prev

next

out of 30

Published on

18-Dec-2015View

14Download

3

Embed Size (px)

DESCRIPTION

My notes on the MIT OCW Calculus Revisited: Multivariate Calculus course

Transcript

<ul><li><p>Calculus Revisited: Multivariable CalculusAs given by Herbert Gross, MIT</p><p>Notes by Aleksandar Petrov</p><p>March 2015</p></li><li><p>Contents</p><p>1 Vector Arithmetic 11.1 The Game of Mathematics . . . . . . . . . . . . . . . . . . . . . 11.2 The Structure of Vector Arithmetic . . . . . . . . . . . . . . . . . . 21.3 Applications to 3-Dimensional Space . . . . . . . . . . . . . . . . . 21.4 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.5 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.6 Equations of Lines and Planes . . . . . . . . . . . . . . . . . . . . . 6</p><p>2 Vector Calculus 82.1 Vector Functions of a Scalar Variable . . . . . . . . . . . . . . . . . 82.2 Tangential and Normal Vectors . . . . . . . . . . . . . . . . . . . . 92.3 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.4 Vectors in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . 11</p><p>3 Partial Derivatives 133.1 n-Dimensional Vector Spaces . . . . . . . . . . . . . . . . . . . . . 133.2 An Introduction to Partial Derivatives . . . . . . . . . . . . . . . . 143.3 Differentiability and the Gradient . . . . . . . . . . . . . . . . . . . 163.4 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.5 Exact Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . 19</p><p>4 Matrix Algebra 214.1 Linearity Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.2 Introduction to Matrix Algebra . . . . . . . . . . . . . . . . . . . . 224.3 Inverting a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.4 Maxima and Minima in Several Variables . . . . . . . . . . . . . . 25</p><p>5 Multiple Integration 265.1 The Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . . 265.2 Multiple Integration and the Jacobian . . . . . . . . . . . . . . . . 265.3 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275.4 Greens Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27</p><p>i</p></li><li><p>Chapter 1</p><p>Vector Arithmetic</p><p>1.1 The Game of Mathematics</p><p>We can define a game to be any system consisting of definitions, rules, andobjectives, where the objectives are carried out as inescapable consequencesof the definitions and the rules by means of strategy. Not all definitions can beclearly set. For example, no definition of number can be given. That meansthat some of the definitions are subjective. However, using only specificobjective facts about these concepts (the rules) allows us to draw inescapableconclusions.</p><p>It is important to distinguish between truth and validity. Truth is sub-jective and can change with the time. Validity means that the conclusion isinescapable result of the definitions and the rules. Simply put, an argumentis valid when it follows logically from the definitions and the rules. If ourpremises are true and our argument is valid, then the conclusions are alsogoing to be true. However, our conclusion may also be true if the premisesare not true or the arguments are not valid. Mathematics deal with the ar-gumentation part of this problem. It draws valid conclusions. However, it isnot necessary that they are true. That will be the case only if the premisesare true. To conclude, one can be sure that a conclusion is true only if theassumptions (definitions and rules) are true and the argumentation is valid.</p><p>That allows us to put a line between pure and applied mathematics.Applied mathematics deal with problems that have definitions and rules thatwe believe to be reality. Pure mathematics focus on the consistency of therules and the validity of the argument. However these rules need not be true.An example could be Lobachevskys geometry which was pure math as itdid not correspond to any physical truths known back then until Einsteinnoticed that it served as a realistic model for his theory of relativity.</p><p>1</p></li><li><p>CHAPTER 1. VECTOR ARITHMETIC 2</p><p>To show that an argument is invalid, all we need do is give one set ofconditions in which the assumptions are obeyed but the conclusion is false.On the other hand, proving that something is true is rather difficult: one hasto find a way to show that the statement is always true.</p><p>1.2 The Structure of Vector Arithmetic</p><p>An important fact to keep in mind is that the operations in vector arithmeticare result of definitions, not nature. We define a vector to be an object thathas a magnitude and direction. A more modern approach will define a vectoras an ordered sequence of numbers. A vector has magnitude (length), direc-tion (orientation) and sense (each direction has two possible senses). Themathematical concept of a vector is geometrically represented by an arrow.Furthermore, as a vector is defined solely by its magnitude, direction andsense, two vectors can be equal even is they do not have the same startingand ending point. The equality of vectors, the zero vector, the summationand subtraction of vectors are all operations that are defined in such a waythat they are easy and useful. Beware in mind that the mathematical struc-ture of vector arithmetic is different from the structure of scalar arithemtic.For example, we talk about summation of vectors but this operation is notthe same operation as summation of scalars. Furthermore, multiplication ofvectors is ambiguous while multiplication of scalars is clearly defined.</p><p>Some of the properties that vectors have are:</p><p> ~a+~b = ~b+ ~a ~a+ (~b+ ~c) = (~a+~b) + ~c ~a+~0 = ~a</p><p> ~a+~b = ~c ~a = ~c~b c(~a+~b) = c~a+ c~b (c+ d)~a = c~a+ d~a</p><p>Keep in mind that these properties were defined. They are not intrinsicto all mathematical structures. For example if you subtract set B from AByou will not get set A (apart from the case when A and B have no commonelements).</p><p>1.3 Applications to 3-Dimensional Space</p><p>When talking about three dimensional vectors we do actually mean vectorswith three coordinates. Of course, a vector is geometrically represented byan arrow and an arrow is a two dimensional element. One extremely usefulproperty (or if you wish definition) of vectors is that the components of a</p></li><li><p>CHAPTER 1. VECTOR ARITHMETIC 3</p><p>vector connecting the origin of a Cartesian coordinate system with a pointare the same as the coordinates of the point. An important note about themathematical structure of vectors is that the recipes stay the same no mat-ter of the dimensionality of the vector. That means that the property statedabove is true for two-, three-, four- and n- dimensional vectors. Furthermore,the definition of magnitude of a vector as the square root of the squares ofits components is also true for any n-dimensional vector provided that thecomponents are given in a Cartesian coordinate system.</p><p>But why do we use Cartesian coordinate system? In Cartesian system~a+~b = (a1+b1, a2+b2). However in polar system ~a+~b 6= (ra+rb, a+b). Thevector properties do not change with different coordinate systems. However,the convenient methods that we mentioned above are result of properties ofthe coordinate system. As a result it is suggested to use Cartesian coordinatesystem as often as possible. A very important consideration, though, is thatas the vector properties do not change with different coordinate systems if agiven property is proven to hold in one coordinate system then it is a vectorproperty that always works in all coordinate systems.</p><p>1.4 The Dot Product</p><p>Lets start with a physical motivation for the dot product. We know thatwork is the product of a path and the component of a force in the directionof the path. That can be written in vector notation as W = | ~A|| ~B| cos()where is the angle between the force vector ~A and the path ~B. Now, wedefine this to be equal to the dot product of two vectors:</p><p>~A ~B = | ~A|| ~B| cos() (1.1)</p><p>As finding the angle between the two vectors is often a pretty hard task,</p><p>~A</p><p>~B</p><p>~A ~B</p><p>Figure 1.1:</p><p>lets try to get rid of the cosine. As can be seen from Figure 1.1 using the</p></li><li><p>CHAPTER 1. VECTOR ARITHMETIC 4</p><p>~A</p><p>~B</p><p>| ~A| cos </p><p>Figure 1.2: Vector projection</p><p>cosine theorem we get the following:</p><p>| ~A ~B|2 = | ~A|2 + | ~B|2 2| ~A|| ~B| cos()</p><p>| ~A|| ~B| cos() = |~A|2 + | ~B|2 | ~A ~B|2</p><p>2</p><p>~A ~B = |~A|2 + | ~B|2 | ~A ~B|2</p><p>2(1.2)</p><p>Note that this result is not dependable on the coordinate system in use. Onlyfor Cartesian coordinate system this simplifies to</p><p>~A ~B = a1b1 + a2b2 + a3b3 (1.3)</p><p>Projections Lets take a look at Figure 1.2. We can see that the projectionof ~A on ~B looks like a dot product but with missing magnitude of anothervector. Furthermore, note that the length of the projection of ~A on ~B doesnot depend on the magnitude of ~B. We define a unit vector ~uB to have thesame direction and sense as ~B but to be with magnitude of unity.</p><p>~uB =~B</p><p>| ~B| (1.4)</p><p>Then as |~uB| = 1, | ~A| cos = |~uB|| ~A| cos . As a result,</p><p>ProjB ~A = ~uB ~A (1.5)</p><p>Structural properties</p><p> ~a ~b = ~b ~a ~a (~b+ ~c) = ~a ~b+ ~a ~c</p></li><li><p>CHAPTER 1. VECTOR ARITHMETIC 5</p><p> (c ~A) ~B = c( ~A ~B)This may seem trivial but one has to always keep in mind which operationsand conclusions are applicable in which situations. For example ~A ~B = 0does not mean that ~A or ~B is zero. It could be the case that they areorthogonal vectors and their cosine is zero.</p><p>1.5 The Cross Product</p><p>Although the cross product has vast applications in the physical sciences,our focus will be on the geometry. The cross product is also called the vectorproduct because the result of it is a vector (contrary to the dot product thatis a scalar). As a result we need to define three parameters of a the cross</p><p>product: magnitude, direction and sense. For ~A ~B the magnitude is definedto be | ~A|| ~B| sin , the direction is perpendicular to both ~A and ~B (that isperpendicular to the plane defined by the two vectors) and sense comes fromthe right hand rule: going from the first vector to the second through thesmaller angle.</p><p>As a result from the definition of the sense of the cross product ~A ~Bis not equal to ~B ~A. They have the same magnitude and direction butopposite sense, thus:</p><p>~A ~B = ~B ~A (1.6)Now, lets consider ~A ( ~B ~C). What is the direction of this vector? Itshould be perpendicular to a vector that is perpendicular to the plane definedby ~B and ~C. That means that ~A ( ~B ~C) is in fact parallel to the planecontaining ~B and ~C. However, there this vector is not equal to ( ~A ~B) ~Cfor the simple reason that each of them is parallel to one of two non-parallelplanes.</p><p>Finally for cross product the following holds:</p><p>~A ( ~B + ~C) = ~A ~B + ~A ~C (1.7)</p><p>The cross product of two vectors can be found through direct multiplicationof their components (keep in mind the sign of the vector products of unitvectors) or through the determinant method.</p><p>An interesting conclusion is that the magnitude of the cross productequals the area of the parallelogram which is enclosed by the two vectors.</p></li><li><p>CHAPTER 1. VECTOR ARITHMETIC 6</p><p>1.6 Equations of Lines and Planes</p><p>Planes are to surfaces what lines are to curves. In the Calculus of singlevariable the topic of tangent line comes up pretty frequently. When doingCalculus of two variables we will use the concept of a tangent plane.</p><p>Lets start with the derivation of the equation of a plane. There areseveral ways to define a plane but for this discussion it is useful to defineit with a point that the plane passes through and the normal vector to theplane. We call the fixed point P0(x0, y0, z0) and the normal vector will be~N = (a, b, c). We want to find an equation for the components of any pointP (x, y, z) lying in the plane. Now, a smart way to approach this problem is</p><p>to see that the vectorPP0 = (x x0, y y0, z z0) is on the plane and thus</p><p>is perpendicular to N . That means that ~N PP0 = 0. As a result:~N PP0 = 0</p><p>(a, b, c) (x x0, y y0, z z0) = 0a(x x0) + b(y y0) + c(z z0) = 0 (1.8)</p><p>Now, several things can be observed from this equation. First, this is theequation for a plane that has a normal vector (a, b, c) and passes throughpoint (x0, y0, z0). Second, a plane can be expressed with an infinite amountof different equations of this kind. This can be easlily deduced from the factthat this equation can be derived for any point P0 on the plane. The normalvector would be the same and only the values of x0, y0 and z0 will change.Third, if we change the coordinates of P0 with those of a point that do not lieon the original plane, will get an equation for another plane that is parallel tothe original one. This is because we keep the normal vector (the componentthat determines the orientation of the plane) and change only its position inspace.</p><p>Finally, there are two important things to note. The equation of a planeis linear. Furthermore, it has two degrees of freedom. That means that wehave to fix two of the variables so that we can calculate the others.</p><p>Next, lets shift our focus to the equation of a line. We choose to fix a linebased on a point P0(x0, y0, z0) that it passes through and a vector parallelto the line (giving its direction) ~v = (a, b, c). Just as we did in the case of aplane, we want to find equation for any arbitrary point P (x, y, z) on the line.</p><p>Then, the vectorPP0 = (x x0, y y0, z z0) is on the line as it connects</p><p>two points that are of the line. As the line is parallel to ~v that means that</p></li><li><p>CHAPTER 1. VECTOR ARITHMETIC 7</p><p>PP0 is a scalar multiple of ~v. As a result:</p><p>t~v =PP0</p><p>t(a, b, c) = (x x0, y y0, z z0)t a = x x0,t b = y y0,t c = z z0</p><p> x x0a = y y0b = z z0c (1.9)One can observe that the components of the position vector</p><p>PP0 are propor-</p><p>tional to the components of the direction vector ~v with the same constantof proportionality t. Furthermore, the equation of a line has one degree offreedom. If we fix one of the three coordinates we can easily find the othertwo.</p><p>A very important point to stress on is that the three parts of the equationdefine a line together. If you use only two parts you will get an equation of aplane (although you will have only two variables). This can be understoodif the deference between the following two sets is understood:</p><p>{(x, y) : 4y 3x = 17}{(x, y, z) : 4y 3x = 17}</p><p>In the second case, which is our case, z is free to take any value. However, ifyou want to define a line, the three variables should be constrained.</p></li><li><p>Chapter 2</p><p>Vector Calculus</p><p>2.1 Vector Functions of a Scalar Variable</p><p>Functions can be divided into four types. They can have a scalar or a vectoras an input and a scalar or a vector as an output - four different combinationsin total. Single variable Calculus deals only with the case of scalar input andoutput. In this section we will discuss functions that have a scalar input anda vector output.</p><p>What is suggested by Mr. Gross is that if there is direct correspondencebetween the definitions and the rules of scalar limits and vector limits, thenall the consequences coming from scalar limits, that use only accepted forvector arithmetic rules, will be also true for vectors. That means that if wedefine limit of a vector in a way that is analogous to the limit of a scalar andwe use only rules (operations)...</p></li></ul>