Vectors, Gradient, Divergence and Curl.
1
Introduction
A vector is determined by its length and direction. They are usually denoted
with letters with arrows on the top ~a or in bold letter a. We will use arrows.
~a
1
Two vectors are equal if they have the same length and the same direction:
*
w
~ *
~v
~v = w
~
If we are given two points in the space (p1 , p2 , p3 ) and (q1 , q2 , q3 ) then we can
compute the vector that goes from p to q as follows:
pq
~ = [q1 − p1 , q2 − p2 , q3 − p3 ]
For example if you are given the point p = (1, 0, 1) and q = (3, 2, 4) then the
vector joining p and q is:
pq
~ = [3 − 1, 2 − 0, 4 − 1] = [2, 2, 3].
The three principal directions (unitary vectors, vectors of length one) in the space
are
~i = [1, 0, 0],
~j = [0, 1, 0]
1
and
~k = [0, 0, 1]
6
~k
~i
~j
Q
Q
Q
Q
Q
s
Q
+
The length of a vector with coordinates [a1 , a2 , a3 ] is
|~a| =
q
a21 + a22 + a23
The sum of two vectors ~a = [a1 , a2 , a3 ] and ~b = [b1 , b2 , b3 ] is obtained by adding
the corresponding components ~a + ~b = [a1 + b1 , a2 + b2 , a3 + b3 ].
We have some properties:
1. Commutative: ~a + ~b = ~b + ~a
2. Associative: ~a + (~b + ~c) = (~a + ~b) + ~c
3. There exists ~0 such that:
~a + ~0 = ~0 + ~a = ~a
and
~a − ~a = ~0
We can multiply a vector by a real number, often called scalar β ∈ R:
β~a = β[a1 , a2 , a3 ] = [βa1 , βa2 βa3 ]
And we have some properties:
2
1. Distributive 1: β(~a + ~b) = β~a + β~b
2. Distributive 2: (β + γ)~a = β~a + γ~a
3. 1~a = ~a,
(−1)~a = −~a,
and
0~a = ~0.
Example 1 If ~a = [1, 2, 3] and ~b = [0, 1, 0], then compute |3~a − ~b|:
3~a − ~b = 3[1, 2, 3] − [0, 1, 0]
= [3, 6, 9 − [0, 1, 0]
= [3, 5, 9]
Then:
|3~a − ~b| =
1.1
√
32 + 52 + 92 =
√
115.
Inner product (dot product)
The inner product of the vectors ~a and ~b is a scalar (real number) defined as
follows:
~a · ~b = |~a||~b| cos α,
where α is the angle between ~a and ~b. If ~a = [a1 , a2 , a3 ] and ~b = [b1 , b2 , b3 ], then
the inner product is:
~a · ~b = a1 b1 + a2 b2 + a3 b3 .
Observe that
|~a| =
√
~a · ~a = a21 + a22 + a23 ,
and then we can compute the angle between ~a and ~b with the formula:
cos α =
~a~b
= q
|~a||~b|
a1 b 1 + a2 b 2 + a3 b 3
q
.
2
2
2
2
2
2
a1 + a2 + a3
b1 + b2 + b3
Two vectors are orthogonal if and only if its inner product is 0:
~a is orthogonal to ~b ⇔ ~a · ~b = 0
3
The inner product has some properties:
1. Commutative: ~a · ~b = ~b · ~a
2. Distributive: (β~a + γ~b) · ~c = β~a · ~c + γ~b · ~c
3. ~a · ~a and ~a · ~a = 0 if and only if ~a = 0.
Recall that there are two important inequalities with vectors:
• Cauchy-Schwarz: |~a · ~b| ≤ |~a||~b|
• Triangle Inequality: |~a + ~b| ≤ |~a| + |~b|
1.2
Vector product (cross-product)
It is denoted by ~v = ~a × ~b and it is a vector with length
|~v | = |~a × ~b| = |~a||~b| sin α,
where α is the angle between ~a and ~b, and the direction of ~v is perpendicular to
both ~a and ~b and such ~a, ~b, ~v , in this order, form a right-handed triple. (This
means that the determinant of ~a, ~b, ~v is positive).
Geometrically speaking |~a × ~b| is the area of the parallelogram with sides ~a
and ~b.
~b |~a × ~b|
-
~a
And we have some properties:
4
1. If β ∈ R then β(~a × ~b) = (β~a) × ~b = ~a × (β~b)
2. Distributives: ~a × (~b + ~c) = ~a × ~b + ~a × ~c and (~a + ~b) × ~c = ~a × ~c + ~b × ~c
3. Anticommutative: ~a × ~b = −~b × ~a
Remark: The vector product is NOT associative. Thus, in general:
~a × (~b × ~c) 6= (~a × ~b) × ~c
Let us see this with an example:
~a = [1, 0, 0], ~b = [1, 0, 0], ~c = [0, 1, 0]
Then
~b × ~c = [0, 0, 1] ⇒ ~a × (~b × ~c) = [0, −1, 0],
but
~a × ~b = [1, 0, 0] × [1, 0, 0] = [0, 0, 0] ⇒ (~a × ~b) × ~c = [0, 0, 0]
1.3
Scalar triple product
It is a scalar (real number) defined as follows:
(~a ~b ~c) = ~a · (~b × ~c)
(Notice that there are no commas in (~a ~b ~c)).
(~a ~b ~c) = ~a · (~b × ~c) = We have:
a1 a2
b1 b2
c1 c2
a3 b3 c3 It is an immediate consequence of the properties of the determinants that:
~a · (~b × ~c) = (~a × ~b) · ~c.
Geometric interpretation: The absolute value of (~a ~b ~c) is the volume of the
parallelepiped (oblique box) with ~a, ~b and ~c as edges.
5
Example 2 Find the area of the parallelogram with vertices (2, 2, 0), (9, 2, 0),(10, 3, 0),(3, 3, 0).
First, we need to find the vectors joining the vertices (2, 2, 0), (9, 2, 0) and
(2, 2, 0), (3, 3, 0) ((10, 3, 0) is not needed!). We will denote by ~a the vector from
(2, 2, 0) to (9, 2, 0) and ~b the vector from (2, 2, 0) to (3, 3, 0). Then:
~a = [9 − 2, 2 − 2, 0 − 0] = [7, 0, 0]
~b = [3 − 2, 3 − 2, 0] = [1, 1, 0].
We know that the area of the parallelogram with edges ~a and ~b is the module of
its vector product:
~ ~ ~ i j k ~
~a × b = 7 0 0 = 7~k = [0, 0, 7],
1 1 0 And therefore:
|~a × ~b| = 7.
Example 3 Find the area of the triangle with vertices (1, 0, 0), (0, 1, 0) and
(0, 0, 1).
To do this, we first compute the area of the the parallelogram that has as
edges the vector joining (1, 0, 0), (0, 1, 0) and the vector joining (1, 0, 0), (0, 0, 1)
and then we divide by 2.
~a = [0 − 1, 1 − 0, 0 − 0] = [−1, 1, 0]
and ~b = [0 − 1, 0 − 0, 1 − 0] = [−1, 0, 1].
Then:
~
i
~a × ~b = −1
−1
~j ~k 1 0 = ~i + ~j + ~k = [1, 1, 1],
0 1 so the area of the parallelogram is:
|~a × ~b| =
√
12 + 12 + 12 =
6
√
3.
√
Hence the area of the triangle is 23 .
Example 4 Find the volume of the parallelepiped determined by the vertices
(1, 1, 1),(4, 7, 2),(3, 2, 1) and (5, 4, 3)
The volume of the parallelepiped is the absolute value of ~a · (~b × ~c), where ~a,~b
and ~c are the vectors joining the points (1, 1, 1) and (4, 7, 2), (1, 1, 1) and (3, 2, 1),
and (1, 1, 1) and (5, 4, 3) respectively:
~a = [4 − 1, 7 − 1, 2 − 1] = [3, 6, 1]
~b = [3 − 1, 2 − 1, 3 − 1] = [2, 1, 0]
~c = [5 − 1, 4 − 1, 3 − 1] = [4, 3, 2].
Then:
3 6 1
~
~a · (b × ~c) = 2 1 0
4 3 2
= −16.
Hence the volume is |~a · (~b × ~c)| = 16.
2
Gradient of a Scalar Field
A vector field F is a function that takes any point in space and assign a vector
to it:
F : From points in the space −→ To vectors in the space
(x, y, z) −→ [F1 (x, y, z), F2 (x, y, z), F3 (x, y, z)]
A scalar field f is a function that takes a point in space and assigns a number to
it:
f : From points in the space −→ To real numbers
(x, y, z) −→ f (x, y, z)
7
Example 5 The function
f (x, y, z) = xyz
is a scalar field that assigns to the point in the space (x, y, z) the real number
xyz but the function
F (x, y, z) = [ p
x
x2 + y 2 + z 2
y
,p
x2 + y 2 + z 2
z
,p
x2 + y 2 + z 2
]
is a vector field that assigns to the point in the space (x, y, z) the unit position
y
x
z
vector [ p
,p
,p
].
x2 + y 2 + z 2
x2 + y 2 + z 2
x2 + y 2 + z 2
Definition 1 The gradient of a given scalar field f (x, y, z) is a vector field de~ and it is defined as follows:
noted by grad f or ∇f
∂f
∂f ~
∂f
∂f
∂f
∂f
~ =
k
,
,
= ~i + ~j +
∇f
∂x ∂y ∂z
∂x
∂y
∂z
This notation means that if we have a point in the space (x0 , y0 , z0 ) then
~ (x0 , y0 , z0 ) = ∂f |(x ,y ,z )~i + ∂f |(x ,y ,z )~j + ∂f |(x ,y ,z )~k
∇f
∂x 0 0 0
∂y 0 0 0
∂z 0 0 0
~ as a linear operator that acts on scalar fields giving vectors fields:
We can think ∇
~ : Smooth Scalar Fields −→ Vector fields
∇
h
i
∂f ∂f ∂f
~
f −→ ∇f =
,
,
∂x ∂y ∂z
∂f ∂f ∂f
,
,
give the
∂x ∂y ∂z
rates of change of f (x, y, z) in the directions of ~i, ~j, ~k respectively. It seems
From Vector Calculus we know that the partial derivatives
natural to ask what is the rate of change in any other direction. This is what the
directional derivatives represent.
Suppose that ~b = [b1 , b2 , b3 ] is a unit vector, thus it has length 1:
|~b| =
q
b21 + b22 + b23 = 1
8
Then the directional derivative of f (x, y, z) in the direction of a unit vector ~b is:
~
D~b f = ~b · ∇f,
(1)
where · is the scalar product.
If ~a = (a1 , a2 , a3 ) is a vector which is not unit then a unit vector with the
same direction than ~a is ~a :
|~a|
q
~a = 1 a21 + a22 + a23 = 1 |~a| = 1.
|~a| |~a|
|~a|
We define the directional derivative of f (x, y, z) in the direction of ~a as
~a ~
· ∇f.
|~a|
D~a f =
(2)
Please note that equation (??) is equivalent to equation (??) when ~a is a unit
vector. (The two definitions of directional derivative are the same.)
Example 6 Given f (x, y, z) = 2x2 + 3y 2 + z 2 and ~a = [1, 0, −2], evaluate the
directional derivative of f in the direction of ~a at the point (2, 1, 3).
First we compute the gradient of f:
~ = [4x, 6y, 2z]
∇f
Since ~a is not a unit vector the directional derivative is:



4x
4x



[1, 0, −2]

 [1, 0, −2] 
√
D~a f = p
·  6y  =
 6y
2
2
2

5 
1 + 0 + (−2) 
2z
2z


1

 = √ (4x − 4z)

5
And evaluating at the point (2, 1, 3) we get:
D~a f (2, 1, 3) =
8 − 12
−4
√
=√ ,
5
5
the minus sign indicates that the function f decreases in the direction of ~a.
9
~ (P ) = grad f (P ) 6= 0,
Remark: If the gradient of f at a point P is not zero ∇f
then it is a vector in the direction of maximum increase of f at P .
Definition 2 A surface is all points in (x, y, z) ∈ R3 that verifies f (x, y, z) = c,
for some smooth scalar field f and constant c.
Consider the unit sphere in R3 , S 2 :
S 2 = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = 1}.
In this case f (x, y, z) = x2 + y 2 + z 2 and c = 1.
The cone with center at (0, 0, 0) is another surface with f (x, y, z) = x2 +y 2 −z 2
and c = 0:
C = {(x, y, z) ∈ R3 : x2 + y 2 − z 2 = 0}.
Theorem 1 Let f be differentiable and S = {(x, y, z) : f (x, y, z) = c} be a
~ at a point P of the surface S is a normal
surface. Then the gradient of f , ∇f
~ (P ) 6= 0).
vector to S at P. (Provided ∇f
Example 7 Find a unit normal vector of the cone of the revolution z 2 = 4(x2 +
y 2 ) at the point P = (1, 0, 2).
In this case:
f (x, y, z) = 4(x2 + y 2 ) − z 2
The theorem ensures that the gradient of f at the point (1, 0, 2) is normal to the
cone:
~ = [8x, 8y, −2z],
∇f
and evaluating it at P
~ (1, 0, 2) = [8, 0, −4].
∇f
10
~ (1, 0, 2) is not unit vector, so we calculate the unit vector
But we observe that ∇f
~ (1, 0, 2):
in the direction of ∇f
~ (1, 0, 2)
∇f
[2, 0, −1]
[8, 0, −4]
[8, 0, −4]
√
=
=p
= √
~ (1, 0, 2)|
80
5
|∇f
82 + (−2)2
And more examples (FOR YOU TO TRY):
1. Find the directional derivative of f at P in the direction of ~a in the following
cases:
• f (x, y, z) = x2 + y 2 − z, P = (1, 1, −2), ~a = [1, 1, 2]
• f (x, y, z) = x2 + y 2 + z 2 , P = (2, 2, −1), ~a = [−1, −1, 0]
• f (x, y, z) = xyz, P = (−1, 1, 3), ~a = [1, −2, 2]
2. Compute the normal to the surface:
• S = {(x, y, z) ∈ R3 : ax + by + cz = d} for any P .
• S = {(x, y, z) ∈ R3 : x2 + 3y 2 + z 2 = 28} and P = (4, 1, 3)
• S = {(x, y, z) ∈ R3 : x2 + y 2 = 25} and P = (4, 3, 8)
~ operator verifies:
Proposition 1 For all f, g smooth scalar fields, the ∇
~ g) = g ∇f
~ + f ∇g,
~
1. ∇(f
~ for any n ∈ N,
~ n = nf n−1 ∇f,
2. ∇f
~ f ) = 12 (g ∇f
~ − f ∇g),
~
3. ∇(
g
g
~ 2 (f g) = g ∇
~ 2 f + 2∇f
~ ∇g
~ + f∇
~ 2 g.
4. ∇
Proof:
11
1. By definition of the gradient:
~ g) = [ ∂(f g) , ∂(f g) , ∂(f g) ]
∇(f
∂x
∂y
∂z
∂f
∂g ∂f
∂g ∂f
∂g
= [ g+f ,
g+f ,
g+f ]
∂x
∂x ∂y
∂y ∂z
∂z
using the product rule for partial derivatives
∂g ∂g ∂g
∂f ∂f ∂f
,
]g + f [ , , ]
= [ ,
∂x ∂y ∂z
∂x ∂y ∂z
using linearity of vectors
~ + f ∇g,
~
= g ∇f
2. It is a consequence of the chain rule for partial derivatives:
n
n
n
~ n = [ ∂(f ) , ∂(f ) , ∂(f ) ]
∇f
∂x
∂y
∂z
∂f
∂f
∂f
~ for any n ∈ N,
= [nf n−1 , nf n−1 , nf n−1 ] = nf n−1 ∇f,
∂x
∂y
∂z
~ f ) = 12 (g ∇f
~ − f ∇g),
~
3. For ∇(
use part 1 with f and g1 and recall that
g
g
∂ 1 = −1 ∂g . (Similarly with the partial derivatives with respect to to y
∂x g
g 2 ∂x
and z).
4. Notice that
2
2
2
~ 2 (f g) = ∇
~ · ∇(f
~ g) = ∂ (f g) + ∂ (f g) + ∂ (f g)
∇
∂x2
∂y 2
∂z 2
Furthermore:
∂ ∂(f g)
∂ ∂f
∂g
∂2f
∂f ∂g
∂g
∂ 2 (f g)
=
=
g+f
=
+f
2
2g + 2
∂x
∂x
∂x ∂x
∂x
∂x ∂x
∂x
∂x
∂x
Similarly:
∂ 2 (f g)
∂2f
∂f ∂g
∂g
=
+f
2
2g +2
∂y ∂y
∂y
∂y
∂y
and
∂2f
∂f ∂g
∂g
∂ 2 (f g)
=
+f
2
2g +2
∂z ∂z
∂z
∂z
∂z
12
The result follows on adding these three equations together and observing
that
2
2
2
~2= ∂ + ∂ + ∂
∇
∂x2 ∂y 2 ∂z 2
and
∂f ∂g ∂f ∂g ∂f ∂g
f~ · ~g =
+
+
∂x ∂x ∂y ∂y ∂z ∂z
(Important) Remark: Some vector fields are gradients of scalar fields. This
means that for a vector field F~ we can find a scalar field f called potential of F~
such that
~ = [ ∂f , ∂f , ∂f ].
F~ = ∇f
∂x ∂y ∂z
In this case we say that F~ is a gradient field.
In the exercises you will be given a vector field and you will be asked to find
a potential for it:
Example 8 Is F~ (x, y, z) = [yz, xz, xy] a gradient field?
If we find a scalar field f (x, y, z) such that:
F1 (x, y, z) = yz =
∂f
∂x
(3)
F2 (x, y, z) = xz =
∂f
∂y
(4)
F3 (x, y, z) = xy =
∂f
∂z
(5)
and
then F~ is a gradient field.
If we consider y and z as constants and we integrate with respect to to x equation
(??), we obtain:
Z
Z
yzdx =
∂f
dx
∂x
xyz + C(y, z) = f (x, y, z)
13
(6)
(7)
This means that if the potential function for F~ exists then it has the shape
of xyz + C(y, z) where C(y, z) is the constant of integration that we have to
determine using equations (??) and (??). (The constant of integration must
depend on y and z because to integrate with respect to x (??) we thought y and
z as constants).
We use now equation (??) to equate with the differential of equation (??)
with respect to y :
∂
∂f
=
(xyz + C(y, z)) = xz
∂y
∂y
or
xz +
∂
C(y, z) = xz,
∂y
and we conclude that for this equality to hold we need
∂
C(y, z) = 0
∂y
That is, C(y, z) does not depend on y or C(y, z) could be written as C(z). Now
we use equation (??) and we equate to the derivative with respect to z of equation
(??):
∂
∂f
=
(xyz + C(z)) = xy
∂z
∂y
or
xy +
∂
C(z) = xy,
∂z
which implies that the constant C(z) does not depend on z so it is just a constant,
C say. We conclude that the potential for F~ is
f (x, y, z) = xyz + C,
C any constant in R.
~ .
You can double check that f (x, y, z) = xyz + C verifies that F~ = ∇f
14
y
x
z
,
,
x2 + y 2 + z 2 x2 + y 2 + z 2 x2 + y 2 + z 2
We need to find a scalar field f (x, y, z) such that:
Example 9 Find a potential for F~ =
F1 (x, y, z) =
∂f
x
2
2 =
∂x
x +y +z
(8)
F2 (x, y, z) =
y
∂f
2
2 =
∂y
x +y +z
(9)
F3 (x, y, z) =
∂f
z
2
2 =
∂z
x +y +z
(10)
2
2
and
2
Integrating with respect to x equation (??) we get that if the potential exists
must it be as follows:
f (x, y, z) =
1
ln(x2 + y 2 + z 2 ) + C(y, z).
2
(11)
Differentiating equation (??) with respect to y and equating to equation (??):
∂
y
y
C(y, z) = 2
2
2 +
∂y
x +y +z
x + y2 + z2
2
which implies that the constant C(y, z) does not depend on y. Similarly differentiating equation (??) with respect to z and equating it to equation (??)
we conclude that the constant does not depend on z either and therefore the
potential is:
f (x, y, z) =
1
ln(x2 + y 2 + z 2 ) + C,
2
C ∈ R.
Give a try to find a potential for:
1. [3x, 5y, −4z]
2. [4x3 , 3y 2 , −6z]
(The solutions are 23 x2 + 52 y 2 − 2z 2 + C and x4 + y 3 − 3z 2 + C respectively).
15
The differential operator
2
2
2
~ 2 f = 4f = ∂ f + ∂ f + ∂ f
∇
∂x2
∂y 2
∂z 2
is called Laplace Operator.
Notice
4 : Smooth Scalar Fields −→ Scalar Fields
f −→ 4f
Example 10 If f (x, y, z) = x3 + y 2 + 3xz 5 then:
∂2f
∂f
= 3x2 + 3z 5 ⇒
= 6x
∂x
∂x2
∂2f
∂f
= 2y ⇒
=2
∂y
∂y 2
∂f
∂2f
= 15xz 4 ⇒ 2 = 60xz 3
∂z
∂z
Hence:
4f = 6x + 2 + 60xz 3 ,
for example:
4f (1, 1, 1) = 6 + 2 + 60 = 68.
3
Divergence of a Vector Field
The divergence of a vector field F~ = [F1 , F2 , F3 ] is a scalar field div F~ defined as
follows:
∂F1 ∂F2 ∂F3
+
+
.
div F~ =
∂x
∂y
∂z
We can understand the divergence as a differential operator acting on smooth
vector fields that produces scalar fields:
div: Smooth Vector Fields −→ Scalar Vector Fields
F~ −→ ∂F1 + ∂F2 + ∂F3
∂x
∂y
∂z
16
Sometimes div F~ is denoted by
~ · F~ = div F~ ,
∇
~ can be considered as the ‘vector’
the dot is the scalar product dot because ∇
~ = [ ∂ , ∂ , ∂ ],
∇
∂x ∂y ∂z
so we have:
~ · F~ = [ ∂ , ∂ , ∂ ] · [F1 , F2 , F3 ] = ∂ F1 + ∂ F2 + ∂ F3 ,
∇
∂x ∂y ∂z
∂x
∂y
∂z
and the ‘products’ ∂ F1 , ∂ F2 and ∂ F3 mean that we have to differentiate F1
∂x
∂y
∂z
with respect to x, F2 with respect to y and F3 with respect to z:
∂F1
∂
F1 =
∂x
∂x
∂F2
∂
F2 =
∂y
∂y
∂
∂F3
F3 =
∂z
∂z
Remark: If f is a scalar field then we have:
div (gradf ) = 4f
or equivalently
~ · ∇f
~ = 4f
∇
Example 11 Compute the divergence of the following vector field:
F~ (x, y, z) = [x3 + y 3 , 3x2 , 3zy 2 ].
By definition:
∂(x3 + y 3 ) ∂(3x2 ) ∂(3zy 2 )
~
~
∇·F =
+
+
= 3x2 + 3y 2
∂x
∂y
∂z
17
4
The Curl of a Vector Field
The Curl of a vector field F~ is another vector defined as the following determinant:
~k ~j
~i
∂F
∂F
∂F
∂F
∂F
∂F
2
3
1
1
2
3
∂
∂
∂
~k
=
~i+
~j+
curl F~ = −
−
−
∂y
∂z
∂z
∂x
∂x
∂y
∂x ∂y ∂z F1 F2 F3 Suppose that F~ represents the velocity of a rotating body then curl F~ has the
direction of the axis of rotation and its module is twice the angular speed of the
rotation.
Definition 3 If the curl of a vector field is zero we say that the vector field is
irrotational.
Remark: The definition of irrotational vector field F~ = [F1 , F2 , F3 ] is equivalent
to have the following three conditions in the partial derivatives of F1 , F2 and F3 :
∂F1
∂F2
=
∂x
∂y
∂F1
∂F3
=
∂z
∂x
∂F2
∂F3
=
∂y
∂z
These three equations can be a short way to prove that something is not irrotational without computing the whole determinant:
Example 12 Consider F~ = [−ωy, ωx, 0], with ω > 0 since ∂F2 = ω and ∂F1 =
∂x
∂y
−ω then F~ is not irrotational.
However if we compute the determinant we obtain curl F~ = [0, 0, 2ω] which
means that the body rotates around the z axis with angular speed ω.
~ for some smooth
Lemma 1 Gradient fields are irrotationals. That is, if F~ = ∇f
scalar field f then curl F~ = 0.
18
Proof: Using the denition of curl:
~k
~j
~i
∂f ∂f ∂f
∂
∂
~
curl ∇f
= curl [ ,
,
] = ∂
∂x ∂y ∂z
∂x
∂y
∂z
∂f ∂f ∂f
∂x ∂y ∂z
=
∂ ∂f − ∂ ∂f ~i + ∂ ∂f − ∂ ∂f ~j + ∂ ∂f − ∂ ∂f ~k
∂y ∂z
∂z ∂y
∂z ∂x ∂x ∂z
∂x ∂y ∂y ∂x
=
2
2
∂2f
∂2f ~
∂ f
∂2f ~
∂ f
∂2f ~
k,
−
i+
−
j+
−
∂y∂z ∂z∂y
∂z∂x ∂x∂z
∂x∂y ∂y∂x
The result follows since the mixed partial derivatives of a smooth scalar field are
equal.
Example 13 The gravitational field p~ is the force of attraction between two
particles at points P0 = (x0 , y0 , z0 ) and P = (x, y, z). It is defined by
p~ = −
where r =
p
c
c
r = − 3 [x − x0 , y − y0 , z − z0 ]
3~
r
r
(x − x0 )2 + (y − y0 )2 + (z − z0 )2 and it is an irrotational field be-
cause the scalar field f (x, y, z) = rc is a potential for it. (Check it!)
Lemma 2 Let F~ be a smooth vector field then
div (curl F~ ) = 0
Proof:
div ( curl F~ ) =
=
∂ ∂F3 − ∂F2 + ∂ ∂F1 − ∂F3 + ∂ ∂F2 − ∂F1
∂x ∂y
∂z
∂y ∂z
∂x
∂z ∂x
∂y
∂ 2 F3 − ∂ 2 F2 + ∂ 2 F1 − ∂ 2 F3 + ∂ 2 F2 − ∂ 2 F1 = 0
∂x∂y ∂x∂z ∂y∂z ∂y∂x ∂z∂x ∂z∂y
Because if F~ is smooth then F1 , F2 , F3 are smooth (scalar fields) and the mixed
second partial derivatives of F1 , F2 , F3 are equal.
19
Example 14 Compute the curl of the following vector field F~ (x, y, z) = [ex cos y, ex sin y, 0]
~k ~j
~i
∂
∂
∂
curl F~ = ∂y
∂z ∂x
x
e cos y ex sin y 0 ∂(ex sin y) ~ ∂(ex cos y)~ (∂ex cos y) ~ (∂ex sin y)~
k+
j−
k−
i
∂x
∂z
∂y
∂z
= (ex sin y − ex (− sin y))~k = [0, 0, 2ex sin y]
=
20