EE2007: Engineering Mathematics II Vector Calculus

EE2007: Engineering Mathematics II Vector Calculus Ling KV School of EEE, NTU ekvling@ntu.edu.sg Rm: S2-B2a-22 Ver: August 28, 2010 Ver 1.6: Martin Adams, Sep 2009 Ver 1.5: Martin Adams, August 2008 Ver 1.4: Martin Adams, October 2007 Ver 1.0: Ling KV, Jul 2005 EE2007/Ling KV/Sep 2010 My part: EE2007: Engineering Mathematics II Vector Calculus Topic 1 Introduction, Dot and Vector Products Topic 2 Differential Calculus Topic 3 Integral Calculus 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/Sep 2010 1

Overview This module deals with vectors and vector functions in 3D-space and extends 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/Sep 2010 2 Introduction Length, temperature and voltages are quantities determined by their magnitudes. They are scalars. Force and velocity are quantities determined by both magnitude and direction. They are vectors. A vector is represented as an arrow or a directed line segment. EE2007/Ling KV/Sep 2010 3

In printed work, we denote vectors by lowercase boldface letters, a, v, etc; in handwritten work, we often denote vectors by underlining a, b or with arrows, a, v. We sometimes also write the vector as an order pair of real numbers and use a boldface letter to emphasise that it represents a vector. Thus r = [x, y, z]. The vector r has magnitude r = x 2 + y 2 + z 2 and the unit vector in the direction of r is ˆr = 1 r r = 1 (xi + yj + zk). x 2 + y 2 + z 2 A unit vector has magnitude = 1, and can be used to indicate a particular direction. We can write r = r ˆr. EE2007/Ling KV/Sep 2010 4 Equality of Vectors Two vectors a and b are equal, written a = b, if they have the same length and the same direction. Hence, a vector can be arbitrarily translated, that is, the initial point can be chosen arbitrarily. This definition is practical in connection with forces and other applications. EE2007/Ling KV/Sep 2010 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/Sep 2010 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/Sep 2010 7

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 scalar. 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/Sep 2010 8 Inner Product The inner product of two nonzero vectors is zero if and only if these vectors are perpendicular. [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/Sep 2010 9

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 length 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/Sep 2010 10 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 a a 0 a a = 0 iff a = 0 } (Symmetry) (Positive-definiteness) (a + b) c = a c + b c (Distributive) a b a b (Schwarz inequality) a + b a + b (Triangle inequality) EE2007/Ling KV/Sep 2010 11

Dot Product Example If a = [1, 2, 0] and b = [3, 2, 1], find the angle between the two vectors. Solution: a = i + 2j; b = 3i 2j + k As shown 3 slides ago: cos γ = a b a b a b = (1 3) + (2 2) + (0 1) = 1; a = 1 2 + 2 2 + 0 2 = 5; b = 3 2 + ( 2) 2 + 1 2 = 14. Therefore: cos γ = 1 5 14 = 1 70 γ = 96.865 o EE2007/Ling KV/Sep 2010 12 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. Solution Displacement vector Force vector p = 2i + 6j + 6k AB = (5 3)i + (8 4)j + (0 0)k = 2i + 4j. Work done by a force = component of force in direction travelled distance travelled = p AB cos θ where θ is the angle between vectors p and AB. Hence: Work done = p AB = (2 2) + (6 4) + (6 0) = 28. EE2007/Ling KV/Sep 2010 13

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/Sep 2010 14 In component form, i j k v = a b = 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/Sep 2010 15

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 commutative) (not associative) EE2007/Ling KV/Sep 2010 16 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 3 (a b c) and a (b c) = c (a b) EE2007/Ling KV/Sep 2010 17

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/Sep 2010 18 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/Sep 2010 19

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. EE2007/Ling KV/Sep 2010 20 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/Sep 2010 21

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. Proof Since x = a cos t and y = b sin t, in other words, cos t = x a and sin t = y b. Since cos 2 t + sin 2 t = 1, we obtain x 2 a 2 + y2 b 2 = 1, z = 0 which is the equation of an ellipse, centre the origin, with major and minor axes lengths a and b. EE2007/Ling KV/Sep 2010 22 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 elliptical helix. EE2007/Ling KV/Sep 2010 23

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 r 1 (t) = r 2 (t). t = 2 2t, 1 + 2t = 1 2t, 3 2t = 1 + t Since there is no common solution, the particles do not collide Their paths cross if they pass through the same point at two possibly different times, t 1 and t 2. t 1 = 2 2t 2, 1 + 2t 1 = 1 2t 2, 3 2t 1 = 1 + t 2 t 1 = 2, t 2 = 2. EE2007/Ling KV/Sep 2010 24 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 a 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 are often, in general denoted as u and v. EE2007/Ling KV/Sep 2010 25

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/Sep 2010 26 Example: Parametric Representation of any point within the surface which is a circle of radius a. r(u, v) = [u cos v, u sin v, 0] or, more usual notation for polar coordinates could be: r(r, θ) = y r [r cos θ, r sin θ, 0] θ x where the r represents the radial distance of the point from the origin and θ the angle it makes with the x axis. r and θ vary in the rectangle R : 0 r a; 0 θ 2π EE2007/Ling KV/Sep 2010 27

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(θ, v) = [a cos θ, a sin θ, v] where the parameters θ, v vary in the rectangle R : 0 θ 2π, 1 v 1 z O θ a r( θ,v) ( v =1) y x (v = 1) EE2007/Ling KV/Sep 2010 28 Example: Parametric Representation of a Sphere A sphere x 2 + 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] or more commonly: r(θ, φ) = [a cos φ cos θ, a cos φ sin θ, a sin φ] where the parameters θ, φ vary in the rectangle z a cos φ R : 0 θ 2π, π/2 φ π/2 θ a Oθ φ r (θ, φ) y x EE2007/Ling KV/Sep 2010 29

Parametric Representations of Common Surfaces 2 2 Problem Set 9.5, p.495 EE2007/Ling KV/Sep 2010 30 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/Sep 2010 31

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/Sep 2010 32 Examples of Vector Functions F(x, y) = 1 ( ) 2 (xi + yj) G(x, y) = 1 yi + xj 2 x2 + y 2 EE2007/Ling KV/Sep 2010 33

Derivative of a Vector Function The derivative of a vector function is defined as v v(t + t) v(t) (t) = lim t 0 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/Sep 2010 34 Note that the usual rules of differentiation carry over to vector functions: (u v) = u v + u v (u v) = u v + u v (u v w) = (u v w) + (u v w) + (u v w ) EE2007/Ling KV/Sep 2010 35

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 = c 2 (v v) = = v v + v v = 2 v v = 0 Thus, the derivative of a vector function v(t) of constant length is either the zero vector or is perpendicular to v(t). EE2007/Ling KV/Sep 2010 36 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/Sep 2010 37

Tangent to a Curve In slide 34, 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/Sep 2010 38 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) = [2 cos t, sin t] Hence r (t) = [ 2 sin t, cos t] and P corresponds to t = π/4. Thus, the tangent to the ellipse is q(w) = r(t) + wr (t) = [ 2, 1/ 2] + w[ 2, 1/ 2]. EE2007/Ling KV/Sep 2010 39

Example, Helix, its Tangent Vector and Derivative of Tangent EE2007/Ling KV/Sep 2010 40 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/Sep 2010 41

Example: Centripetal Acceleration The vector function r(t) = R cos ωti + 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 = Rω sin ωti + Rω cos ωtj is tangent to C, and its magnitude (speed ) is v = r r = Rω is constant. The angular speed is equal to ω. The acceleration vector is a = v = Rω 2 cos ωti Rω 2 sin ωtj = ω 2 r which is of a constant magnitude ω 2 R, with a direction towards the origin. The vector a is the centripetal acceleration. EE2007/Ling KV/Sep 2010 42 Surface Normal Recall that (in slide 34), 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/Sep 2010 43

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/Sep 2010 44 Hence n = 1 N N = [cos v cos u, cos v sin u, sin v] = 1 [x, y, z] a EE2007/Ling KV/Sep 2010 45

Gradient of a Scalar Field For a given 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 = 4xi + (z 6y)j + yk. 3 assuming, of course, f is differentiable EE2007/Ling KV/Sep 2010 46 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 EE2007/Ling KV/Sep 2010 47

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 = f x i + f y j + f zk = 4xi + 6yj + 2zk = [4x, 6y, 2z] and at P, f = 8i + 6j + 6k = [8, 6, 6]. Thus, D b f = f b b = [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. Geometrically, D b f represents the slope of surface f(x, y, z) in the direction of b. EE2007/Ling KV/Sep 2010 48 Example: Gradient and Directional Derivative See Problem Set 8.9. 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/Sep 2010 49

f Characterises Maximum Increase Let f(p ) = f(x, y, z) be a scalar function having continuous first partial derivatives. Then f exists and its length and direction are independent of the particular choice of Cartesian coordinates in space. If at 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/Sep 2010 50 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 in 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/Sep 2010 51

where r (t) = [x (t), y (t), z (t)], which is a tangent vector of C. Since C lies in S, this vector is tangent to S. What this means is that the vector f = f x i + f y j + f z k = [ 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 if this vector is not the zero vector. EE2007/Ling KV/Sep 2010 52 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 a is a point on the plane. Thus, the required equation is [2x, 9y 2, 6z 2 ] (r [1, 1, 2]) = 0 EE2007/Ling KV/Sep 2010 53

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 44 where the surface normal was computed using n = 1 N N where N = r u r v EE2007/Ling KV/Sep 2010 54 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/Sep 2010 55

Visualising Surface Normal, Contour, and Gradient EE2007/Ling KV/Sep 2010 56 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 both gravitational and electrostatic fields. EE2007/Ling KV/Sep 2010 57

(See Problem Set 8.9. p.452) Conservative Vector Field Find f for a given v or state that v has no potential. EE2007/Ling KV/Sep 2010 58 Example If v = [yz, xz, xy], find the corresponding potential function f(x, y, z), or state that v has no potential. Solution v = f therefore: yzi + xzj + xyk = f x i + f y j + f z k i.e. f x = yz f = xyz + f 1(y, z) f y = xz f = xyz + f 2(x, z) f z = xy f = xyz + f 3(x, y) For all 3 versions of f to agree, f 1 (y, z) = f 2 (x, z) = f 3 (x, y). The only possible solution is that f 1 = f 2 = f 3 = c (Constant). Hence f(x, y, z) = xyz + c. EE2007/Ling KV/Sep 2010 59

Example If v = [ye x, e x, 1], find the corresponding potential function f(x, y, z), or state that v has no potential. Solution v = f therefore: ye x i + e x j + k = f x i + f y j + f z k i.e. f x = yex f = ye x + f 1 (y, z) f y = ex f = ye x + f 2 (x, z) f z = 1 f = z + f 3(x, y) For all 3 versions of f to agree, f 1 (y, z) = z + c, f 2 (x, z) = z + c and f 3 (x, y) = ye x + c. Hence f(x, y, z) = ye x + z + c. EE2007/Ling KV/Sep 2010 60 Can you show that v = [x, y]/(x 2 + y 2 ) gives f = 1 2 ln(x2 + y 2 ) + c where c is a constant? EE2007/Ling KV/Sep 2010 61

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/Sep 2010 62 Physical example: The Gradient Vector (x,y ) 2 2 f (x,y ) 2 2 f(x,y) f * 140 100 60 20 (x,y ) 1 1 f (x,y ) 1 1 Bukit Timah Hill 164m Singapore (!) Ordnance Survey Map f(x 1, y 1 ) = steepest descent at point (x 1, y 1 ) etc. EE2007/Ling KV/Sep 2010 63

Physical example: The Gradient Vector f(x,y) = 1020 f(x,y) = Isobars 1030 1040 1050 H f (x,y) = Wind! 1000 990 980 970 L f(x, y) = wind vector (strength and direction) at (x, y). EE2007/Ling KV/Sep 2010 64 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 = 3z + 2x 2yz. Note that v is a scalar. EE2007/Ling KV/Sep 2010 65

Physical Meaning of 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 the surface perpendicular to the y axis is v 1 x z EE2007/Ling KV/Sep 2010 66 Visualising 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/Sep 2010 67

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/Sep 2010 68 Summary, the Operator operates on scalar field and gives f - a vector. operates on vector field gives F (a scalar) or F (a vector). 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 (eg. the wind!). The Gravitational and Electrostatic potentials are examples of scalar fields for which the gradient field is physically significant. EE2007/Ling KV/Sep 2010 69

Identities about Curl and Divergence curl (grad f) = ( f) = 0 div (curl F) = ( F) = 0 EE2007/Ling KV/Sep 2010 70 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/Sep 2010 71

Example: Line Integral [pp.466] Find the value of the line integral F dr when F(r) = [ y, xy] and C C is the circular arc from A to B. The curve C is represented by r = [cos θ, sin θ], 0 θ π 2 Hence, dr dθ = [ sin θ, cos θ] F = [ y, xy] = [ sin θ, cos θ sin θ] C F dr = π 2 θ=0 F dr π dθ dθ = 2 [ sin θ, cos θ sin θ] [ sin θ, cos θ]dθ θ=0 If F represents a force, then the line integral gives the work done by F in the displacement along path C. EE2007/Ling KV/Sep 2010 72 [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). r(t) = [cos t, sin t, 3t], 0 t 2 π dr dt = [ sin t, cos t, 3 ] F = [z, x, y] = [3t, cos t, sin t] C F dr = 2π t=0 F dr dt dt = 2 π t=0 [3t, cos t, sin t] [ sin t, cos t, 3 ]dt = 7π EE2007/Ling KV/Sep 2010 73

[pp.467] Dependence of line integral on path F = [5z, xy, x 2 z] C1 : r 1 (t) = [t, t, t] dr 1 (t) = [1, 1, 1]dt C2 : r 2 (t) = [t, t, t 2 ] dr 2 (t) = [1, 1, 2t]dt C1 C2 F dr = F dr = 1 t=0 1 t=0 [5t, t 2, t 3 ] [1, 1, 1 ]dt = 37 12 [5t 2, t 2, t 4 ] [1, 1, 2t]dt = 28 12 EE2007/Ling KV/Sep 2010 74 A line integral Line Integrals and Independence of Path 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. Hence, if f can be found, then F dr = f(b) f(a) C ie. There is no need to evaluate the integral - just substitute b and a into the scalar function f and subtract them!!! EE2007/Ling KV/Sep 2010 75

Example: Independence 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. Hence - find f(x, y, z) according to the method on page 58. Then just substitute B : (1, 1, 7) into f(x, y, z) and then A : (0, 1, 2) into f(x, y, z) and subtract them to give the answer. Note that any constants in f(x, y, z) will cancel each other out after subtraction. EE2007/Ling KV/Sep 2010 76 Try it!!! EE2007/Ling KV/Sep 2010 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/Sep 2010 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/Sep 2010 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/Sep 2010 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/Sep 2010 81

Surface Area of a Sphere Have you ever wondered why the surface area of a sphere of radius a is accepted as being 4πa 2? It can be proved with a surface Integral! EE2007/Ling KV/Sep 2010 82 Surface Area of a Sphere (Cont d) In the limit, as δθ 0 and δφ 0, area of shaded region da aδφ (a cos φδθ). Hence total surface area A of sphere is: A = 2π θ=0 π 2 φ= π 2 a 2 cos φdφdθ 2π = a 2 [sin φ] π 2 πdθ 2 θ=0 = 2a 2 [θ] 2π 0 = 4πa 2 x δφ z θ O θ a cos φ δθ a φ r( θ, φ) δθ y EE2007/Ling KV/Sep 2010 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 Volumetric and surface integrals are related through the divergence theorem of Gauss. This is of practical interest because one of the two integrals is often much simpler to solve than the other. EE2007/Ling KV/Sep 2010 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: Easier if we use the VOLUMETRIC integral as F = 3x 2 + x 2 + x 2 = 5x 2 (a scalar). Hence T (F ) dv = From diagram: dv = rdθdrdz, x = r cos θ. T 5x 2 dv x a z O θ r dr dz dθ ( z =b) y a EE2007/Ling KV/Sep 2010 85

Hence by the Divergence Theorem: z = S b z=0 F n da = a r=0 2π θ=0 T 5x 2 dv 5(r cos θ) 2 rdθdrdz x a O θ r dr dz dθ ( z =b) y a = 5 b a 2π z=0 r=0 θ=0 r 3 cos 2 θdθdrdz...and solve! EE2007/Ling KV/Sep 2010 86 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/Sep 2010 87

Stokes s Theorem and Path Independence If F = 0, then by Stokes s theorem, S ( F) n da = C F dr = 0 i.e., if F is conservative, the line integral around any closed path will be zero, and between any two different points will be independent of path. EE2007/Ling KV/Sep 2010 88 Verification of Stokes s Theorem Let F = [y, z, x] and S the paraboloid z = 1 (x 2 + y 2 ), z 0 Line Integral: The curve C is the circle r = [cos s, sin s, 0], 0 s 2π and therefore dr = [ sin s, cos s, 0]. Consequently, the line integral is simply 2π F dr = [sin s, 0, cos s] [ sin s, cos s, 0] ds = π C C F dr = 0 2π 0 sin 2 s ds = π EE2007/Ling KV/Sep 2010 89

Surface Integral: F = i j k x y z y z x = [ 1, 1, 1] and note that if we write the equation of the surface in the form f(x, y, z) = 0 we have z 1 + (x 2 + y 2 ) = 0 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/Sep 2010 90