LECTURE 3 3.1Rules of Vector Differentiation

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1 LETURE 3 3.1Rules of Vector Differentiation We hae defined three kinds of deriaties inoling the operator grad( ) i j k, x y z 1 3 di(., x y z curl( i x 1 j y k z 3 d The good news is that you can apply all the usual formualae for differentiation with dx replced by proided you are careful. This is because grad and curl are ectors, wherer as di is a scalar. Also di and curl apply to ector fields, where as grad applies to scalar fields. In this lecture we look at more complicated identities inoling ector operators. The main thing to appreciate it that the operators behae both as ectors and as differential operators, so that the usual rules of taking the deriatie of, say, a product must be obsered. Let (r) u and (r) be ector fields, f (r) and g (r) be scalar fields and and be constants. 1 Differentiation is Linear grad( f g) f g grad ( f ) grad( g) di( u di( di( curl( u curl( curl( grad( fg) fgrad( g) ggrad( f ) di( f fdi( grad( f ).( curl( f fxu ( f ) xu fcurl ( grad( f ) x( Vector Product Rule di( ux. curl( u. curl( curl ( u u( di( ) ( di( ) (. ) u ( u. ) grad( u. ( u. ) (. ) u u curl( curl( 3. Vector Multiple Operations di( grad( f )). ( f ) f curl ( grad( f )) x( f ) 0

2 di ( curl( ). x( 0 curl( curl( ) (.( u Example: James lerk Maxwell established a set of four ector equations which are fundamental to working out how electromagnetic waes propagate. The entire telecommunications industry is built on these. did di B 0 curle B t curlh J D t In addition, we can assume the following, B r 0 H, J E, D r 0E, where all the scalars are constants. Now show that in a material with zero free charge density, 0, and with zero conductiity, 0, the electric field E must be a solution of the wae equation E E r0 r 0( ) t Example: Let 3 3xyz i xy j x yzk, 3x yz Find (i) di ( (ii) curl ( (iii). grad ( ) and hence (i di( ( curl(. 3. Line Integrals We are used to calculating integrals with respect to a set of coordinate axes, usually artesian. For b example Area under a cure f (x) between a and b = f ( x) dx a In this section we will generalize this idea to line integrals which are entirely analogous to the aboe idea. Here, we allow the axis of the integral to be an arbitrary cure in space and the function f will ary along that line. As an example of this, suppose that the temperature of our body at a particular time is T(x,y, z). Now suppose we wish to calculate the aerage temperature of the blood in a particular ein. This ein traces out a cure through our body and we wish to aerage the temperature along that cure. This inoles a line integral like the ones we will consider here. We will consider two types of line integrals: integrals of scalar functions and integrals of ector functions. As we will see, once we hae parameterized the integral using t on [a,b], the integral in both cases reduces to an ordinary integral oer [a,b]. This can be ealuated using integration rules. Line integrals of scalar functions Suppose we wish to integrate f (x,y, z) along a cure. Letting s be the arclength along the cure, we wish to find

3 f (x,y,z)ds. The arc length s has units of length and the integral aboe is the area under the cure along that cure. Howeer, the integral as it is written aboe cannot be ealuated since we do not hae an expression for x, y and z along. To get this, suppose that the cure is parameterized using the parameter t, then we can write the integral aboe as 1 t t0 ds f ( x(, y(, z( ). Where we hae used the hain Rule, and t ranges from t0 to t1. Where r( = x(i +y(j+z(k is the parametric form of the cure. Examples of line integrals of scalar functions An important special case of the line integral aboe is where f (x,y, z) =1, which defines the arc length of the cure. Another important application is when f (x,y, z) represents the linear density (mass per unit distance), for example along a wire. Then the line integral defines the total mass of the wire. To ensure the integral is defined, we assume that the cure is smooth, or piecewise smooth, ie continuous with finitely many smooth pieces. The alue of the scalar line integral is independent of the parameterization used to define. In particular, the integral has the same alue if the direction (orientation) of the parameterization is reersed. If is a closed path, sometimes is written instead of Following results hold for line integral. 1. kfds k fds. ( f g) ds fds gds 3. ( f g ) ds fds c 1 gds 1. The line integral of a ector function Suppose a particle is moed along a cure from r(a) to r(b) by a force F(r). What is the work done W by the force F on the particle? W b a F. dr This expression holds for any ector function, not just force as in the deriation aboe. Some notes about this line integral are listed below. To ensure the integral is defined, we assume that is smooth or piecewise smooth. The alue of the line integral of a ector function is independent of the parameterization used to define the cure as long as the direction (orientation) is the same. If the direction is reersed, then the alue of the integral is multiplied by 1. If is a closed path then is often written The line integral has the following properties:

4 a. kfds k Fds b. ( F G) ds Fds Gds c. ( F G ) ds Fds c Parametrisation of ures 1 1 Gds The key to ealuating such integrals is to define a single co-ordinate t that parametrises the cure. onsider the cures in D. For some cures it is obious how to do this, eg. Use the x-coordinate as the parameter: Straight line y a bx x t, y a bt or r( ( t, a b Parabola y a bx x t, y a bt or r( ( t, a bt ) For other cures one can use an angular formulation ircle x y a x a cost, y asin t or r( ( acost, asin and dr( ( asin t, acos x y Ellipse 1 x a cost, y bsin t or a b r( ( acost, bsin and dr( ( asin t, bcos x a x b x c Straight line r = a + bt or t(say) l m n Therefore x a lt, y b mt, z c nt Helix (shape of light spring) r ( acost, asin t, c Here a is the radius of the helix and a b = tan, where is the helix angle. Ealuation of work Integrals 1 parametrise the cure as r( ( x(, y(, z( ) work the limits a and b on t. 3 ealuate the ector field along the (x,y,z)=(x(,y(,z(), form the dot product and integrate w.r.t t: ( r). dr tb ta dx( dy( dz( ( x(, y(, z(. i j k Example1: Find the work done moing a particle from (0,0,0) to (1,1,1) in the field F (x y ) i 3xy j k along the straight path 1, the straight line joining (0,0,0) to (1,1,1) and parametrically gien by x= t, y =t, z = t 3. Example: Find the work done moing a particle from (0,0,0) to (1,1,1) in the field 3 F ( xy z ) i x j 3xz k along the following paths a) the straight path 1, the straight line joining (0,0,0) to (1,1,1). b) the path composed of the three straight lines joining (0,0,0) to (1,0,0) to (1,1,0) to (1,1,1).

5 Note that both answers are same as they would be for any cure joining (0,0,0) and (1,1,1). This is because the aboe force field F is conseratie. 3.5 onseratie Vector fields and independence of path Recall from lecture, a ector field is conseratie in a region, if its circulation along eery closed cure in the region is zero. i.e. curl( 0 The Fundamental Theorem of Line Integrals Let F be a conseratie ector field with potential function f, and be any smooth cure starting at the point A and ending at the point B. Then To proe the fundamental theorem of line integrals we will use the following outcome of the chain rule: If r( = x(i + y(j is a ector alued function, then d f(r() = fx x'( + f y y'(we are now ready to proe the theorem. We hae Theorem: The necessary and sufficient condition for a continuous ector field F to be conseratie (or irrotational) in a simply connected region R is F that it is the gradient of a scaler field. Example: Find the work done by the ector field F(x,y) = (x -3y)i + (3y - 3x)j along the cure indicated in the graph below

6 Independence of Path and losed ures Example: Find the work done by the ector field F(x,y) = (cos x + y)i + (x+ e sin y )j + (sin(cos z))k along the closed cure shown below Theorem: onseratie Vector Fields and losed ures Let F be a ector field with components that hae continuous first order partial deriaties and let be a piecewise smooth cure. Then the following three statements are equialent 1. F is conseratie.. is independent of path. 3. for all closed cures. Example: alculate curl (F) for the force field F taken in worked in preious example, hence show that the field is conseratie. Find the scalar field and hence calculate the work done in moing from (1,-,1) to (3,1,4). Remarks The course is also true; if the work integral is independent of path taken between any two points then there must exist a scalar potential function such that F.