Scalar Fields and Vector fields - IIT Ropar

1 Scalar Fields and Vector fieldsDefinition A Scalar field is an assignment of a Scalar toeach point in region in the space. thetemperature at a point on the earth is a scalarfield. A Vector field is an assignment of a Vector toeach point in a region in the space. thevelocity field of a moving fluid is a Vector fieldas it associates a velocity Vector to each pointin the A Scalar field is a map from D to , where D is a subset of n. A Vector field is a map from D to n, where D is a subset of a subset of n. For n=2: Vector field in plane, for n=3: Vector field in space Example: Gradient fieldLine integral Line integral in a Scalar field Line integral in a Vector fieldLINE INTEGRAL IN A Scalar FIELDMOTIVATIONA rescue team follows a path in a danger area where for eachposition the degree of radiation is defined.

2 Compute the totalamount of radiation gathered by the rescue team along Smooth CurvesPiecewise Smooth CurvesA classic property of gravitational Fields is that, subject to certain physical constraints, the work done by gravity on an object moving between two points in the field is independent of the path taken by the of the constraints is that the pathmust be a piecewise smooth curve. Recall that a plane curve Cgiven bysmooth curve. Recall that a plane curve Cgiven byr(t) = x(t)i+ y(t)j, a t bis smooth ifare continuous on [a, b] and not simultaneously 0 on (a, b).Piecewise Smooth CurvesSimilarly, a space curve Cgiven byr(t) = x(t)i+ y(t)j + z(t)k, a t bis smooth ifis smooth ifare continuous on [a, b] and not simultaneously 0 on (a, b).

3 A curve Cis piecewise smoothif the interval [a, b] can be partitioned into a finite number of subintervals, on each of which Cis 1 Finding a Piecewise Smooth ParametrizationFind a piecewise smooth parametrization of the graph of Cshown in Cconsists of three line segments C1, C2, and C3, you can construct a smooth parametrizationfor each segment and piece them together by making the last t-value in Cicorrespond to the first t-value in Ci+ 1, as 1 Solutionvalue in Ci+ 1, as , Cis given byExample 1 SolutionBecause C1, C2, and C3are smooth, it follows that Cis piecewise of a curve induces anorientationto the ,inExample1,thecurveisPiecewise Smooth CurvesForinstance,inExample1.

4 Thecurveisoriented such that the positive direction isfrom (0, 0, 0), following the curve to (1, 2, 1).Line IntegralsYou will study a new type of integral called aline integralfor which you integrate over a piecewise smooth introduce the concept of a line integral, consider themass of a wire of finite length, given by a density (mass per unit length) of the wire at the point(x,y,z) is given byf(x,y,z).Line IntegralsPartition the curve Cby the pointsP0, P1, .., Pnproducing nsubarcs, as shown in nsubarcs, as shown in IntegralsThe length of theith subarc is given by , choose a point (xi,yi,zi) in each the length of each subarc is small, the total massofthewirecanbeapproximatedbythesumof thewirecanbeapproximatedbythesumIf you let || || denote the length of the longestsubarc and let || || approach 0, it seemsreasonable that the limit of this sum approachesthe mass of the IntegralsLine IntegralsTo evaluate a line integral over a plane curve Cgiven by r(t) =x(t)i+y(t)j, use the fact thatA similar formula holds for a space IntegralsNote that if f(x, y, z)

5 = 1, the line integral gives the arc length of the curve C. That is,Evaluatewhere Cis the line segment shown in 2 Evaluating a Line IntegralBegin by writing a parametric form of the equation of the line segment:x = t, y =2t, and z = t, 0 t 2 SolutionTherefore, x'(t) = 1, y'(t) = 2, and z'(t) = 1, which implies thatSo, the line integral takes the following 2 SolutionLine IntegralsFor parametrizations given by r(t) =x(t)i+y(t)j +z(t)k, it is helpful to remember the form of dsas Just as for an ordinary single integral, we can interpret the line integral of a positive function as an area. In fact, if f(x, y) 0, represents the area of one side of the fence or curtain shown here, ( ),Cf x y ds shown here, whose: Base is C.

6 Height above the point (x, y) is f(x, y). Now, let Cbe a piecewise-smooth curve. That is, Cis a union of a finite number of smooth curves C1, C2, .., Cn, where the initial point of Ci+1is the terminal point of Ci. Then, we define the integral of falong Cas the sum of the integrals of falong each of the smooth pieces of C: (),f x y ds ()( )( )( )12,,,..,nCCCCf x y dsf x y ds f x y dsf x y ds=++ + LINE INTEGRAL IN A Vector FIELDMOTIVATIONA ship sails from an island to another one along a fixedroute. Knowing all the sea currents, how much fuel willbe needed ?One of the most important physical applications of line integrals is that of finding the workdone on an object moving in a force example, Figure shows an inverse square force field similar to the gravitational field of the Integrals of Vector FieldsTo see how a line integral can be used to findwork done in a force fieldF, consider an objectmoving along a pathCin the field, as shown Integrals of Vector FieldsTo determine the work done by the force, you needconsider only that part of the force that is acting in thesame direction as that in which the object is means that at each point onC.

7 You can consider theprojectionF ToftheforcevectorFontotheunittangentproj ectionF a small subarc of length si, the increment of work is Wi= (force)(distance) [F(xi,yi,zi) T(xi,yi,zi)] siwhere(xi,yi,zi)is a point in theith Integrals of Vector FieldsConsequently, the total work done is given by the following line integral appears in other contexts and is the basis of the following definition of the line integral of a Vector in the definition thatLine Integrals of Vector FieldsFind the work done by the force fieldon a particle as it moves along the helix given byExample Work Done by a Forcefrom the point (1, 0, 0) to ( 1, 0, 3 ), as shown in (t) = x(t)i+ y(t)j + z(t)k= costi+ sin tj+ tkit follows that x(t) = cost, y(t) = sin t, and z(t) = SolutionSo, the force field can be written asTo find the work done by the force field in moving a particle along the curve C, use the fact thatr'(t) = sin ti+ costj+ kand write the SolutionLine Integrals of Vector FieldsFor line integrals of Vector functions, the orientation of the curve Cis the orientation of the curve is reversed, the If the orientation of the curve is reversed, the unit tangent Vector T(t) is changed to T(t)

8 , and you obtainLine Integrals in Differential FormLine Integrals in Differential FormLine Integrals in Differential FormA second commonly used form of line integrals isderived from the Vector field notation used in thepreceding Fis a Vector field of the form F(x, y) = Mi+ Nj, and Cis given by r(t) = x(t)i+ y(t)j, then F dris often written as given by r(t) = x(t)i+ y(t)j, then F dris often written as M dx+ N Integrals in Differential FormThis differential form can be extended to three variables. The parentheses are often omitted, as Evaluating a Line Integral in Differential FormLet Cbe the circle of radius 3 given byr(t) = 3 costi+ 3 sin tj, 0 t 2 as shown in Figure.

9 Evaluate the line integralExample SolutionBecause x= 3 costand y= 3 sin t, you have dx= 3 sin tdtand dy= 3 cost dt. So, the line integral isExample SolutionSuppose instead of being a force field, suppose thatF representsthe velocity field of a fluid flowingthrough a region in these circumstances, the integral ofF .Talong a curve inthe region gives the fluid s flow along the : Finding Circulation Around a CircleFlux Across a Plane CurveTo find the rate at which a fluid is entering or leaving aregion enclosed by a smooth curveC in the xy-plane, wecalculate the line integral over C , the scalarcomponent of the fluid s velocity field in the direction ofthe curve s outward-pointing normal the difference between flux andcirculation: Flux is the integral of the normalcomponent ofF.

10 Circulation is the integral to evaluate Flux of F acrossCwechooseasmoothparameterizationth at traces the curveC exactlyonce as t increases from a to find the outward unit normalvectornbycrossingthecurve svectornbycrossingthecurve sunit tangent vectorTwith thevectork. If the motion is clockwise, k Tpoints outward; if the motion is counterclockwise, T kpoints outwardWe choose: n= T kNow,Here the circle on the integral shows that the integration around the closedcurveC is to be in the counterclockwise : Finding Flux Across a CircleNote that the flux ofFacross the circle is positive, implies the net flow across the curveis outward. A net inward flow would have given a negative IndependenceUnder differentiability conditions, a field F is conservative iff it is the gradient field of a Scalar function ; , iff for some.