Session 6
Vector Analysis: Gradient, Divergence, Curl
1
Introduction
Concepts from vector analysis are useful when investigating functions of more than one
variable. Vector operators such as the divergence or the curl are physically meaningful and
widely employed in applications. This session reviews the basics of vector analysis.
A scalar is a quantity a characterized by its magnitude. Specification of a vector v
requires stating its direction as well as its magnitude |v| = v. Vectors u and v can be added
w =u+v
Vector addition is commutative and associative. Vectors can also be multiplied by scalars.
Unit vectors are vectors of unit length while the zero vector has zero length and arbitrary
direction. A set of linearly independent unit vectors in 3D Euclidean space are called unit
vectors. In Cartesian coordinates these are written as i, j and k. Any other vector can then
be expressed by stating its scalar components, vx = v cos α, vy = v cos β and vz = v cos γ,
(the projections of the vector on the directions of the unit vectors), i.e.
v = vx i + vy j + vz k
Here α, β and γ are the direction cosines of v while the magnitude of v is
v=
q
vx2 + vy2 + vz2
The direction cosines satisfy the equation
cos2 α + cos2 β + cos2 γ = l2 + m2 + n2 = 1
and the unit vector in the direction of v, v1 is
v1 =
vy
vz
vx
i+ j+ k
v
v
v
1
2
Scalar, Vector and Multiple Products
The scalar (or dot) product of two vectors a and b is
a · b = ax bx + ay by + az bz = ab cos θ
is a scalar quantity. It is equal to the length of b times the magnitude of the projection of
a onto b and it is also equal to the length of a times the magnitude of the projection of b
onto a. Here θ is the angle between the vectors.
The dot product is commutative and distributive. Also, the dot product of orthogonal
(i.e. linearly independent) vectors is zero. The dot product of a vector with itself is the
square of its magnitude
The vector (or cross) product of two vectors a and b is
a × b = (ay bz − az by )i + (az bx − ax bz )j + (ax by − ay bx )k = c
This vector c is directed perpendicularly to the plane of the two vectors and has the magnitude
|c| = c = |a × b| = ab| sin θ|
The vector product is not commutative but it is distributive. Also, the vector product of
two parallel vectors is zero.
The tangential velocity vector in rotating systems and the moment vector in mechanics
are examples of the above.
Three multiple products are important (a · b)c, the product of the scalar ab cos θ and the
vector c; (a × b) · c, the triple scalar product (and the volume of the parallelepiped formed
by the three vectors); and (a × b) × c, a vector in the plane of a and b, perpendicular a c.
3
Differentiation of Vectors
The derivative of a vector with respect to a parameter t is defined as
dv(t)
v(t + ∆t) − v(t)
= lim
∆t→0
dt
∆t
If v = f (t)i + g(t)j + h(t)k then
df
dg
dh
dv
= i+ j+ k
dt
dt
dt
dt
Derivative formulae for vector products are readily obtained from the above definition. Also,
note that the derivative of a vector of constant length but changing direction is perpendicular
to the vector.
2
4
Space Curves
A curve in 3D space is defined by the parametric equations
x = x(t),
y = y(t),
z = z(t)
The position vector r from the origin to a point P (x, y, z), corresponding to a specific value
of t is
r = xi + yj + zk
As the value of the parameter changes from t to t + ∆t the point P traverses a space curve.
Let s be the arc length along the curve. In the limit when ∆t → 0,
dx
dy
dz
dx
dy
dz ds
dr ds
ds
dr
= ( i + j + k) = ( i + j + k) =
=u
dt
dt
dt
dt
ds
ds
ds dt
ds dt
dt
u is the unit tangent vector to the curve and pointing in the direction of increasing arc length,
therefore
ds
=
dt
s
(
dx 2
dy
dz
) + ( )2 + ( )2
dt
dt
dt
where ds is the element of arc length of the space curve.
Further, since u is a unit vector, the rate of change of u with s,
d2 r
d2 x
d2 y
d2 z
du/dt
du
= 2 = 2i + 2j + 2k =
ds
ds
ds
ds
ds
ds/dt
is a vector perpendicular to the tangent vector. The length of this new vector is
du
| |=
ds
s
(
d2 x 2
d2 y 2
d2 z 2
)
+
(
)
+
(
)
ds2
ds2
ds2
This is called the curvature of the curve and it measures the rate of change of direction
of the curve with distance along it. The reciprocal of the curvature is the radius of curvature
ρ. Introducing the principal unit normal vector
n=
du
ds
du
| ds |
=
du
ρ
ds
yields
s
n
d2 x
d2 y
d2 z
du
= = n ( 2 )2 + ( 2 )2 + ( 2 )2
ds
ρ
ds
ds
ds
3
The two vectors u and n define a third vector b = u × n called the binormal vector.
Differentiating with respect to s
db
2
=ρ ds
dx dy dz
ds ds ds
d2 x d2 y d2 z
ds2 ds2 ds2
d3 x d3 y d3 z
ds3 ds3 ds3
1
n = − n
τ
where 1/τ is the torsion and |τ | is the radius of torsion of the curve. Finally
1
1
dn
= b− u
ds
τ
ρ
The above set of differential relations involving the three mutually perpendicular unit vectors
associated with the curve are called Frenet’s formulae.
Specific instances of the above occur in problems in kinematics of points and rigid bodies.
5
The ∇ Operator: Gradient, Divergence and Curl
Consider a scalar function of position ψ(x, y, z). The gradient vector of ψ is
grad ψ = ∇ψ = i
∂ψ
∂ψ
∂ψ
+j
+k
∂x
∂y
∂z
So, the component of ∇ψ in any direction is the rate of change of ψ with respect to distance
along that direction and grad ψ is a vector function associated with ψ in a form which is
independent of the system of coordinates being used.
The dot product of the tangent unit vector to a curve u = dr/ds and ∇ψ is
u · ∇ψ =
dψ
dr
· ∇ψ =
ds
ds
Therefore, ∇ψ points in the direction of greatest rate of change of ψ and its length is equal
to the value of that maximum derivative.
It is often convenient to think of ∇ as a vector operator, i.e.
∇=i
∂
∂
∂
+j
+k
∂x
∂y
∂z
Consider now the vector function of position F(x, y, z) = Fx i + Fy j + Fz k. The scalar
and vector products of the operator ∇ into F are
∇·F=i·
∂F
∂F
∂Fx ∂Fy ∂Fz
∂F
+j·
+k·
=
+
+
= div F
∂x
∂y
∂z
∂x
∂y
∂z
4
which is called the divergence of F and
∂F
∂F
∂F
+j×
+k×
=
∂x
∂y
∂z
∂Fx ∂Fz
∂Fy ∂Fx
∂Fz ∂Fy
−
) + j(
−
) + k(
−
) = curl F
i(
∂y
∂z
∂z
∂x
∂x
∂y
∇×F=i×
which is called the curl of F.
The derivative of F in the direction of a unit vector u is
dF
= (u · ∇)F
ds
where
u · ∇ = ux
∂
∂
∂
+ uy
+ uz
∂x
∂y
∂z
The following is a collection of useful differentiation formulae.
∇ · ψu = ψ∇ · u + u · ∇ψ
∇ × ψu = ψ∇ × u + ∇ψ × u
∇·u×v =v·∇×u−u·∇×v
∇ × (u × v) = v · ∇u − u · ∇v + u(∇ · v) − v(∇ · u)
∇(u · v) = u · ∇v + v · ∇u + u × (∇ × v) + v × (∇ × u)
∇ × (∇ψ) = curl (grad ψ) = 0
∇ · (∇ × u) = div (curl u) = 0
∇ · (∇ψ1 × ∇ψ2 ) = 0
∇ × (∇ × u) = curl (curl u) = grad (div u) − ∇2 u
where
∇2 =
∂2
∂2
∂2
+
+
∂x2 ∂y 2 ∂z 2
5
is the Laplacian operator and
∂2
∂2
∂2
+
+
)(ux i + uy j + uz k) =
∂x2 ∂y 2 ∂z 2
∂ 2 ux ∂ 2 ux ∂ 2 ux
∂ 2 uy ∂ 2 uy ∂ 2 uy
∂ 2 uz ∂ 2 uz ∂ 2 uz
( 2 +
+
)i
+
(
+
+
)j
+
(
+
+
)k
∂x
∂y 2
∂z 2
∂x2
∂y 2
∂z 2
∂x2
∂y 2
∂z 2
∇2 u = (
Finally, for the position vector r the following are valid
∇·r=3
∇×r=0
u · ∇r = u
where u is any vector.
6
Practice Problems
6.1
The temperature at a point in space is T = xy + yz + zx.
a) Find the direction in which the temperature changes most rapidly with distance from
(1, 1, 1). What is the maximum rate of change?
b) Find the derivative of T in the direction of the vector 3i − 4k at (1, 1, 1).
Answer: a) Here ∇T =√(y + z)i + (x + z)j + (y + x)k. The maximum rate of change at
(1, 1, 1) is |∇T (1, 1, 1)| = 2 3 and direction cosines are
1
1
1
∇T
= √ i + √ j + √ k = cos αi + cos βj + cos γk
|∇T |
3
3
3
b) The required derivative is
∇T (1, 1, 1) ·
3i − 4k
2
=−
|3i − 4k|
5
6
6.2
For each of the following vector functions F, determine whether ∇φ = F has a solution and
determine it if it exists.
a) F = 2xyz 3 i − (x2 z 3 + 2y)j + 3x2 yz 2 k
b) F = 2xyi + (x2 + 2yz)j + (y 2 + 1)k
Answer:
a) Here ∇φ = F requires ∇ × F = 0 which is not the case here, so no solution.
b) Here ∇ × F = 0 so that
φ(x, y, z) = x2 y + y 2 z + z + c
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