EE2007: Engineering Mathematics II Vector Calculus Ling KV School of EEE, NTU ekvling@ntu.edu.sg Rm: S2-B2b-22 Ver 1.1: Ling KV, October 22, 2006 Ver 1.0: Ling KV, Jul 2005 EE2007/Ling KV/Aug 2006
EE2007: Engineering Mathematics II Vector Calculus My part: week 10 Introduction, Dot and Vector Products, Scalar Triple Product week 10-11 Differential Calculus week 11-12 Integral Calculus week 13 Revision Classes Text: Kreyszig, E. (1999). Advanced Engineering Mathematics, Chapters 8-9, 8th Ed. John Wiley & Sons, Inc. [NTU Library, QA401.K92] Reference: 1. H. M. Schey (1997). Div, Grad, Curl and all that, an informal text on vector calculus, W.W. Norton & Company, Inc. [QA433S328] EE2007/Ling KV/Aug 2006 1
Overview This module deals with vectors and vector functions in 3D-space and extends the differential calculus to these vector functions. Forces, velocities and various other quantities are vectors. This makes vector calculus a natural instrument for engineers. In 3D space, geometrical ideas become influential, and many geometrical quantities (tangents and normals, for example) can be given by vectors. We will first review the basic algebraic operation with vectors in 3D space. Vector functions, which represent vector fields and have various physical and geometrical applications, will be introduced next. Three physically and geometrically important concepts related to scalar and vector fields, namely, the gradient, divergence and curl will be discussed. Integral theorems (Vector Integral Calculus) involving these concepts will then follow. EE2007/Ling KV/Aug 2006 2
Introduction Length, temperature and voltages are quantities determined by its. They are. Force and velocity are quantities determined by both and. They are. A vector is represented as an arrow or a. EE2007/Ling KV/Aug 2006 3
In printed work, we denote vectors by lowercase boldface letters, a, v, etc; in handwritten work, we often denote vectors by arrow, a, v. We sometimes also write the vector as an order pair of real numbers and use boldface letter to emphasize that it represents a vector. Thus r = [x, y, z]. The vector r has magnitude r = and the unit vector in the direction of r is ˆr = 1 r r =. A unit vector has magnitude particular director. We can write r = r ˆr., and can be used to indicate a EE2007/Ling KV/Aug 2006 4
Equality of Vectors Two vectors a and b are equal, written a = b, if they have the same and the same. Hence, a vector can be arbitrarily, that is, the initial point can be chosen arbitrarily. This definition is practical in connection with forces and other applications. EE2007/Ling KV/Aug 2006 5
Position Vector A Cartesian coordinate system being given, the position vector, r of a point A : (x, y, z) is the vector with the origin (0, 0, 0) as the initial point and A as the terminal point. Thus, a position vector can be written as r = xi + yj + zk, or r = [x, y, z] where i, j, k are standard Cartesian unit vectors along the x, y, z directions. EE2007/Ling KV/Aug 2006 6
Examples Find the components of the vector v, given the initial point P : (1, 0, 0) and the terminal point Q : (4, 2, 0). Find also v. If p = [1, 3, 1], q = [5, 0, 2], and u = [0, 1, 3] represent forces. What is the resultant force? Determine the force p such that p, q = [0, 3, 4], and u = [1, 1, 0] are in equilibrium. If p = 6 and q = 4, what can be said about the magnitude and direction of the resultant? EE2007/Ling KV/Aug 2006 7
Geometrical Application Using vectors, prove that the diagonals of a parallelogram bisect each other. The vector r from O to P is of the form r = k(a + b). We also have r = a + l(b a). Thus, ka + kb = (1 l)a + lb. From this, k = 1 l, k = l, hence k = 1 2, l = 1 2, and the proof is complete. EE2007/Ling KV/Aug 2006 8
Inner Product (Dot Product) of Vectors The inner product or dot product, a b of two vectors a = [a 1, a 2, a 3 ] and b = [b 1, b 2, b 3 ] is defined as a b = a b cos γ if a 0, b 0 a b = 0 if a = 0 or b = 0 where 0 γ π is the angle between a and b (measured when the vectors have their initial points coinciding). Note that the dot product is a. In components 1, a b = a 1 b 1 + a 2 b 2 + a 3 b 3. 1 see text pp.410 for a derivation EE2007/Ling KV/Aug 2006 9
Inner Product The inner product of two nonzero vectors is zero if and only if these vectors are. [Length and angle in terms of inner product] a a = a 2, so that we have a = a a We can also obtain the angle γ between two nonzero vectors cos γ = a b a b = a b a a b b EE2007/Ling KV/Aug 2006 10
Component or Projection The concept of the component or projection of a vector a in the direction of another vector b( 0) is an important one. From the figure, p = a cos γ Thus, p is the of the orthogonal projection of a on a straight line l. Multiply p by b / b, we have p = a b, (b 0) since a b = a b cos γ. b EE2007/Ling KV/Aug 2006 11
General Properties of Inner Products For any vectors, a, b, c and scalars q 1, q 2, [q 1 a + q 2 b] c = q 1 a c + q 2 b c (Linearity) a b = b a (Symmetry) } a a 0 (P ositive def initeness) a a = 0 iff a = 0 (a + b) c = a c + b c (Distributive) a b a b (Schwarzinequality) a + b a + b (T riangleinequality) a + b 2 + a b 2 = 2( a 2 + b 2 ) (P arallelogramequality) EE2007/Ling KV/Aug 2006 12
Dot Product Example If a = [1, 2, 0] and b = [3, 2, 1], find the angle between the two vectors. [ 96.865 o ] EE2007/Ling KV/Aug 2006 13
Applications of Inner Products [Work done by a force] Find the work done by a force p = [2, 6, 6] acting on a body if the body is displaced from point A : (3, 4, 0) to point B : (5, 8, 0) along the straight line segment AB. Sketch p and AB. [ 28 ] EE2007/Ling KV/Aug 2006 14
[Component of a force in a given direction] What force in the rope will hold a car of 5000kg in equilibrium? [ 2113kg ] EE2007/Ling KV/Aug 2006 15
Vector Product (Cross Product) The vector product (cross product) v = a b of two vectors a = [a 1, a 2, a 3 ] and b = [b 1, b 2, b 3 ] is a vector whose magnitude is given by v = a b sin γ and the direction of v is perpendicular to both a and b and such that the three vectors a, b, v, in this order, form a right-handed triple. Thus, if a and b have the same or opposite direction or if one of the these vectors is the zero vector, then a b = 0. EE2007/Ling KV/Aug 2006 16
In component, v = a b = i j k a 1 a 2 a 3 b 1 b 2 b 3 Example: Let a = [4, 0, 1] and b = [ 2, 1, 3]. Then i j k v = a b = 4 0 1 = i 10j + 4k = [1, 10, 4]. 2 1 3 EE2007/Ling KV/Aug 2006 17
General Properties of Vector Products For any vectors a,b, c and scalar l, (la) b = l(a b) = a (lb) a (b + c) = (a b) + (a + c) (a + b) c) = (a c) + (b + c) b a = (a b) a (b c) (a b) c } (distributive) (anti communtative) (not associative) EE2007/Ling KV/Aug 2006 18
Typical Applications of Vector Products Find the moment of a force p about a point Q. EE2007/Ling KV/Aug 2006 19
Find the velocity of the rotating disk, whose angular speed is ω(> 0). Let P be any point on the disk and d its distance from the axis. Then P has a speed ωd. Let r be the position vector of P referred to a coordinate system with the origin on the axis of rotation. If we define a vector w whose direction is that of the axis of rotation and such that the rotation appears clockwise if one looks from the initial point of w to its terminal point. Then length of w is equal to the angular speed ω. EE2007/Ling KV/Aug 2006 20
Then ωd = w r sin γ = w r where γ is the angle between w and r. Thus we see that the velocity vector v of P can be represented in the form v = w r EE2007/Ling KV/Aug 2006 21
Scalar Triple Product The scalar triple product or mixed triple product of three vectors a = [a 1, a 2, a 3 ], b = [b 1, b 2, b 3 ], c = [c 1, c 2, c 3 ] is denoted by (a b c) and is defined by (a b c) = a (b c) = a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 It can be shown that (ka kb kc) = k(a b c) and a (b c) = c (a b) EE2007/Ling KV/Aug 2006 22
Fact: a (b c) = c (a b) = b (c a) Proof: a (b c) = = = a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 c 1 c 2 c 3 b 1 b 2 b 3 a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 a 1 a 2 a 3 = c (a b) = b (c a) EE2007/Ling KV/Aug 2006 23
Geometric Interpretation of Scalar Triple Product Show that the absolute value of the scalar triple product is the volume of the parallelepiped with a, b, c as edge vectors. EE2007/Ling KV/Aug 2006 24
Parametric Representation of Curves A curve in space is determined by three equations of the form x = x(t), y = y(t), z = z(t) Alternatively, a curve C in space can be represented by a position vector r(t) = [x(t), y(t), z(t)] = x(t)i + y(t)j + z(t)k and t is a parameter specifying points (x(t),y(t),z(t)) along the curve. The parameter t may be time or something else. For example, a straight line is represented by x(t) = 1 + 3t, y(t) = 2 5t, z(t) = 3 + 3t EE2007/Ling KV/Aug 2006 25
Examples: Parametric Representation of Curves Straight line. A straight line L through a point A with position vector a in the direction of a constant vector b can be represented in the form r(t) = a + tb = [a 1 + tb 1, a 2 + tb 2, a 3 + tb 3 ]. EE2007/Ling KV/Aug 2006 26
Ellipse and Circle. The vector function r(t) = [a cos t, b sin t, 0] represents an ellipse in the xy-plane with centre at the origin and principal axes in the direction of the x and y axis. In fact, since cos 2 t + sin 2 t = 1, we obtain x 2 a 2 + y2 b 2 = 1, z = 0 EE2007/Ling KV/Aug 2006 27
Circular Helix The twisted curve C represented by the vector function r(t) = [a cos t, b sin t, ct], (c 0) is called a circular helix. EE2007/Ling KV/Aug 2006 28
Application: Parametric Representation of Curves We can use parametric equations to find the intersection of two curves. Example: Two particles move through space with equations r 1 (t) = [t, 1 + 2t, 3 2t] and r 2 (t) = [ 2 2t, 1 2t, 1 + t]. Do the particles collide? Do their paths cross? Solution: The particle will collide if. Their paths cross if EE2007/Ling KV/Aug 2006 29
Representation of Surfaces A surface S in the xyz-space can be represented as z = f(x, y) or g(x, y, z) = 0 For example, z = + a 2 x 2 y 2 or x 2 + y 2 + z 2 = 0, z 0 represents a hemisphere of radius a and centre at the origin. Just like the parametric representation of a curve in space, we could also represent a surface parametrically. Surfaces are two-dimensional, hence we need two parameters, which we denote as u and v. EE2007/Ling KV/Aug 2006 30
Thus a parametric representation of a surface S in space is of the form r(u, v) = [x(u, v), y(u, v), z(u, v)], (u, v) in R. where R is some region in the uv-plane. EE2007/Ling KV/Aug 2006 31
Example: Parametric Representation of a Cylinder x 2 + y 2 = a 2, 1 z 1, represent a cylinder of radius a, height 2 in the z-direction. A parametric representation of this cylindrical surface is r(u, v) = [a cos u, a sin u, v] where the parameters u, v vary in the rectangle R : 0 u 2π, 1 v 1 EE2007/Ling KV/Aug 2006 32
Example: Parametric Representation of a Sphere A sphere x y + y 2 + z 2 = a 2 can be represented in the form r(u, v) = [a cos v cos u, a cos v sin u, a sin v] where the parameters u, v vary in the rectangle R : 0 u 2π, π/2 v π/2 EE2007/Ling KV/Aug 2006 33
Parametric Representations of Common Surfaces 2 2 Problem Set 9.5, p.495 EE2007/Ling KV/Aug 2006 34
Vector and Scalar Functions and Fields Vector functions are functions whose values are vectors v = v(p ) = [v 1 (P ), v 2 (P ), v 3 (P )] depending on the point P in space. A curve or surface in space which are described parametrically in vector form, are examples of vector functions. Scalar functions are functions whose values are scalars f = f(p ) depending on P. An example of scalar functions is the spatial distribution of temperature in a room. EE2007/Ling KV/Aug 2006 35
A vector field is defined by a vector function within region of space or surface in space or a curve in space. Field of tangent vectors of a curve Field of normal vectors of a surface Similarly, a scalar function defines a scalar field in a region on a surface or curve. Examples of scalar field are temperature field in a body and the pressure field of the air in the earth s atmosphere. Vector fields can be used to describe fluid flows, electromagnetism and heat transfer in space. Vector and scalar functions may also depend on time t or on other parameters. EE2007/Ling KV/Aug 2006 36
Examples of Scalar and Vector Functions The distance f(p ) of any point P from a fixed point P o in space is a scalar function f(p ) = f(x, y, z) = (x x o ) 2 + (y y o ) 2 + (z z o ) 2 The velocity vector v(p ) of a rotating body constitute a vector field, the so-called velocity field of the rotation. v(x, y, z) = w r = w [x, y, z] EE2007/Ling KV/Aug 2006 37
According to Newton s law of gravitation, a particle of mass M at a fixed point P o will have an attractive force p on a particle in space at point P. Its magnitude is proportional to 1/r 2 where r is the distance between P and P o, p = GMm ˆr r 2 where ˆr is the unit vector in the (outward) radial direction. EE2007/Ling KV/Aug 2006 38
Examples of Vector Functions F(x, y) = xi + yj G(x, y) = yi + xj x2 + y 2 EE2007/Ling KV/Aug 2006 39
Derivative of a Vector Function The derivative of a vector function is defined as v (t) = lim t 0 v(t + t) v(t) t In components, v (t) = [v 1(t), v 2(t), v 3(t)] A vector function is said to be differentiable at a point t if the above limit exists. If the vector function v(t) represents a curve C is space, then v (t) is a vector tangent to the curve C at the point P : (x(t), y(t), z(t)), provided v(t) is differentiable and v (t) 0. EE2007/Ling KV/Aug 2006 40
Note that the usual rules of differentiation carry over to vector functions: (u v) = (u v) = (u v w) = EE2007/Ling KV/Aug 2006 41
Derivative of a Vector Function of Constant Length Let v(t) be a vector function whose length is constant, v(t) = c Then and v 2 = (v v) = Thus, the derivative of a vector function v(t) of constant length is either or is EE2007/Ling KV/Aug 2006 42
Partial Derivatives of a Vector Function Let r(t 1, t 2 ) = a cos t 1 i + a sin t 1 j + t 2 k. Then r t 1 = a sin t 1 i + a cos t 1 j r t 2 = k EE2007/Ling KV/Aug 2006 43
Tangent to a Curve In slide 40, we mentioned that if curve C is represented by r(t), then the tangent to the curve C is r (t). The unit tangent vector is then given by u = 1 r r. The tangent to C at an arbitrary point P is given given by q(w) = r + wr, where w is a scalar parameter. EE2007/Ling KV/Aug 2006 44
Example: Tangent to an Ellipse Find the tangent to the ellipse 1 4 x2 + y 2 = 1 at P : ( 2, 1/ 2). Solution: The ellipse can be represented as r(t) = Hence r (t) = and P corresponds to t =. Thus, the tangent to the ellipse is q(w) =. EE2007/Ling KV/Aug 2006 45
Example, Helix, its Tangent Vector and Derivative of Tangent EE2007/Ling KV/Aug 2006 46
Velocity and Acceleration Curves as just discussed from the viewpoint of geometry also play an important role in mechanics paths of moving bodies. Consider a path C given by r(t), where t is now time. Then, the derivative r (t) = d dt r(t) is the velocity vector of the motion, and the acceleration vector is a(t) = r (t). EE2007/Ling KV/Aug 2006 47
Example: Centripetal Acceleration The vector function r(t) = R cos ω(t)i + R sin ωtj represents a circle C of radius R with centre at the origin of the xy-plane and describes a motion of a particle P in the counter-clockwise sense. The velocity vector v = r = is to C, and its magnitude ( ) is v = r r = is. The angular speed is equal to. The acceleration vector is a = v = which is of a magnitude, with a direction. The vector a is the. EE2007/Ling KV/Aug 2006 48
Tangent Plane and Surface Normal Recall that (in slide 40), we mentioned that if curve C is represented by r(t), then the tangent to the curve C is the derivative dr(t) dt. Thus, given the parametric representation of the surface S : r(u, v), we could obtain the surface normal as N = r u r v where r u and r v are partial derivatives of r(u, v) wrt u and v respectively. The unit normal vector is n = 1 N N EE2007/Ling KV/Aug 2006 49
Example: Normal Vector to a Spherical Surface Since a parametric representation of the sphere is r(u, v) = [a cos v cos u, a cos v sin u, a sin v] so, r u = [ a cos v sin u, a cos v cos u, 0] r v = [ a sin v cos u, a sin v sin u, a cos v] and this gives N = r u r v = a 2 cos v[cos v cos u, cos v sin u, sin v], and N = a 2 cos v EE2007/Ling KV/Aug 2006 50
Hence n = 1 N N = [cos v cos u, cos v sin u, sin v] = 1 [x, y, z] a EE2007/Ling KV/Aug 2006 51
Gradient of a Scalar Field For a give scalar function f(x, y, z), the gradient of f is the vector function 3 grad f = f x i + f y j + f z k. Introducing the differential operator (read nabla or del), we write = x i + y j + z k grad f = f = f x i + f y j + f z k. For example, if f = 2x 2 + yz 3y 2, then f = 2i + (z 6y)j + yk. 3 assuming, of course, f is differentiable EE2007/Ling KV/Aug 2006 52
f and the Directional Derivative Gradient vector ( f) arises when we attempt to define a rate of change for a scalar function of several variables, f(x, y, z). The rate of change of f(x, y, z) at a point P in the direction of a vector b is called the directional derivative of f at P in the direction of a vector b: D b f = f b b Geometrically, D b f represents the slope of surface f(x, y, z) in the direction of b. Example: Given f = x 2 + y 2, then f = 2xi + 2yj. EE2007/Ling KV/Aug 2006 53
Example: Directional Derivative Find the directional derivative of f(x, y, z) = 2x 2 + 3y 2 + z 2 at the point P : (2, 1, 3) in the direction of the vector b = i 2k. Solution: We obtain f = [4x, 6y, 2z] and at P, f = [8, 6, 6]. Thus, D b f = [8, 6, 6] [1, 0, 2] [1, 0, 2] = 4 5 The minus sign indicates that f decreases at P in the direction of b. EE2007/Ling KV/Aug 2006 54
Example: Gradient and Directional Derivative See Problem Set 18.9. p.452 The force in an electrostatic field f(x, y, z) has the direction of grad f = f. Hence, if f(x, y) = x 2 + 9y 2, then the force at P : ( 2, 2) is f = [2x, 18y] = [2( 2), 18(2)] = [ 4, 36]. The flow of heat in a temperature field takes place in the direction of maximum decrease of temperature T. Hence if the temperature field is T (x, y) = x/y, then the heat will flow in the direction T = [1/y, x/y 2 ]. EE2007/Ling KV/Aug 2006 55
f Characterizes Maximum Increase Let f(p ) = f(x, y, z) be a scalar function having continuous first partial derivaties. Then f exists and its length and direction are independent of the particular choice of Cartesian coordinates in space. If a a point P, the gradient of f is not the zero vector, it has the direction of maximum increase of f at P. Proof: see text pp.448. Example: If on a mountain the elevation above sea level is z(x, y) = 1500 3x 2 5y 2 (meters), what is the direction of steepest ascent at P : ( 0.2, 0.1)? Solution: The direction of the steepest ascent is given by z = [ 6x, 10y] and at P, z = [1.2, 1]. EE2007/Ling KV/Aug 2006 56
Gradient ( f) as a Surface Normal Vector Consider a surface S in space given by f(x, y, z) = c = const Recall that a curve C in space can be represented by r(t) = [x(t), y(t), z(t)] If we want C to lie of S, then its components must satisfy f(x(t), y(t), z(t)) = c If we differentiate the above with respect to t, we get by chain rule, f x x (t) + f y y (t) + f z z (t) = [ f x, f y, f z ] r (t) = 0 EE2007/Ling KV/Aug 2006 57
where r (t) = [x (t), y (t), z (t)], which is a tangent vector of C. Since C lie of S, this vector is tangent to S. What this means is that the vector f = [ f x, f y, f z ] is perpendicular to tangent vector of S at P. Thus, we conclude: Let f be a differentiable scalar function that represents a surface S : f(x, y, z) = c. The gradient of f (i.e. f) at a point P of S is a normal vector of S at P is this vector is not the zero vector. EE2007/Ling KV/Aug 2006 58
Example: Normal Vector to a Surface Example: Find an equation of the plane tangent to the surface x 2 + 3y 3 + 2z 3 = 12 at the point (1, 1, 2). Solution: A vector normal to the surface S : f(x, y, z) = c, where f(x, y, z) = x 2 + 3y 3 + 2z 3 is f = [2x, 9y 2, 6z 2 ] Recall that the equation of a plane has the form N (r a) = 0 where N is a normal vector to the plane, and is a point on the plane. Thus, the required equation is [2x, 9y 2, 6z 2 ] (r [1, 1, 2]) = 0 EE2007/Ling KV/Aug 2006 59
Surface Normal of a Spherical Surface Re-visited The sphere g(x, y, z) = x 2 + y 2 + z 2 a 2 = 0 has a unit normal vector given by n = 1 g g 1 = 2 x 2 +y 2 +z = 1 a [x, y, z] 2[2x, 2y, 2z] This agrees with the answer given in slide 50 where the surface normal was computed using n = 1 N N where N = r u r v EE2007/Ling KV/Aug 2006 60
Example: Gradient ( f) as Surface Normal Vector Find a unit normal vector n of the cone of revolution z 2 = 4(x 2 + y 2 ) at the point P : (1, 0, 2). Solution: The surface of the cone can be represented by f(x, y, z) = 4(x 2 + y 2 ) z 2 = 0 Thus, f = [8x, 8y, 2z] and at P, f = [8, 0, 4]. Hence, a unit normal vector of the cone at P is n = 1 f f = 1 [2, 0, 1]. 5 The other normal is -n. EE2007/Ling KV/Aug 2006 61
Visualising Surface Normal, Contour, and Gradient EE2007/Ling KV/Aug 2006 62
Conservative Vector Fields Some vector fields can be obtained from scalar fields. Such a vector field is given by a vector function v, v = f The function f is called a potential function of v. Such a field is called conservative because in such a vector field, energy is conserved. Examples of conservative vector fields are gravitational field and electrostatic field. EE2007/Ling KV/Aug 2006 63
Example: Conservative Vector Field Problem Set 8.9. p.452 Find f for a given v or state that v has no potential. v = [yz, xz, xy] gives f = xyz + c where c is a constant. v = [ye x, e x, 1] gives f = ye x + z + c where c is a constant. v = [x, y]/(x 2 + y 2 ) gives f = 1 2 ln(x2 + y 2 ) + c where c is a constant. EE2007/Ling KV/Aug 2006 64
A Physical Example: The Gradient Vector If f(x, y, z) is a scalar-valued function describing, for example, temperatures in a room, then Duf, the rate of change of f along a fixed line whose direction is given by the unit vector u, is called the directional directive of f in the direction of u. This derivative is a spatial rate of change, not a temporal one. It gives the temperature gradient, the change of temperature per unit length in a given direction. This gradient vector turns out to be orthogonal to the isotherms of f, the surfaces along which the temperature described by f are constant. The gradient vector points in the direction of increasing temperatures, and the greatest rate of change in f is the length of grad f. EE2007/Ling KV/Aug 2006 65
Divergence of a Vector Field Let v(x, y, z) = [v 1, v 2, v 3 ] be a differentiable vector function, then the function div v = v 1 x + v 2 y + v 3 z is called the divergence of v. Another common notation for the divergence of v is v, div v = v = ( x i + y j + zk) (v) = v 1 x + v 2 y + v 3 z Example: If v = [3xz, 2xy, yz 2 ], then v =. Note that v is a. EE2007/Ling KV/Aug 2006 66
Physical Meaning of the Divergence Roughly speaking, the divergence measures outflow minus inflow. See also text pp.454, Examples 1 and 2. More on this later when we talk about Divergence Theorem. Example: Let v = [v 1, v 2, v 3 ] represents the velocity of the fluid flow. Then the rate of fluid volume flow across surface A is v 1 (A) y z.... EE2007/Ling KV/Aug 2006 67
Visualizing the Divergence 4 4 (Problem Set 8.10, p.456) Plot the given velocity field v of a fluid flow in a square centered at the origin. Recall that the divergence measures outflow minus inflow. By looking at the flow near the sides of the square, can you see whether div v must be positive, or negative, or zero? EE2007/Ling KV/Aug 2006 68
Curl of a Vector Field Let v(x, y, z) = [v 1, v 2, v 3 ] be a differentiable vector function. Then the function curl v = v = i j k x y z v 1 v 2 v 3 is called the curl of the vector function v. Example: Let v = [yz, 3xz, z], then v = i j k x y z yz 3xz z = [ 3x, y, 2z] EE2007/Ling KV/Aug 2006 69
Example: Curl We have seen earlier that the velocity field of a rotating rigid body can be represented in the form v = w r where wωk, and r = [x, y, z]. Thus, v = w r = [ ωy, ωx, 0] Now, let compute v = i j k x y z ωy ωx 0 = Hence the curl of the velocity field of a rotating rigid body has the direction in the axis of rotation, and its magnitude equals to twice the angular speed of the rotation. EE2007/Ling KV/Aug 2006 70
Summary, the Operator operates on field gives f. operates on field gives f or f. If F = f, then F = 0. Since the curl characterises the rotation in a field, we say that F is irrotational. If F is not associated with velocity, we usually say F is conservative. Gradient vector ( f) arises when we attempt to define a rate of change for a scalar function of several variables (f(x, y, z)). This vector is orthogonal to the level set of the scalar function (Surface f(x, y, z) = c) and is tangent to the lines of flow of the gradient field. The Gravitational and Electrostatic potentials are examples of scalar fields for which the gradient field is physically significant. EE2007/Ling KV/Aug 2006 71
Line Integrals The concept of a line integral is a simple and natural generalisation of a definite integral b a f(x)dx. If we represent the curve C by a parametric representation r(t) = [x(t), y(t), z(t)], (a t b) then the line integral of a vector function F(r) over a curve C is defined by b F(r) dr = F(r(t)) dr(t) dt C a dt In terms of components, with dr = [dx, dy, dz], the above becomes C F(r) dr = b a [F 1, F 2, F 3 ] [ dx dt, dy dt, dz dt ] dt EE2007/Ling KV/Aug 2006 72
Example: Line Integral [pp.466] Find the value of the line integral when F(r) = [ y, xy, 0] and C is the circular arc from A to B. If F represents a force, then the line integral gives the in the displacement along path C. by F EE2007/Ling KV/Aug 2006 73
[pp.466] Find the value of the line integral when F(r) = [z, x, y] and C is the helix r(t) = [cos t, sin t, 3t], (0 t 2π). EE2007/Ling KV/Aug 2006 74
[pp.467] Dependence of line integral on path EE2007/Ling KV/Aug 2006 75
Line Integrals and Independent of Path A line integral C F(r) dr = b a [F 1, F 2, F 3 ] [ dx dt, dy dt, dz dt ] dt with continuous F 1, F 2, F 3 in a domain D in space is independent of path in D if and only if F is the gradient of some function f in D, i.e. F = f. Proof: see text, pp.472. EE2007/Ling KV/Aug 2006 76
Example: Independent of Path Evaluate the integral F dr = [3x 2, 2yz, y 2 ] [dx, dy, dz] C C from A : (0, 1, 2) to B : (1, 1, 7) by showing that F is conservative. Solution: F is conservative if there exists a scalar function f such that F = f.... EE2007/Ling KV/Aug 2006 77
Surface Integrals Let the surface S be given by a parametric representation r(u, v) = [x(u, v), y(u, v), z(u, v)], (u, v) in R and S is piecewise smooth so that S has a normal vector N = r u r v and unit normal n = 1 N N Then, for a given vector function F, the surface integral over S is defined as F nda = F(u, v) N(u, v) du dv S Note that the integrand is a scalar because we take dot product. Indeed, F n is the component of F normal to the surface. This R EE2007/Ling KV/Aug 2006 78
integrand arises naturally in flow problems where F represents the velocity vector of the fluid and we sometimes call the surface integral the flux integral. EE2007/Ling KV/Aug 2006 79
Example: Flux Through a Surface Compute the flux of water through the parabolic cylinder S : y = x 2, 0 x 2, 0 z 3 if the velocity vector is F = [3z 2, 6, 6xz]. EE2007/Ling KV/Aug 2006 80
Solution: Let x = u and z = v, we have y = u 2 and hence a parametric representation of S is S : r(u, v) = [u, u 2, v], 0 u 2, 0 v 3 Thus, r u = [1, 2u, 0] r v = [0, 0, 1] N = r u r v = [2u, 1, 0] F(u, v) = [3v 2, 6, 6uv] Hence S F n da = F(u, v) N(u, v) du dv R = 3 2 v=0 u=0 [3v2, 6, 6uv] [2u, 1, 0] du dv = 72 EE2007/Ling KV/Aug 2006 81
Example: Surface Integral If F = [x 2, 0, 3y 2 ] and S is the portion of the plane x + y + z = 1 in the first octant. Evaluate the surface integral of F over S. EE2007/Ling KV/Aug 2006 82
Solution: Let u = x and v = y, we have z = 1 u v. Hence a parametric representation of S is S : r(u, v) = [u, v, 1 u v], 0 u 1 v, 0 v 1 and N = r u r v = [1, 0, 1] [0, 1, 1] = [1, 1, 1] Hence S F n da = F(u, v) N(u, v) du dv R = 1 v=0 = 1 3 1 v u=0 [u2, 0, 3v 2 ] [1, 1, 1] du dv Note that the normal vector can also be computed from the surface function: g(x, y, z) = x + y + z 1 = 0, i.e. N = g = [1, 1, 1]. EE2007/Ling KV/Aug 2006 83
Divergence Theorem of Gauss Let T be a closed bounded region in space whose boundary is a piecewise smooth orientable surface S. Let F(x, y, z) be a vector function that is continuous and has continuous first partial derivatives in some domain containing T. Then T (F ) dv = where n is the outer unit normal vector of S. S F n da Volume integral and surface integral are related through the divergence theorem of Gauss. This is of practical interest because one the two kinds of integral is often simpler than the other. EE2007/Ling KV/Aug 2006 84
Evaluation of Surface Integral by the Divergence Theorem Evaluate S F n da where F = [x3, x 2 y, x 2 z], S is the closed surface consisting of the cylinder x 2 + y 2 = a 2, 0 z b and the circular disks at z = 0 and z = b. Solution: EE2007/Ling KV/Aug 2006 85
Stokes s Theorem Let S be a piecewise smooth oriented surface in space and let the boundary of S be a piecewise smooth simple closed curve C. Let F(x, y, z) be a continuous vector function that has continuous first partial derivatives in a domain in space containing S. Then ( F) n da = F dr where n is a unit normal vector of S and, depending on n, the integration around C is taken in the sense shown in the figure. S C EE2007/Ling KV/Aug 2006 86
Stokes s Theorem and Path Independence If F = 0, then by Stokes s theorem, ( F) n da = F dr = 0 S C i.e., if F is conservative, the line integral is independent of path. EE2007/Ling KV/Aug 2006 87
Verification of Stokes s Theorem Let F = [y, z, x] and S paraboloid z = 1 (x 2 + y 2 ), z 0 the The curve C is the circle r = [cos s, sin s, 0], 0 s 2π. Consequently, the line integral is simply 2π F dr = [sin s, 0, cos s] [ sin s, cos, 0] ds = π C 0 On the other hand, F = i j k x y z y z x = [ 1, 1, 1] EE2007/Ling KV/Aug 2006 88
and N = (z 1 + (x 2 + y 2 )) = [2x, 2y, 1] so that = = 1 = = π S R R r=0 ( F) n da [ 1, 1, 1] [2x, 2y, 1] dx dy ( 2x 2y 1) dx dy ( 2r cos θ 2r sin θ 1)r dθ dr 2π θ=0 EE2007/Ling KV/Aug 2006 89