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EE2007: Engineering Mathematics IIVector 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 IIVector Calculus
My part:
week 10 Introduction, Dot and Vector Products, Scalar Triple Productweek 10-11 Differential Calculusweek 11-12 Integral Calculusweek 13 Revision Classes
Text: Kreyszig, E. (1999). Advanced Engineering Mathematics, Chapters 8-9, 8thEd. John Wiley & Sons, Inc. [NTU Library, QA401.K92]
Reference:
1. H. M. Schey (1997). Div, Grad, Curl and all that, an informal text on vectorcalculus, 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
r̂ =1|r|
r = .
A unit vector has magnitude , and can be used to indicate a
particular director. We can write r = |r|̂r .
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 = 12, 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 = [a1, a2, a3] and b = [b1, b2, b3] is defined as
a · b = |a||b| cos γ if a 6= 0, b 6= 0a · 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 components1, a · b = a1b1 + a2b2 + a3b3 .1see 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(6= 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| , (b 6= 0) since a · b = |a||b| cos γ.
EE2007/Ling KV/Aug 2006 11
General Properties of Inner Products
For any vectors, a,b, c and scalars q1, q2,
[q1a + q2b] · c = q1a · c + q2b · c (Linearity)a · b = b · a (Symmetry)
a · a ≥ 0a · a = 0 iff a = 0
}(Positive− definiteness)
(a + b) · c = a · c + b · c (Distributive)|a · b| ≤ |a||b| (Schwarzinequality)
|a + b| ≤ |a|+ |b| (Triangleinequality)|a + b|2 + |a− b|2 = 2(|a|2 + |b|2) (Parallelogramequality)
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.
[ 96.865o ]
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 = [a1, a2, a3] and b = [b1, b2, b3] is a vector
whose magnitude is given by
|v| = |a||b| sin γ
and the direction of v is
perpendicular to both a and band 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
a1 a2 a3
b1 b2 b3
∣∣∣∣∣∣∣Example: Let a = [4, 0,−1] and b = [−2, 1, 3]. Then
v = a× b =
∣∣∣∣∣∣∣i j k4 0 −1−2 1 3
∣∣∣∣∣∣∣ = i− 10j + 4k = [1,−10, 4].
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)
}(distributive)
b× a = −(a× b) (anti− communtative)a× (b× c) 6= (a× b)× c (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 wand 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 = [a1, a2, a3], b = [b1, b2, b3], c = [c1, c2, c3]
is denoted by (a b c) and is defined by
(a b c) = a · (b× c) =
∣∣∣∣∣∣∣a1 a2 a3
b1 b2 b3
c1 c2 c3
∣∣∣∣∣∣∣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) =
∣∣∣∣∣∣∣a1 a2 a3
b1 b2 b3
c1 c2 c3
∣∣∣∣∣∣∣= −
∣∣∣∣∣∣∣c1 c2 c3
b1 b2 b3
a1 a2 a3
∣∣∣∣∣∣∣ = c · (a× b)
=
∣∣∣∣∣∣∣b1 b2 b3
c1 c2 c3
a1 a2 a3
∣∣∣∣∣∣∣ = 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= [a1 + tb1, a2 + tb2, a3 + tb3].
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
cos2 t + sin2 t = 1, we obtain
x2
a2+
y2
b2= 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 6= 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 equationsr1(t) = [t, 1 + 2t, 3− 2t] and r2(t) = [−2− 2t, 1− 2t, 1 + t]. Do the particlescollide? 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 = +√
a2 − x2 − y2 or x2 + y2 + z2 = 0, z ≥ 0represents 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
x2 + y2 = a2, −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 sinu, 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 xy + y2 + z2 = a2 can be represented in the form
r(u, v) = [a cos v cos u, a cos v sinu, 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 Surfaces2
2Problem 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 ) = [v1(P ), v2(P ), v3(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 Po in space
is a scalar function
f(P ) = f(x, y, z) =√
(x− xo)2 + (y − yo)2 + (z − zo)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 Po will have an
attractive force p on a particle in
space at point P . Its magnitude
is proportional to 1/r2 where r is
the distance between P and Po,
p = −GMm
r2r̂
where r̂ 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 + y2
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) 6= 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
|v|2 =
and
(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(t1, t2) = a cos t1i + a sin t1j + t2k. Then
∂r
∂t1= −a sin t1i + a cos t1j
∂r
∂t2= 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
14x2 + y2 = 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 Derivativeof 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) = ddtr(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 sinωtj 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 vectorv = 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 = ru × rv
where ru and rv 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 sinu, a sin v]
so,
ru = [−a cos v sinu, a cos v cos u, 0]
rv = [−a sin v cos u,−a sin v sinu, a cos v]
and this gives
N = ru × rv = a2 cos v[cos v cos u, cos v sinu, sin v], and
|N| = a2 cos v
EE2007/Ling KV/Aug 2006 50
Hence
n =1|N|
N = [cos v cos u, cos v sinu, sin v] =1a[x, y, z]
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
function3
grad f = ∂f∂xi +
∂f∂y j + ∂f
∂zk .
Introducing the differential operator (read nabla or del),
∇ = ∂∂xi +
∂∂yj + ∂
∂zk
we write
grad f = ∇f = ∂f∂xi +
∂f∂y j + ∂f
∂zk .
For example, if f = 2x2 + yz − 3y2, then ∇f = 2i + (z − 6y)j + yk.3assuming, 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:
Dbf =∇f · b|b|
Geometrically, Dbf represents the slope of surface f(x, y, z) in the
direction of b.
Example: Given f = x2 + y2, then ∇f = 2xi + 2yj.
EE2007/Ling KV/Aug 2006 53
Example: Directional Derivative
Find the directional derivative of f(x, y, z) = 2x2 + 3y2 + z2 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,
Dbf =[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) = x2 + 9y2, 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/y2].
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− 3x2 − 5y2 (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
∂xx′(t) +
∂f
∂yy′(t) +
∂f
∂zz′(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
x2 + 3y3 + 2z3 = 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) = x2 + 3y3 + 2z3 is
∇f = [2x, 9y2, 6z2]
Recall that the equation of a plane has the form N · (r− a) = 0where N is a normal vector to the plane, and is a point on the plane.
Thus, the required equation is
[2x, 9y2, 6z2] · (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) = x2 + y2 + z2 − a2 = 0 has a unit normal
vector given by
n = 1
|∇g|∇g
= 1
2√
x2+y2+z2[2x, 2y, 2z]
= 1a[x, y, z]
This agrees with the answer given in slide 50 where the surface
normal was computed using
n =1|N|
N where N = ru × rv
EE2007/Ling KV/Aug 2006 60
Example: Gradient (∇f) as Surface Normal Vector
Find a unit normal vector n of the cone of revolution
z2 = 4(x2 + y2) at the point P : (1, 0, 2).
Solution: The surface of the cone
can be represented by
f(x, y, z) = 4(x2 + y2)− z2 = 0Thus, ∇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√5[2, 0,−1].
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 = [yex, ex, 1] gives f = yex + z + c where c is a constant.
v = [x, y]/(x2 + y2) gives f = 12 ln(x2 + y2) + 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) = [v1, v2, v3] be a differentiable vector function, then
the function
div v = ∂v1∂x + ∂v2
∂y + ∂v3∂z
is called the divergence of v.
Another common notation for the divergence of v is ∇ · v,
div v = ∇ · v = ( ∂∂xi +
∂∂yj + ∂
∂zk) · (v)= ∂v1
∂x + ∂v2∂y + ∂v3
∂z
Example: If v = [3xz, 2xy,−yz2], 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 = [v1, v2, v3]represents the velocity of the
fluid flow.
Then the rate of fluid volume
flow across surface A is
v1(A)∆y∆z. ...
EE2007/Ling KV/Aug 2006 67
Visualizing the Divergence4
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) = [v1, v2, v3] be a differentiable vector function. Then
the function
curl v = ∇× v =i j k∂∂x
∂∂y
∂∂z
v1 v2 v3
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 velocityfield of a rotating rigid body can berepresented in the form v = w×r wherewωk, 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
af(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 ∫C
F(r) · dr =∫ b
a
F(r(t)) · dr(t)dt
dt
In terms of components, with dr = [dx, dy, dz], the above becomes∫C
F(r) · dr =∫ b
a
[F1, F2, F3] · [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 by Fin the displacement along path C.
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
[F1, F2, F3] · [dx
dt,dy
dt,dz
dt] dt
with continuous F1, F2, F3 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∫C
F · dr =∫
C
[3x2, 2yz, y2] · [dx, dy, dz]
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 = ru × rv and unit normal n =1|N|
N
Then, for a given vector function F, the surface integral over S is
defined as ∫ ∫S
F · ndA =∫ ∫
R
F(u, v) ·N(u, v) du dv
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
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 = x2, 0 ≤ x ≤ 2, 0 ≤ z ≤ 3
if the velocity vector is F = [3z2, 6, 6xz].
EE2007/Ling KV/Aug 2006 80
Solution: Let x = u and z = v, we have y = u2 and hence a
parametric representation of S is
S : r(u, v) = [u, u2, v], 0 ≤ u ≤ 2, 0 ≤ v ≤ 3
Thus,ru = [1, 2u, 0]rv = [0, 0, 1]N = ru × rv = [2u,−1, 0]F(u, v) = [3v2, 6, 6uv]
Hence∫ ∫SF · n dA =
∫ ∫R
F(u, v) ·N(u, v) du dv
=∫ 3
v=0
∫ 2
u=0[3v2, 6, 6uv] · [2u,−1, 0] du dv
= 72
EE2007/Ling KV/Aug 2006 81
Example: Surface Integral
If F = [x2, 0, 3y2] 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 isS : r(u, v) = [u, v, 1− u− v],
0 ≤ u ≤ 1− v, 0 ≤ v ≤ 1
and
N = ru × rv = [1, 0,−1]× [0, 1,−1] = [1, 1, 1]
Hence∫ ∫SF · n dA =
∫ ∫R
F(u, v) ·N(u, v) du dv
=∫ 1
v=0
∫ 1−v
u=0[u2, 0, 3v2] · [1, 1, 1] du dv
= 13
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 =∫ ∫
S
F · n dA
where n is the outer unit normal vector of S.
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 DivergenceTheorem
Evaluate∫ ∫
SF · n dA where F = [x3, x2y, x2z], S is the closed
surface consisting of the cylinder x2 + y2 = a2, 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∫ ∫S
(∇× F) · n dA =∫
C
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.
EE2007/Ling KV/Aug 2006 86
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 is independent of path.
EE2007/Ling KV/Aug 2006 87
Verification of Stokes’s Theorem
Let F = [y, z, x] and S the
paraboloid
z = 1− (x2 + y2), z ≥ 0
The curve C is the circle r = [cos s, sin s, 0], 0 ≤ s ≤ 2π.
Consequently, the line integral is simply∫C
F · dr =∫ 2π
0
[sin s, 0, cos s] · [− sin s, cos, 0] ds = −π
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 + (x2 + y2)) = [2x, 2y, 1]
so that ∫ ∫S(∇× F) · n dA
=∫ ∫
R[−1,−1,−1] · [2x, 2y, 1] dx dy
=∫ ∫
R(−2x− 2y − 1) dx dy
=∫ 1
r=0
∫ 2π
θ=0(−2r cos θ − 2r sin θ − 1)r dθ dr
= −π
EE2007/Ling KV/Aug 2006 89
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