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3.1
Introduction to Vectors, Gradient and Directional Derivative
Introduction
The underlying elements in vector analysis are vectors and scalars. We
use the notation β to denote the real line which is identified with the set
of real numbers, β2 to denote the Cartesian plane, β3 to denote the
ordinary 3-space. We denote vectors by bold face Roman letters.
Vectors
There are qualities in physics and science characterized by both
magnitude and direction, such as displacement, velocity, force, and
acceleration. To describe such quantities, we introduce the concept of a
vector as a directed line segment ππ from one point π to another point
π. Here P is called the initial point or origin of ππ , and Q is called the
terminal point, end, or terminus of the vector.
We will denote vectors by bold-faces letters or letters with an arrow
over them. Thus the vector ππ may be denoted by π¨ or π΄ as in following
figure. The magnitude or length of the vector is then denoted by
ππ , π΄ , π΄ , ππ π΄.
The following comments apply.
(a) Two vectors π¨ and π© are equal if they have the same magnitude
and direction regardless of their initial point. Thus π¨ = π© in
following 1st figure.
(b) A vector having direction opposite to that of a given vector π¨ but
having the same magnitude is denoted by β π¨ [see 2nd following
figure] and is called the negative of π¨.
Scalars
Other quantities in physics and science are characterized by
magnitude only, such as mass, length, and temperature. Such quantities
are often called scalars to distinguish them from vectors. However, it must
be emphasized that apart from units, such as feet, degrees, etc., scalars are
nothing more than real numbers. Thus we can denote them, as usual, by
ordinary letters. Also, the real numbers 0 and 1 are part of our set of
scalars.
Vector Algebra
There are two basic operations with vectors:
(a) Vector Addition; (b) Scalar Multiplication.
π¨ π¨
π© -A
(a) Vector Addition
Consider vectors π¨ and π©, pictured in Fig(a). The sum or resultant of π¨
and π©, is a vector πͺ formed by placing the initial point of π© on the
terminal point of π¨ and them joining the initial point of π¨ to the terminal
point of π©, pictured in fig(b). The sum πͺ is written πͺ = π¨ + π©. This
definition here is equivalent to the Parallelogram Law for vector addition,
pictured in fig(c).
Extensions to sums of more than two vectors are immediate. Consider,
for example, vectors π¨, π©, πͺ, π« in fig(a). Then fig(b) shows how to obtain
the sum of resultant E of the vectors π¨, π©, πͺ, π«, that is, by connecting the
end of each vector to the beginning of the next vector.
π©
π¨
(π) (π)
π¨
π©
C=A+B
(π)
π¨
π©
(π)
π¨
πͺ
π©
π«
(π)
π¨
π©
π«
πͺ
π¬ = π¨ + π© + πͺ + π«
The difference of vectors π¨ and π©, denoted by π¨ β π©, is that vector πͺ,
which added to π©, gives π¨. equivalently, π¨ β π©, may be defined as
π¨ + βπ© .
If π¨ = π©, then π¨ β π© is defined as the null or zero vector; it is
represented by the symbol π. It has zero magnitude and its direction is
undefined. A vector that is not null is a proper vector. All vectors will be
assumed to be proper unless otherwise stated.
(b) Scalar Multiplication
Multiplication of a vector π¨ by a scalar π produces a vector ππ¨ with
magnitude π times the magnitude of π¨ and the direction of ππ¨ is in
the same or opposite of π¨ according as m is positive or negative. If
π = 0, then ππ¨ = 0, the null vector.
Laws of Vector Algebra
The following theorem applies
Theorem: Suppose π¨, π©, πͺ are vectors and π and π are scalars. Then the
following laws hold:
π΄1 π¨ + π© + πͺ = π¨ + π© + πͺ Associative Law for
Addition
π΄2 There exists a zero vector π such that, for every vectorπ΄.
π¨ + π = π + π¨ = π¨ Existence of Zero Element
π΄3 For ever vector π¨, there exists a vector β π¨ such that
π¨ + βπ¨ = βπ¨ + π¨ = π Existence of Negatives
π΄4 π¨ + π© = π© + π¨ Commutative Law for
Addition
π1 π(π¨ + π©) = ππ¨ + ππ© Distributive Law
π2 π + π π¨ = ππ¨ + ππ¨ Distributive Law
π3 π ππ¨ = (ππ)π¨ Associative Law
π4 1 π¨ = π¨ Unit Multiplication
The above eight laws are the axioms that define an abstract structure
called a Vector space.
The above laws split into two sets, as indicated by their labels. The first
four laws refer to vector addition. One can then prove the following
properties of vector addition.
a) Any sum π¨π + π¨π + β― + π¨π of vectors requires no parentheses
and does not depend on the order of the summands.
b) The zero vector π is unique and the negative β π¨ of a vector π¨ is
unique
c) ( Cancellation Law) If π¨ + πͺ = π© + πͺ, then π¨ = π©.
The remaining four laws refer to scalar multiplication. Using these
additional laws, we can prove the following properties.
PROPOSITION:
a) For any scalar π and zero vector π, we have ππ = π.
b) For any vector π¨ and scalar 0, we have 0π¨ = 0.
c) If ππ¨ = 0, then π = 0 or π¨ = π.
d) For any vector π¨ and scalar π, we have βπ π¨ = π βπ¨ =
β ππ¨
Unit vectors
Unit vectors are vectors having unit length. Suppose A is any vector with
length π¨ > 0. Then π¨/ π¨ is a unit vector, denoted by a, which has the
same directions as A. Also, any very A may be represented by a unit
vectors a in the direction of A multiplied by the magnitude of A .That is,
π¨ = π¨ π.
Example: Suppose π¨ = 3. Then π = π¨ /3 is a unit vector in the
directions of A. Also , π¨ = 3π.
Rectangular Unit Vectors i, j, k
An important set of unit vectors, denoted by i, j and k, are those having
the directions, respectively, of the positive π₯, π¦, πππ π§ axes of three-
dimensional rectangular coordinate system. The coordinate system shown
in fig, which we use unless otherwise stated, is called a right handed
coordinate system. The system is characterized by the following property.
If we curl the fingers of the right hand in the direction of a 900rotation
from the positive π₯ β ππ₯ππ to the positive π¦ β ππ₯ππ , then the thumb will
point in the direction of the positive π§ β ππ₯ππ .
Generally speaking, suppose nonzero vectors A,B,C have the same initial
point and are not coplanar. Then A,B,C are said to form a right-hand
system or dextral system if a right-threaded screw rotated through an angle
less than 1800 from A to B will advance in the direction C as shown in
fig.
Components of a vector
Any vector A in three dimensions can be represented with an initial point
at the origin 0 = 0,0,0 and its end point at some point, say,
(π΄1 , π΄2 , π΄3), then the vectors, π΄1π, π΄2π, π΄3π are called the component
vectors of A in the π₯, π¦, π§ directions, and the scalars π΄1 , π΄2 , π΄3 are called
the components of A in the x, y, z directions , respectively
The sum of π΄1π, π΄2π, πππ π΄3π is the vector A , so we may write
π¨ = π΄1π + π΄2π + π΄3π
π₯
π¦
π§
π π
π π
(π)
π¨ π©
πͺ
(π)
π
π₯
π§
π¦
π¨
π΄3π
π΄2π
π΄1π
(π)
The magnitude of A follows
π¨ = π΄12 + π΄2
2 + π΄32
Consider a point π π₯, π¦, π§ in space. The vector r form the origin 0 to the
point π is called the position vector (or radius vector). Thurs r may be
written
π = π₯π + π¦π + π§π
It has magnitude π = π₯2 + π¦2 + π2
The following proposition applies,
Proposition: Suppose π¨ = π΄1π + π΄2π + π΄3π and π© = π΅1π + π΅2π + π΅3π.
Then
i. π¨ + π© = π΄1 + π΅1 π + π΄2 + π΅2 π + π΄3 + π΅3 π
ii. ππ¨ = π π΄1π + π΄2π + π΄3π = ππ΄1 π + ππ΄2 π + ππ΄3 π
Example: Suppose π¨ = 3π + 5π β 2π and π© = 4π β 8π + 7π
a. To find π¨ + π© and corresponding components , obtaining π¨ + π© =
7π β 3π + 5π
b. To find 3π¨ β 2π©, first multiply by the scalars and then add:
3π¨ β 2π© = 9π + 15π β 6π + β9π + 16π β 14π
= π + 31π β 20π
c. To find π¨ and π© , take the square root of the sum of the square
of the components:
π¨ = 9 + 25 + 4 = 38 and π© = 16 + 64 + 49 = 129
The DOT and CROSS Product
Definition: Let π and π be two vectors. The scalar product or dot product
of π and π is defined to be π. π = ππcosπ where π is the angle between
the two vectors when drawn from a common origin.
Note:
π. π = π. π.
π. π = π 2 = π2.
π. π = 0 if π and π are perpendicular vectors.
π. π + π = π. π + π. π.
π. π = π. π = π. π = 1.
π. π = π. π = π. π = 0.
If π = π1π + π2π + π3π and π = π1π + π2π + π3π then π. π =
π1π1 + π2π2 + π3π3.
Definition: Let π, π be two nonzero vectors. Then the vector product or
cross product of π and π is a vector perpendicular to both π and π with
magnitude ππ sin π where 0 β€ π β€ π is the angle between π and π and
whose direction is along a unit vector π such that π, π, π from a right
handed system. Thus π Γ π = ππ sin π π
Note:
π Γ π = Area of the parallelogram with π and π as adjacent sides.
π Γ π = β(π Γ π).
π Γ π = 0 if π and π are parallel.
π Γ π + π = π Γ π + π Γ π.
π Γ π = π Γ π = π Γ π = 0.
π Γ π = π = βπ Γ π, π Γ π = π = βπ Γ π, π Γ π = π = βπ Γ π.
If π = π1π + π2π + π3π and π = π1π + π2π + π3π then π Γ π =
π2π3 β π3π2 π + π3π1 β π1π3 π + π1π2 β π2π1 π.
= π π ππ1 π2 π3
π1 π2 π3
.
Definition: The scalar triple product or box product of three vectors
π, π, π is defined to be the scalar π. π Γ π . It is sometimes denoted by
π π π .
π. π Γ π = π π π =
π1 π2 π3
π1 π2 π3
π1 π2 π3
.
Note:
π. π Γ π =Volume of the parallelepiped turned by the
coterminous edges π, π, π.
π π π = π π π = π π π .
π π π = β π π π = β π π π = β π π π .
The vectors π, π, π are coplanar if any only if π π π = 0.
π Γ π Γ π = π. π π β π. π π.
π Γ π Γ π = π. π π β π. π π.
π Γ π . π Γ π = π. π π. π π. π π. π
.
π Γ π Γ π Γ π = π π π π β π π π π .
Scalar and vector fields:
Scalar filed: A scalar function πΉ(π₯, π¦, π§) defined over some region of
space π· is a function that assigns to each point π0 in π· with coordinates
π₯0 , π¦0 , π§0 the number πΉ π0 = πΉ π₯0 , π¦0 , π§0 . The set of all numbers
πΉ(π) for all points π in π· are said to form a scalar field over π·.
Example:
The temperature at any point within or on the earthβs surface at a
certain time defines a scalar field.
The scalar function of position πΉ π₯, π¦, π§ = π₯π¦π§2 for (π₯, π¦, π§)
inside the unit sphere π₯2 + π¦2 + π§2 = 1 defines a scalar field
throughout the unit sphere.
Vector field: More general than a scalar field πΉ(π₯, π¦, π§) is a vector field
defined by a vector function π(π₯, π¦, π§) over some region of space π· that
assigns to each point π0 in π· with coordinates π₯0 , π¦0 , π§0 the vector
π π0 = π π₯0 , π¦0 , π§0 with its tail at π0.Functions of this type are called
either vector functions or vector-valued functions.
Example:
The vector-valued function π π₯, π¦, π§ = π₯ β π¦ π β (π¦ β π§)π +
π₯π¦π§ β 2 π, for (π₯, π¦, π§) inside the ellipsoid π₯2
π2 +π¦2
π2 +π§2
π2 = 1,
defines a vector field throughout the ellipsoid.
Suppose the velocity at any point within a moving fluid is known at a
certain time defines a vector field.
Limits and continuity of vector functions of a single real variable
A vector function of a single real variable π π‘ = π1 π‘ π + π2 π‘ π +
π3 π‘ π is said to have π as its limit at π‘0, written limπ‘βπ‘0
π π‘ = π, where π =
π1π + π2π + π3π.If lim π‘βπ‘0
π1 π‘ = π1 , limπ‘βπ‘0
π2 π‘ = π2 , and limπ‘βπ‘0
π3 π‘ = π3. If,
in addition, the vector function is defined at π‘0 and lim π‘βπ‘0
π π‘ = π(π‘0),
then π(π‘) is said to be continuous at π‘0 . A vector function π(π‘) that is
continuous for each π‘ in the interval π β€ π‘ β€ π is said to be continuous
over the interval. A vector function of a single real variable that is not
continuous at a point π‘0 is said to be discontinuous at π‘0 .
Differentiability and the derivative of a vector function of a single real
variable
The vector function of a single real variable π π‘ = π1 π‘ π + π2 π‘ π +
π3 π‘ π defined over the interval π β€ π‘ β€ π and π‘0π π π . Then
limπ‘βπ‘0
π π‘ βπ(π‘0)
π‘βπ‘0, it exists, is called the derivative of π(π‘) at π‘0 and is denoted
by πβ² π‘0 or ππ
ππ‘ at π‘ = π‘0 . We also say that π π‘ is differentiable at a
point π‘0 in the interval if its components are differentiable at π‘0. It is said
to be differentiable over the interval if it is differentiable at each point of
the interval, and when π π‘ is differentiable its derivative with respect to π‘
is
ππ
ππ‘=
ππ1
ππ‘π +
ππ2
ππ‘π +
ππ3
ππ‘π.
Note: Every differentiable function is continuous.
When ππ
ππ‘ is differentiable, the second order derivative π2π/ππ‘2 is
defined as π2π
ππ‘2 =π
ππ‘
ππ
ππ‘ and, in general, provided the derivates exist,
ππ π
ππ‘π =π
ππ‘
ππβ1π
ππ‘πβ1 , πππ π β₯ 2.
Properties:
Let π π‘ and π π‘ be differentiable function of π‘ over some interval
π β€ π‘ β€ π.
ππΆ
ππ‘= 0 ( C is constant vector).
π
ππ‘ ππ = π
ππ
ππ‘ ( an arbitrary constant scalar).
π
ππ‘ π Β± π =
ππ
ππ‘Β±
ππ
ππ‘.
π
ππ‘ π. π =
ππ
ππ‘. π + π.
ππ
ππ‘.
π
ππ‘ π Γ π =
ππ
πππ‘Γ π + π Γ
ππ
ππ‘.
It π π‘ is a differentiable function of π‘ and π‘ = π‘ π is a
differentiable function of π , then
ππ
ππ =
ππ
ππ‘.ππ‘
ππ .
Partial derivatives of a vector function of real variables
Let π be a vector function of scalar variable π,q, π‘. Then we write π =
π π, π, π‘ . Treating π‘ as a variable and π, π as constants, we define
limπΏπ‘β0π π ,π ,π‘+πΏπ‘ βπ π ,π ,π‘
πΏπ‘ if exists, is called partial derivative of π with
respect to π‘ and is donated by ππ
ππ‘.
Similarly, we can define ππ
ππ,ππ
ππ also.
It ππ
ππ‘ exists
π2π
ππ‘2 and π2π
ππππ‘ are defined as
π2π
ππ‘2 =π
ππ‘
ππ
ππ‘ ,
π2π
ππππ‘=
π
ππ
ππ
ππ‘ etc.
Properties:
π
ππ‘ ππ =
ππ
ππ‘π + π
ππ
ππ‘ (π is a scalar differential function ).
π
ππ‘ ππ = π
ππ
ππ‘ .
π
ππ‘ π Β± π =
ππ
ππ‘Β±
ππ
ππ‘.
π
ππ‘ π. π =
ππ
ππ‘. π + π.
ππ
ππ‘.
π
ππ‘ π Γ π =
ππ
πππ‘Γ π + π Γ
ππ
ππ‘.
Let π = πΉ1π + πΉ2π + πΉ3π, where πΉ1 , πΉ2 , πΉ3 are differential scalar
functions of more than one variable, then
ππ
ππ‘=
ππΉ1
ππ‘ π +
ππΉ2
ππ‘ π +
ππΉ3
ππ‘ π.
Level surfaces:
Let π π₯, π¦, π§ be a single valued continuous scalar function defined at
every point π β π·. Then π π₯, π¦, π§ = π (constant), defines the equation
of a surface and is called a level surface of the function. For different
values of π, we obtain different surface, no two of which intersect. For
example, if π π₯, π¦, π§ represent temperature in a medium, then
π π₯, π¦, π§ = π represents a surface on which the temperature is a
constant π. Such surfaces are called isothermal surfaces.
Example:
Find the level surface of the scalar fields in space, defined by π π₯, π¦, π§ =
π§ β π₯2 + π¦2.
Solution: We find that π π₯, π¦, π§ = π
β π§ β π₯2 + π¦2 = π β π₯2 + π¦2 = π§ β π 2
The level surfaces are cones.
Directional Derivatives and the Gradient Operator
Consider a scalar function π€ = π π₯, π¦, π§ with continuous first order
partial derivatives with respect to π₯, π¦, and π§ that is defined in same region
D of Space, and let a space curve Ξ in D have the parametric equations
π₯ = π₯ π‘ , π¦ = π¦ π‘ , πππ π§ = π§ π‘ . Then from the chain rule
ππ€
ππ‘=
ππ
ππ₯
ππ₯
ππ‘+
ππ
ππ¦
ππ¦
ππ‘+
ππ
ππ§
ππ§
ππ‘
= ππ
ππ₯π +
ππ
ππ¦π +
ππ
ππ§π .
ππ₯
ππ‘π +
ππ¦
ππ‘π +
ππ§
ππ‘π .
The first vector, denoted by gradπ =ππ
ππ₯π +
ππ
ππ¦π +
ππ
ππ§π is called the
gradient of the scalar function π expressed in terms of Cartesian
coordinates, the second vector ππ
ππ‘=
ππ₯
ππ‘π +
ππ¦
ππ‘π +
ππ§
ππ‘π is seen to be a
vector that is tangent to the space curve Ξ,consequently ππ€
ππ‘ is the scalar
product of grand π and ππ
ππ‘ at the point π₯ = π₯(π‘), π¦ = π¦ π‘ ,and π§ = π§ π‘
for any given value of π‘.
Another notation for gradπ is βπ =ππ
ππ₯ π +
ππ
ππ¦π +
ππ
ππ§ π, where the symbol
βπ is either read βdel fβ or βgrad fβ. In this notation, the vector operator
β= ππ
ππ₯+ π
π
ππ¦+ π
π
ππ§ is the gradient operator expressed in terms of
Cartesian coordinates, and if π is a suitably differentiable scalar function
of π₯, π¦, and π§.
β΄ βπ =ππ
ππ₯π +
ππ
ππ¦π +
ππ
ππ§π.
Geometrical representation of the gradient:
Let π π = π π₯, π¦, π§ be a differentiable scalar field. Let π π₯, π¦, π§ = π
be a level surface and π0 π₯0, π¦0 , π§0 be a point on it. There are infinite
numbers of smooth curves on the surface passing through the point π0.
Each of these curves has a tangent at π0 the totality of all these tangent
lines from a tangent plane to the surface at the point π0. A vector normal
to this plane at π0 is called the normal vector to the surface at this point.
Consider now a smooth curve πΆ on the surface passing through a point π
on the surface. Let π₯ = π₯ π‘ . π¦ = π¦ π‘ , π§ = π§ π‘ be the parametric
representation of the curve πΆ. Any point π on πΆ has the position vector
π = π₯ π‘ π + π¦ π‘ π + π§ π‘ π . Since the curve lies on the surface , we have
π(π₯ π‘ , π¦ π‘ , π§ π‘ = π
Then π
ππ‘π(π₯ π‘ , π¦ π‘ , π§ π‘ = 0
By chain rule, we have ππ
ππ₯
ππ₯
ππ‘+
ππ
ππ¦
ππ¦
ππ‘+
ππ
ππ§
ππ§
ππ‘= 0
Or π ππ
ππ₯+ π
ππ
ππ¦+ π
ππ
ππ§ . π
ππ₯
ππ‘+ π
ππ¦
ππ‘+ π
ππ§
ππ‘ = 0
Or βπ. πβ² π‘ = 0
Let βπ π β 0 and πβ² π‘ β 0.Now, πβ² π‘ is a tangent vector to πΆ at the
point π and lies in the tangent plane to the surface at π.Hence βπ π is
orthogonal to every tangent vector atπ. Therefore, βπ π is the vector
normal to the surface π π₯, π¦, π§ = π at the point π.
The unit normal vector is π =grad π
grad π .
Example: Find a unit normal vector to the surface π₯π¦2 β 2π₯π¦π§ = 3 at the
point (1, 4, 3).
Solution: Let π π₯, π¦, π§ = π₯π¦2 β 2π₯π¦π§ = 3
βππ
ππ₯= π¦2 β 2π¦π§,
ππ
ππ¦= 2π₯π¦ β 2π₯π§ and
ππ
ππ§= β2π₯π¦.
β΄ βπ =πβ
ππ₯π +
πβ
ππ¦π +
πβ
ππ§π
= π¦2 β 2π¦π§ π + 2π₯π¦ β 2π₯π§ π β 2π₯π¦π
βπ 1,4,3 = β8π + 2π β 8π.
The unit normal vector at (1,4,3) is
π =β8π+2πβ8π
64+4+64=
2 β4π+πβ4π
132.
Note: The angle between two surfaces π1 π₯, π¦, π§ and π2 π₯, π¦, π§ at a
point is the angle between their norals at that point.
π
π
πβ²(π‘)
π
βπ(π0)
Ξ
π
i.e. cos π = ππ .ππ
ππ . ππ . Where ππ =
βπ1
βπ1 , ππ =
βπ2
βπ2 .
Properties:
Let π and π be any two differentiable scalar fields. Then
β f Β± g = βf Β± βg.
β π1π + π2π = π1βπ + π2βπ, π1 , π2 are arbitrary constants.
β fg = fβg + gβf.
β f
g =
gβfβfβg
g2 , g β 0.
Directional derivative:
Consider a scalar function π π₯, π¦, π§ with continuous first order partial
derivatives with respect to x, y, and z that is defined in some region of
space. The partial derivatives πβ
ππ₯,πβ
ππ¦ and
πβ
ππ§ can be interpreted as the slope
(or rate of change) of π π₯, π¦, π§ along π, π and π directions, respectively.
We can also evaluate derivatives along intermediate between π, π and π.
The result is a directional derivative. The directional derivative of π at a
point π(π₯, π¦, π§) in the direction of a unit vector π is denoted by π·π π .
The directional derivative is the slope (or rate of change) of π π₯, π¦, π§ in
the direction of π.
π·π π = βπ. π.
Example: Find the directional derivative of π(π₯, π¦, π§) = π₯2 + 3π¦2 + 2π¦2
in the direction of the vector 2π β π β 2π and determine its value at the
point 1, β3,2 .
Solution: βπ = 2π₯π + 6π¦π + 4π§π and the unit vector is
π =2πβπβ2π
4+1+4=
2πβπβ2π
9=
1
3(2π β π β 2π).
Directional derivative of π in the direction π
π·π π = ββ . π
= 2π₯π + 6π¦π + 4π§π .1
3 2π β π β 2π
=4
3π₯ β 2π¦ β
8
3π§.
Directional derivative π·π π at the point 1, β3,2 is
π·π π 1,β3,2 =4
3+ 6 β
16
3= 2.
Properties:
The most rapid increase of a differentiable function π π₯, π¦, π§ at a
point π in space occurs in the direction of the vector ππ =
gradπ π . The directional derivative at π is
π·ππ π = gradπ π = ππ
ππ₯
π
2+
ππ
ππ¦
π
2+
ππ
ππ§
π
2
1
2
.
The most rapid decrease of a differentiable function π π₯, π¦, π§ at a
point π in space occurs when the vector ππ just defined in π and
gradπ are oppositely directed , so that ππ = βgradπ π the
directional derivative at π is
π·ππ π = β gradπ π = β ππ
ππ₯
π
2+
ππ
ππ¦
π
2+
ππ
ππ§
π
2
1
2
.
There is a zero local rate of change of a differentiable function
π π₯, π¦, π§ at a point π in space in the direction of any vector ππ that
is orthogonal to gradπ at π , so that ππ .grad π π = 0.