apuntes zurita

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Propaedeutic Mathematics: Vector Analysis

Instituto N acional de AstrofisicaOptica y Electr6nica

Jorge R. Zurita SanchezMarch 24, 2011


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Contents

Preface

1 Vector Analysis1.1 Directional Derivat ive and Gradient1.2 Flux and Divergence . . . . .1.3 Work and Curl . . . . . . . .1.4 Conservative Forces (Fields) .Bibliography

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Preface

You have already taken vector analysis during the first years of college. Instead of reviewing thedefinitions of vector operations, the course focuses on the geometrical and physical understandingof the diverse vector operations. It is strongly recommended that you solve th e exercises. Someexercises of these notes were adapted from exercises belonging to th e books written by Boas [1],and Arfken and Weber [2].

Boas says that "To use mathematics effectively in applications. you need not just knowledge,bu t skill. Skill can be only obtained only through practice. You can obtain a certain superficialknowledge of mathematics by listening to lectures, bu t cannot obtain skill this way. How manystudents have I heard say: It looks to easy when you do it or I understand it but I cannot solveproblems! Such statements show lack of practice and consequent lack of skill. .. . Th e only way tolearn to solve problems is to solve problems" [1]. In this vein, if you need to solve more problemsthan the ones that are contained in these notes, there are plenty of calculus books in the librarywhere you can find a lot of them.

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CHAPTER 1. VECTOR ANALYSIS INAOE Jorge R. Zurita Sanchez

Also, Eq. (1.1) can be written as r = r 0 +sn8 . Recalling your first course of calculus, th e derivative(one dimension) is a measure how steep is the function f (x) at an arbitrary point X 0 . From this,we can extend th e analysis to the 2D case and define the derivative,

df(x, y)ds (1.2)

At first instance you notice that the change of the function j(x, y) depends on the direction inwhich you decide to step forward. For this reason, the Eq. (1.2) is called directional derivative.We require to find dj(x,y)jds explicitly. Since x andy depend on s [see Eq. (1.1)], we apply thechain rule to Eq. (1.2). This renders

df(x, y)ds

aj(x, y) dx aj(x, y) dyax ds + ay ds '

aj(x ,y) aj(x ,y)ax Ux + ay Uy (1.3)Therefore you need the partial derivatives and the components of ns in the x and y directions toobtain the directional derivative. Back to your friend's question, you will tell him that he is goingto ascend or descend = dj(xa, Yo)

ds 8 ' (1.4)that is, the directional derivative times the magnitude of the the displacement. Th e directionalderivative in the x-direction (ux = 1 and uy = 0) is a j jax (as you might have expected).

Th e directional derivative usually is expressed differently from Eq. (1.3) asdf(x, y) _ " f( ) .ds - v x, y n 8 , (1.5)

where the V-operator is defined as(1.6)

Notice that the result of the V-operator acting on the scalar function f(x, y) is a vector denominatedgradient of f(x, y) . Next we will discuss the gradient.

The height of a mountain is still represesnted by f(x, y). Now let us consider thatf(x, y) = c, (1.7)

where c is an arbitrary constant. Equation (1. 7) defines th e set of points (x, y) of the the mountainwith height c and represents a curve in the xy plane which is called an isoheight curve. You knowthat the topography of a mountain in a map is described by plotting several isoheight curves [seeth e isoheight curves for our montain in Fig. 1.1 (b)]. In this way, we can represent 3D informationon a 2D plot.

Let the points r 0 and r 1 be located at the same isoheigth curve. Both are close to each other.When you depart from r a displacement s to reach r1 th e change vanishes. The reason is thatyou move into the same isoheight curve. As the point r 1 approaches to r 0 . along the isoheightcurve, the displacement vectors becomes a tangential vector St [see Fig. 1.2(a)] with respect to this3


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(a)/f(x,y)=c

(b)

Figure 1.2: (a) Directional derivative along the isoheight curve. (b) Th e gradient \1f is perpendicular to the isoheight curve and points toward the direction in which the function increases.curve. Consequently, the directional derivative must vanish along the direction of this tangentialvector, that is

dj(xo, Yo) _ " f( ) . _ Od - v Xo, Yo llt- ,St (1.8)where ll t is the unit tangential vector. When the dot product of two vectors vanishes, the vectorsare orthogonal. Therefore \1 f(x 0 , Yo) is a normal vector with respect to the isoheight curve.

What direction does the function at an arbitrary point (xo,Yo) increase th e most, (or equivalently the function's steepness is the largest)? To answer this, you need to get back to Eq. (1.5).From this equation,

d j ( ~ ; Yo) = \1 f(x 0 , Yo) l l 8 = IV f( x 0 , Yo) I ns ICOS a= IV f(xo, Yo) ICOS a, (1.9)where a is the angle between \1 f( x 0 , Yo) and n 5 . Now you can notice that the directional derivativeat an arbitrary point (x 0 , Yo) is the largest when a= 0. Therefore, also, \1 f(x 0 , Yo) indicates thedirection in which the function increases the most.

So far, we have dealt with functions f : IR 2 -+ R You can extend the analysis for the casef : JR3 -+ R For example, f(x, y, z) could represent the temperature inside a chamber. Similarly,the directional derivative indicates the steepness of a function f(x, y, z), bu t in th e direction of aunit vector in the 3D space [ns = uxnx + uyny + U z l l z , + + = 1]. Also, the directionalderivative is expressed as in Eq. (1.5), bu t the \!-operator is

a a a\1= -a nx+-a ny+-a lz .X y Z (1.10)

I f f(x, y, z) is the temperature in a chamber, thenf(x, y, z) = c, (1.11)

are points (x, y, z) in th e 3D space such that the temperature is c and represents and isothermalsurface. The vector \1 f(x, y, z) is a normal vector with respect to the isothermal surface.

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EXERCISES1. Find a normal vector to the surface z+9x 2 +2 y2 = 0 at point (1/3, 1, -3). Find the equation

of the tangent plane and the normal line at this point.2. Let the temperature of a plate be T(x,y) = axy - bx where a= 10 -4 C/cm2 and b =

10 -2 C/cm. (a) Draw the isothermal curves forT= 0,1,2 C. (b) Heat flows in thedirection - \ IT; explain the negative sign and add arrows of heat flux on the isothermal plotdepicted previously. (c) Find the direction in which the temperature changes most rapidlyfrom (1m, 1m). (d) Find the directional derivative ofT at (1m, 1m) in the direction ofvector v'Snx + 2ny.

3. Show that "'!( e ) - aj(r, e) 8f(r, e)v T, - Dr l lr + T ae ne,where (T, e) are the polar coordinates.

1.2 . Flux and DivergenceThe flux S(r) tell us how much of a particular quantity per unit area per unit time passes througha point. At this moment, we restrict ourselves to the case in which the flux is constant, thatis, S(r) = Sanx. As an example, let S be a flux of fishes. You have rectangular net with .ctreaA= ax b to catch them as illustrated in Fig. 1.3. You can rotate the net an angle e [see Fig. 1.3].

~ ~ t> - < ; : Z Y A- ~ j ~ t>t>


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We can establish a vector notation where the area of the net is represented by a normalvector to the surface with magnitude A, namely, A = A(cosBnx +sinBny) Therefore the amountof fishes per unit of time can be expressed as

=SA. (1.13)You have seen that it is very easy to obtain the amount of fishes per unit of time crossing

the net when the flux of fishes is constant and the net is planar. How can you ubtain the amountof fishes per unit time passing through an arbitrarily folded net and inhomogeneous flux S(r)-inhomogeneous stands for a spatial dependent flux? Imagine that you divide the net into verysmall patches dA, so the flux can be considered uniform inside the patch. Then you can useEq. (1.13), to calculate the amount of fishes passing through a small patch as

d = S(r) dA. (1.14)This can be seen in Fig. 1.4 Finally the total amount of fishes per unit of time passing through

Figure 1.4: An arbitrary folded net is divided into small patches.this net is obtained by adding all contributions d of each one of the patches. Consequently,

= Is S(r) dA. (1.15)Equation (1.15) is surface integral where S denotes the shape of the net. Therefore, a surfaceintegral must be evaluated to obtain the amount of fishes that undergoes an arbitrarily folded net.

Let us calculate the amount of fishes per unit of time ,0. coming ou t from a small cubewith volume ,0.x ,0.y ,0.z [see Fig. 1.5]. Th e partial contribution of each of th e cube's faces is:

Face 1 Sx(x + ,0.x, y, z),0.y,0.z Face 2 -Sx(x, y, z),0.y,0.zFace 3 Sy(x, y + ,0.y, z),0.x,0.z Face 4 -Sy(x, y, z),0.x,0.zFace 5 Sz (x, y, z + ,0.z ),0.x,0.y Face 6 -Sz(x, y, z),0.x,0.y

Thus, the total amount of fishes per unit of time coming out of the cube isSx(x+,0.x,y,z)-Sx(x,y,z),0. A ,0. Sy(x,y+,0.y,z)-Sy(x,y,z)A A A,0.x xuy z + ,0.y uxuyuzSz(x,y,z+,0.z)-Sz(x,y,z)A A A+ ,0.z uxuyuz. (1.16)

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Then, in the limit of an infinitesimal cube with volume dV = xdydz, the total total amount offishes per unit of time that leave this infinitesimal cube isdel:>= (8Sx(x,y,z) 8Sy(x,y,z) fJSz(x,y,z)) dV& + + fu . (1.17)

Notice that you can rewritten Eq. (1.17) asdel:>= v S(r)dV. (1.18)

V S(r) is the called the divergence of S(r). Thus, the amount of fishes per unit of time that leavethis infinitesimal cube is proportional to the divergence of the flux S(r).

Suppose that a surface S encloses a volume V, thus S = fJV [fJV denotes the boundary ofthe enclosed volume]. You would like to calculate the amount of fishes per unit of time c[> passingthrough this closed surface. As you already know, c[> is obtained from

c!:> = 1 S(r) dA,fav (1.19)where th e circle of the integral denotes a close surface. However, you are going to calculate c[> byusing another method. You divide the volume that encloses S into very small cubes with volumedV. J\ext you calculate the amount of fishes per unit time del:> crossing each cube. You alreadyknow that del:> for a small cube is given by Eq. (1.18). Finally, the total amount of fishes per unitof time undergoing th e closed surface S is obtained by adding the partial contributions del:> of eachof the cube, thus

c[> = 1 S (r) dA = j V S (r )dV.fav v (1.20)Why can c!:> be obtained from a volume integral? Consider the face shared by two adjacent cubes.Let dc!:>1 be the amount of fishes per unit time passing through the face of cube 1. Relative tocube 2, the amount of fishes per unit time passing through this face must be dc!:>2 = -dc!:> 1 [seeFig. 1.6]. Consequently, only the faces of the cubes that do not share a face with another cube willcontribute as we sum all the cubes occupying the volume. Precisely, these contributing faces mustlye only at the boundary of the volume, namely, the surface. This is how Eq. (1.20) works.

Sx(x+L'.x,y,z)

(x,y,z)Sx(x,y,z)

Figure 1.5: The amount of fishes per unit time undergoing a small cube.

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I Sxd

Figure 1.6: d

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