident. This is true and we will see this it once we learn Green s theorem. Avoiding actuall computing an integral like this should be motivation enough for us to learn Green s theorem. 8. challenge. ompute the integral for I 2. Hints: You can use the substitution u = tan(t/2) and partial fractions. It s best to use smmetr and compute 2 times the integral from to π.