9
Vector Calculus
Vector calculus is the calculus of functions that assign vectors to points in space. It can
be applied to problems such as ¿nding the work done by a force ¿eld in moving an object
along a curve or ¿nding the rate of Àuid Àow across a surface. Following is information
about vectors and the use of vectors in calculus.
Vectors
The term vector is used to indicate a quantity that has both magnitude and direction. A
vector is often represented by an arrow or a directed line segment. The length of the
arrow represents the magnitude of the vector and the arrow points in the direction of
the vector. Two directed line segments are considered equivalent in the sense that they
have the same length and point in the same direction. In other words, a vector y can
be thought of as a set of equivalent directed line segments. A two-dimensional vector
is an ordered pair d @ +d4 > d5 , of real numbers. A three-dimensional vector is an
ordered triple d @ +d4 > d5 > d6 , of real numbers. An q-dimensional vector is an ordered
q-tuple d @ +d4 > d5 > = = = > dq , of real numbers.The numbers d4 > d5 > = = = > dq are called the
components of d=
Notation for Vectors
You can enter vectors in any one of the following ways.
q-tuples within parentheses or brackets: +5> 4> 3,, +{4 > {5 > {6 ,, ^6> 5> 4`, ^{4 > {5 > {6 `
4 q matrices: 4 5 6 , 8 4 6 4: ; 5 , {4 {5 {6 {7
5
6
4
68
{4
q 4 matrices: 7 3 8,
,
7
{5
4
This Àexibility allows you to use the output of previous work as input, without undue
worry about the shape of that output. For purposes of clear exposition, you will ¿nd it
preferable to use consistent notation for vectors.
L To create a vector in matrix form
1. Click
or choose Insert + Matrix.
306
Chapter 9 Vector Calculus
2. Set the number of rows (or columns) to 4 and the number of columns (or rows) to
the dimension of the vector.
3. Enter the values for the components in the input boxes.
Vector Sums and Scalar Multiplication
The sum of two vectors ^{4 > {5 > ===> {q ` and ^|4 > |5 > ===> |q ` is de¿ned by
^{4 > {5 > ===> {q ` . ^|4 > |5 > ===> |q ` @ ^{4 . |4 > {5 . |5 > ===> {q . |q `
The product of a scalar d and a vector ^{4 > {5 > ===> {q ` is de¿ned by
d ^{4 > {5 > ===> {q ` @ ^d{4 > d{5 > ===> d{q `
L To evaluate a vector sum
1. Type the expression in mathematics mode
2. Choose Evaluate.
L Evaluate
+{4 > {5 > {6 , . +|4 > |5 > |6 , @ +{4 . |4 > {5 . |5 > {6 . |6 ,
+5> 4> 3, . +5> 6> 8, @ +7> 5> 8,
^4> 5> 6> 7=6` ^6> 8=4> 9> 3` @ ^5> := 4> 6> 7= 6`
d
f
d.f
4
6
7
.
@
@
e
g
e.g
5
4
4
L To evaluate the product of a scalar with a vector
1. Type the expression in mathematics mode
2. Choose Evaluate.
L Evaluate
d +{4 > {5 > {6 , @ +d{4 > d{5 > d{6 ,
9+5> 6> 8, @ +45> 4;> 63,
Dot Product
The dot product (or inner product) of two vectors +d4 > d5 > ===> dq , and +e4 > e5 > ===> eq , is
de¿ned by
+d4 > d5 > ===> dq , +e4 > e5 > ===> eq , @ d4 e4 . d5 e5 . . dq eq
Vectors
307
The dot is available on the Common Symbols toolbar and in the Binary Operations
drop-down panel below
.
For the following examples of dot products with q @ 6, de¿ne
3
4
4
d @ 4 5 6 , e @ C 3 D , f @ ^6> 5> 4`, and g @ +5> 4> 3,
4
with New De¿nition from the De¿ne submenu.
L Evaluate
+x> y> z, +{> |> }, @ x{ . y| . z}
d f @ 43
+4> 5> 6, +6> 5> 4, @ 43
d e @ 5
fg@ 7
Cross Product
The cross product of three-dimensional vectors d @ +d4 > d5 > d6 , and e @ +e4 > e5 > e6 , is
de¿ned by
d e @ +d5 e6 d6 e5 > d6 e4 d4 e6 > d4 e5 d5 e4 ,
To enter the cross symbol, click the cross on the Common Symbols toolbar or in the
drop-down panel below
.
For the following examples, use the vectors d @
3
4 5 6 , e @ C
f @ ^6> 5> 4`, and g @ +5> 4> 3,, as de¿ned previously.
4
4
3 D,
4
L Evaluate
d e @ +5> 7> 5,
d f @ +7> ;> 7,
3
4 3
4 3
4
=68
=;8
= 4:; 4
C =:6 D C =65 D @ C 4= 5;< 8 D
4=5
=::
= :65 8
4 5 8
8 6 8
@
f g @ +4> 5> :,
8 63 46
Triple Products
Since the cross product of two vectors produces another vector, it is possible to string
cross products together. Use the same vectors d, e, f, and g as before for these triple
vector products. Note that different choices of parentheses generally produce different
results. The default order of operations for products is from left to right.
308
Chapter 9 Vector Calculus
L Evaluate
d e f @ ^;> 7> 49`
+d e, f @ +;> 7> 49,
d +e f, @ +49> 7> ;,
d ++e f, g, @ +3> 3> 3,
+d +e f,, g @ +;> 49> 57,
4 5 8 8 6 8 : 5 ; @ 547 84 533
4 5 8 8 6 8
: 5 ; @ 547 84 533
4 5 8 8 6 8 : 5 ;
@ 6: 46< 96
To obtain intermediate results, select a subexpression that is surrounded by parentheses and press the CTRL key down while evaluating. This technique does an in-place
computation, as illustrated in the following examples.
L
CTRL
+ Evaluate, Evaluate
d +e f, @ d +5> 7> 5, @ +49> 7> ;,
+d e, f @ +5> 7> 5, f @ +;> 7> 49,
4 5 8 8 6 8
: 5 ;
@ 8 63 46 : 5 ; @ 547 84 533
L
4 5 8 8 6 8 : 5 ;
@ 4 5 8 67 8 64 @ 6: 46< 96
CTRL
+ Evaluate, CTRL + Evaluate, Evaluate
d ++e f, g, @ d ++5> 7> 5, g, @ d +5> 7> 9, @ +3> 3> 3,
+d +e f,, g @ +d +5> 7> 5,, g @ +49> 7> ;, g @ +;> 49> 57,
+d e, +f g, @ +5> 7> 5, +4> 5> :, @ +57> 49> ;,
++d e, f, g @ ++5> 7> 5, f, g @ +;> 7> 49, g @ +49> 65> 3,
Vectors
309
3
4 333
4 3
44 3
44
=68
=;8
4=68
4=;8
C =:6 D CCC =65 D C =56 DD C
=8: DD
4=5
=::
4=59
=6:8
3
4 33
4 3
44
=68
= 559 4
4=;8
=8: DD
@ C =:6 D CC 5= 443 8 D C
4=5
= 95: 8
=6:8
3
4 3
4 3
4
=68
= 766 :9
4= 797 ;
@ C =:6 D C 4= 3:9 4 D @ C = ;33 <4 D
4=5
6= ::8 8
=3 8<< <
Tip Parentheses are important. As always, careful and consistent use of mathematical
notation is in order. When in doubt, add extra parentheses to clarify an expression.
When mixing cross products with scalar products, use parentheses for clarity.
L Evaluate
+4> 3> 4, ++4> 5> 6, +6> 5> 4,, @ ;
++4> 3> 4, +4> 5> 6,, +6> 5> 4, @ ;
Without these parentheses, you obtain different results for the ¿rst product.
L Evaluate
5
6
6
+4> 3> 4, +4> 5> 6, +6> 5> 4, @ 7 7 5 8
4
+4> 3> 4, +4> 5> 6, +6> 5> 4, @ ;
+4> 5> 6, +6> 5> 4, +4> 3> 4, @ ;
Note The triple scalar product has an interesting geometric interpretation. The volume
of the parallelepiped spanned by three vectors D, E, and F is equal to mD +E F,m.
Thus, the volume of the parallelepiped spanned by +4> 4> 3,, +4> 3> 4,, and +3> 4> 4, is
given by
m+4> 4> 3, ^+4> 3> 4, +3> 4> 4,`m @ 5
In particular, this value does not depend on the order of the vectors in the triple scalar
product.
m+4> 3> 4, ^+4> 4> 3, +3> 4> 4,`m @ 5
m+3> 4> 4, ^+4> 4> 3, +4> 3> 4,`m @ 5
310
Chapter 9 Vector Calculus
Vector Norms
You can compute vector norms for every positive integer q, where
4
[
q q
nynq @
myl m
and for 4, with
nyn4 @ pd{ +myl m,
as illustrated by the following examples.
L Evaluate
n +d> e> f, n4 @ mdm . mem . mfm
n +4> 5> 4, n8 @
s
8
67
n ^;> 43> 5` n4 @ 43
n +d> e> f, n6 @
n +8> 4<> :, n; @
u
6
mdm6 . mem6 . mfm6
s
s
8
67 ; 49 <;< :4; 79: @ 6;= 798
n ^d> e> f` n4 @ pd{ +mdm > mem > mfm,
u
d 5
5
@
mdm
.
mem
e 5
d e f @
7
u
7
mdm7 . mem7 . mfm7
L To enter the two pairs of vertical lines used in the norm symbol
Click
or choose Insert + Brackets, and select the norm symbols.
Before doing the next set of examples, make the following de¿nition.
L De¿ne + New De¿nition
y @ ^6> 5> 4`
L Evaluate, Evaluate Numerically
nyn4 @ 9
nyn53 @
nyn5 @
s
53
s
47 @ 6=:74:
67;:; 65<:; @ 6=333378436
nyn9 @
s
9
:<7 @ 6=376
nyn4 @ 6
This series of examples suggests that for a vector y,
olp nynq @ nyn4
$4
q
The default nyn is the 5-norm, which is also known as the Hxfolghdq qrup.
Example 82 The area of the parallelogram in the plane with vertices +3> 3,, +d4 > d5 ,,
+e4 > e5 ,, and +d4 . e4 > d5 . e5 , is given by
n +d4 > d5 > 3, +e4 > e5 > 3, n
In particular, the area of the parallelogram spanned by the two vectors +4> 5, and +5> 4,
Gradient, Divergence, and Curl
311
is given by
n+4> 5> 3, +5> 4> 3,n @ 6
Since
D E @ nDn nEn frv where is the angle between the vectors D and E, you can use the dot product to ¿nd
the angle between two vectors.
Example 83 De¿ne D @ +4> 5> 6, and E @ +5> 4> 5, and solve the equation DE @
nDn nEn frv , to get
4s
47
Solution is: @ duffrv
:
Apply Evaluate Numerically to get @ 5=467:=
Gradient, Divergence, and Curl
Gradient, divergence, and curl are notable omissions from the Vector Calculus menu,
because they are implemented as “u”, “u”, and “u” followed by Evaluate. For
the symbol u, choose the nable
from the binary-operations symbol panel under
, or choose it directly from Common Symbols toolbar.
The Gradient
If i +{4 > {5 > = = = > {q , is a scalar function of q variables, then the vector
Ci
Ci
Ci
+f4 > f5 > = = = > fq , >
+f4 > f5 > = = = > fq , > = = = >
+f4 > f5 > = = = > fq ,
C{4
C{5
C{q4
is the judglhqw of i at the point +f4 > f5 > = = = > fq , and is denoted ui. For q @ 6, the
vector ui at +d> e> f, is normal to the level surface i+{> |> }, @ i +d> e> f, at the point
+d> e> f,.
L To compute the gradient of the function i+{> |> }, @ {|}
Evaluate u{|}.
L Evaluate
u{|} @ +|}> {}> {|,
You can also operate on the function value after de¿ning the function. For example,
if i is de¿ned by the equation i +{> |> }, @ {|} , then you can evaluate ui +{> |> },.
L Evaluate
ui +{> |> }, @ +|}> {}> {|,
312
Chapter 9 Vector Calculus
Since gradients can be calculated for scalar functions of q arbitrary variables, the
gradient operator uses all the variables that appear in the function, assuming that the
variables are ordered lexicographically. For this reason, you may have to edit the result
if you are representing arbitrary constants with letters. In the following example, if f is
a constant parameter, take the answer:
L Evaluate
u fxy . y5 z @ xy> fy> fx . 5yz> y5
and delete the ¿rst entry to get:
u fxy . y5 z @ fy> fx . 5yz> y5
In the preceding example, we regarded a function of four variables as a function of
three variables by treating one of the letters as a constant parameter. You can also regard
a function of two variables as a function of three variables. In the following example,
we regard u{| as the value of a function of three variables.
L To ¿nd the gradient for i +{> |> }, @ u{|
1. Leave the insertion point in the expression u{|.
2. Apply Evaluate.
u{| @ +|> {,
3. Place the insertion point after the { and enter zero as the third coordinate.
u{| @ +|> {> 3,
Note In physics, i represents potential energy, and ui represents force.
Divergence
A vector ¿eld is a vector-valued function. If I +{> |> }, @ ^s+{> |> },> t+{> |> },> u+{> |> },`
is a vector ¿eld, then the scalar
Ct
Cu
Cs
+d> e> f, .
+d> e> f, .
+d> e> f,
uI @
C{
C|
C}
is the glyhujhqfh of I at the point +d> e> f,. The dot product notation is used because
the symbol u can be thought of as the vector operator
C C C
> >
u@
C{ C| C}
The default is that the ¿eld variables are {, |, and }, in that order. If you wish to label
the ¿eld variables differently, reset the default with Set Basis Variables on the Vector
Gradient, Divergence, and Curl
313
Calculus submenu.
For the following example, use De¿ne + New De¿nition to de¿ne the following
vector ¿elds.
I @ ^|}> 5{}> {|`
J @ +{}> 5|}> } 5 ,
3 5 4
{
K @ |} 5{} {|
M @ C {| D
5{}
where I and J are represented as 6-tuples, K is represented as a 4 6 matrix, and M as
a 6 4 matrix. Compute divergence with Evaluate.
L Evaluate
uI @ 3
u +{|> {> 3, @ |
u +d> e> f, @ 3
u J @ 8}
u +{}> 5|}> } 5 , @ 8}
uK
u M @ 8{
@3
u d{> e{|> f} 5 @ d . e{ . 5f}
Curl
If I +{> |> }, @ +s+{> |> },> t+{> |> },> u+{> |> },, is a vector ¿eld, then the vector
Ct Cs Cu Ct
Cs
Cu
>
>
uI @
C| C} C} C{ C{ C|
is called the fxuo of I . The default is that the ¿eld variables are {, |, and }, in that
order. If you wish to label the ¿eld variables differently, reset the default with Set
Basis Variables on the Vector Calculus submenu. The vector ¿eld I in the following
example is de¿ned as in the previous section.
L Evaluate
u +|}> 5{}> {|, @ +{> 3> },
u I @ ^{> 3> }`
u |} 5{} {| @ { 3 }
3
4 3
4
{5
3
u C {| D @ C 5} D
5{}
|
5
6 5
6
d{5
3
u 7 e{| 8 @ 7 5f} 8
5f{}
e|
Laplacian
The Odsodfldq of a scalar ¿eld i+{> |> }, is the divergence of ui and is written
u5 i @ u ui
Ci Ci Ci
>
>
@ u
C{ C| C}
314
Chapter 9 Vector Calculus
C 5i
C 5i
C5i
.
.
C{5
C| 5
C} 5
The default is that the ¿eld variables are {, |, and }, in that order. If you wish to label
the ¿eld variables differently, reset the default with Set Basis Variables on the Vector
Calculus submenu.
@
L Evaluate
u5 { . | 5 . 5} 6 @ 5 . 45}
u { . | 5 . 5} 6 @ 4> 5|> 9} 5
u u { . | 5 . 5} 6 @ 5 . 45}
Directional Derivatives
The gluhfwlrqdo ghulydwlyh of a function i at the point +d> e> f, in the direction x @
+x4 > x5 > x6 , is given by the inner product of ui and x at the point +d> e> f,. That is, for
a vector x of unit length and a scalar function i,
Gx i +d> e> f, @ ui +d> e> f, x
Ci
Ci
Ci
+d> e> f, x4 .
+d> e> f, x5 .
+d> e> f, x6
@
C{
C|
C}
L To compute the directional derivative of i +{> |> }, @ {|} in the direction
x @ frv vlq > vlq vlq > frv
;
<
;
<
<
1. Select u{|}.
2. While holding down the CTRL key, choose Evaluate.
3. Choose Evaluate (or Evaluate Numerically).
L
CTRL
+ Evaluate, Evaluate, Evaluate Numerically
u{|} frv ; vlq < > vlq ; vlq < > frv <
@ +|}> {}> {|, frv ; vlq < > vlq ; vlq < > frv <
t
t
s s @ 45 |} 5 . 5 vlq 4< . 45 {} 5 5 vlq 4< . {| frv 4< @ =648<<|} . =463;<{} . =<6<9<{|
Plots of Vector Fields and Gradients
A function that assigns a vector to each point of a region in two- or three-dimensional
Plots of Vector Fields and Gradients
315
space is called a yhfwru hog. The gradient of a scalar-valued function of two variables
is a vector ¿eld.
Plots of Vector Fields
The operation Plot 2D + Vector Field requires a pair of expressions in two variables
representing the horizontal and vertical components of the vector ¿eld.
L To plot a two-dimensional vector ¿eld
1. Type a pair of two-variable expressions, representing the horizontal and vertical components of a vector ¿eld, into a vector.
2. Leave the insertion point in the vector, and from the Plot 2D submenu, choose Vector Field.
Example 84 To visualize the vector ¿eld I +{> |, @ ^{ . |> { |`
Place the insertion point in the vector ^{ . |> { |`, and from the Plot 2D submenu,
choose Vector Field.
L Plot 2D + Vector Field
^{ . |> { |`
4
y
2
-4
-2
0
2
x
4
-2
-4
g|
@
g{
i +{> |, the curve has slope i +{> |,. You can get an idea of the appearance of the graphs
of the solution of a differential equation from the direction ¿eld–that is, a plot depicting
short line segments with slope i+{> |, at points +{> |,. This can be done using Plot 2D
+ Vector Field and the vector-valued function
4> g|
gw
I +w> |, @ g|
4> gw At a point +{> |, on a solution curve of a differential equation of the form
316
Chapter 9 Vector Calculus
that assigns to each point +w> |, a vector of length one in the direction of the derivative
at the point +w> |,.
5
5
Example 85 The direction ¿eld for the differential equation g|
gw @ | w is the twodimensional vector ¿eld plot of the vector valued function
5
4> | w5
I +w> |, @
n+4> |5 w5 ,n
L Plot 2D + Vector Field
4
3
E
Eu
C 4
4 . m|5 w5 m5
> u
| 5 w5
4 . m|5 w5 m5
F
F
D
4
y
2
-4
-2
0
2
t
4
-2
-4
Several of the solution curves are depicted in the following plot.
4
2
-4
-2
0
2
t
4
-2
-4
L To plot a three-dimensional vector ¿eld
1. Type three three-variable expressions, representing the {-, |-, and }-components of
a vector ¿eld, into a vector.
Plots of Vector Fields and Gradients
317
2. Leave the insertion point in the vector.
3. From the Plot 3D submenu, choose Vector Field.
L Plot 3D + Vector Field
^|}> {}> {|`
The three-dimensional version is often a challenge to visualize. Multiple views can
be helpful. Three or four views of the vector ¿eld i +{> |> }, @ +|}> {}> {|, can provide
a reasonable graphical representation. Boxed axes can also help the visualization.
L To change the view
1. Click the frame until a small box appears in the upper-right corner of the frame.
2. With the left mouse button held down, rotate the plot.
4
2
4
2
z0
-2
-4
5
z0
z0
-4
-4
-2
-2
y0
2
2
4
4
2
-5
0x
4
-4
-2
0y
-2
0x
2
4
-2
-4
-4
-4
-2
0y
2
4
-5
5 0x
Plots of Gradient Fields
Scalar-valued functions of two variables can be visualized in several ways. Given the
function i +{> |, @ {| vlq {|, choosing Rectangular from the Plot 3D submenu pro-
318
Chapter 9 Vector Calculus
duces a surface represented by the function values. Another way to visualize such a
function is to choose Gradient from the Plot 2D submenu. This choice produces a plot
of the vector ¿eld that is the gradient of this expression, plotting vectors at grid points
whose magnitude and direction indicate the steepness of the surface and the direction of
steepest ascent.
The vector ¿eld that assigns to each point +{> |, the gradient of i at +{> |, is called
the judglhqw hog associated with the function i .
L To plot a gradient ¿eld
1. Type an expression i+{> |,.
2. Leave the insertion point in the expression, and from the Plot 2D submenu, choose
Gradient.
For example, type the expression {5 . 5| 5 , and choose Gradient from the Plot 2D
submenu. This procedure produces a plot of the vector ¿eld that is the gradient of this
expression. The following shows the relative steepness on the left, the surface on the
right.
L Plot 2D + Gradient, Plot 3D + Rectangular
{5 . 5| 5
4
y
2
-4
-2
0
-2
2
x
4
70
60
50
40
30
20
10
0
-4
-4
-4
-2
-2
y0
2
2
4
0x
4
The gradient ¿eld for a scalar-valued function i+{> |> }, of three variables is a threedimensional vector ¿eld where each vector represents the direction of maximal increase.
The surface represented by the function values is embedded in four-dimensional space,
so you must use indirect methods such as plotting the gradient ¿eld to help you visualize
this surface.
L Plot 3D + Gradient
{} . {| . |}
Scalar and Vector Potentials
319
4
2
4
2
z0
-2
-4
z0
-2
-4
-4
-2
-2
y0
2
4
4
2
0x
-4
-4
-2
0
y
2
4
-5
5 0x
As with other three-dimensional vector ¿elds, multiple views convey graphical information more effectively than does a single view.
Scalar and Vector Potentials
Scalar Potential on the Vector Calculus menu is the inverse of the gradient in the
sense that it ¿nds a scalar function whose gradient is the given vector ¿eld, or it tells you
that such a function does not exist. The vector potential has an analogous interpretation
in terms of the curl.
Scalar Potentials
The following are examples of scalar potential with the standard basis variables.
L Vector Calculus + Scalar Potential
+{> |> },, Scalar potential is 45 } 5 . 45 |5 . 45 {5
+{> }> |,, Scalar potential is |} . 45 {5
+|> }> {,, Scalar potential does not exist
In the next example, choose Evaluate, and then from the Vector Calculus submenu
choose Scalar Potential. Because the vector ¿eld is a gradient, it has the original
function as a scalar potential.
L Evaluate, Vector Calculus + Scalar Potential
u {| 5 . |} 6 @ |5 > 5{| . } 6 > 6|} 5 , Scalar potential is {| 5 . |} 6
Note When the vector ¿eld is a gradient, it has the original function as a scalar potential.
320
Chapter 9 Vector Calculus
You would normally expect the scalar potential of the vector ¿eld fy> fx . 5yz> y5
to be xfy . y 5 z that is, you expect f to be treated as a constant. When the number of
variables differs from the number of components in the ¿eld vector, a dialog box asks
for the ¿eld variables. In this case, you can enter u,v,w to get the expected result. The
dialog box also appears when you ask for the scalar potential of a vector ¿eld that speci¿es fewer than three variables, such as +|> {> 3,. Enter x,y,z in the dialog box to get the
expected result {| for the scalar potential of this vector ¿eld.
Vector Potential
A vector potential exists for a vector ¿eld I if and only if
gly I @ u I @ 3
That is, the vector ¿eld is solenoidal.
Unless otherwise speci¿ed, the operators curl and vector potential apply to scalar or
vector functions of a set of exactly three standard basis variables. The default is {> |> },
but you can use other sets of basis variables by choosing Set Basis Variables from the
Vector Calculus submenu and changing
the
5
6 default variables.
|
Start with u +{|> |}> }{, @ 7 } 8 to get the following.
{
L Vector Calculus + Vector Potential
5
6
5 4 5
6
|
5 } . {|
7 } 8, Vector potential is 7
8
|}
{
3
Notice that we did not get the original vector ¿eld when we asked for a vector potential of its curl. That is because the vector potential is determined only up to a ¿eld
whose curl is zero. You can verify that this is the case. First, calculate the difference of
the two vectors.
L Evaluate
5
6 5 4 5
6 5 4 5 6
{|
5 } . {|
}
5
7 |} 8 7
8@ 7 3 8
|}
{}
}{
3
Then compute the curl of the difference.
L Evaluate
5 4
6 5 6
3
}5
u7 3 8 @ 7 3 8
{}
3
5
Matrix-Valued Operators
321
Try the same experiment after changing the basis variables to x> y> z with Vector
Calculus + Set Basis Variables.
L Vector Calculus + Vector Potential
5
6
5 4 5
6
y
5 z . xy
7 z 8, Vector potential is 7
8
yz
x
3
A vector ¿eld can be written either as the triple +y> z> x, or as a column matrix. Try
choosing Vector Potential for +y> z> x,.
L Vector Calculus + Vector Potential
+y> z> x,, Vector potential is
4
5
z5 xy> yz> 3
Matrix-Valued Operators
Matrix-valued operators include the Hessian and the Jacobian.
Hessian
The Khvvldq is the q q matrix
5
C5 i
9 C{54
9
9 C5 i
9
9 C{ C{
9
5
4
9
..
9
.
9
7 C5 i
C 5i
C{4 C{5
C 5i
C{55
..
.
..
C 5i
C{4 C{q
C 5i
C{5 C{q
..
.
6
:
:
:
:
:
:
:
:
:
8
.
C 5i
C 5i
C{q C{4 C{q C{5
C{5q
of second partial derivatives of a scalar function i +{4 > {5 > = = = > {q , of q variables. The
order of the variables affects the ordering of the rows and columns of the Hessian, but it
is natural to require that the list of variables be ordered lexicographically, and this choice
is the one that is made.
322
Chapter 9 Vector Calculus
L Vector Calculus + Hessian
5
6
} |
3 { 8
{ 3
3
{|}, Hessian is 7 }
|
d . e , Hessian is
6
6
9d 3
3 9e
5
3
9 |}
z{|}, Hessian is 9
7 {}
{|
5
3
9 }|
{}z|, Hessian is 9
7 {}
{|
{5 . | 6 , Hessian is
e . d , Hessian is
6
|}
3
z}
z|
6
{} {|
z} z| :
:
3 z{ 8
z{ 3
}|
3
}z
z|
{}
}z
3
{z
6
5 3
3 9|
9d 3
3 9e
6
{|
z| :
:
{z 8
3
A possible source of confusion is the inclusion of something in the variable list that
was intended to be a constant parameter, such as f in f{|}. In this case, f will be
interpreted as a variable, and you will obtain a 7 7 matrix. Since f occurs ¿rst in the
alphabet, partial derivatives involving f occur in the ¿rst row and column. Delete the
¿rst row and ¿rst column of the resulting matrix to get the desired result.
L Vector Calculus + Hessian
5
3
9 |}
f{|}, Hessian is 9
7 {}
{|
|}
3
f}
f|
{}
f}
3
f{
6
{|
f| :
:
f{ 8
3
5
3
or 7 f}
f|
6
f} f|
3 f{ 8 for f constant
f{ 3
After de¿ning i +{> |> }, @ 6{|5 }, you can ¿nd the Hessian for the expression
i +{> |> },.
Matrix-Valued Operators
323
L Vector Calculus + Hessian
5
3
i+{> |> },, Hessian is 7 9|}
6| 5
5
3
i+}> |> {,, Hessian is 7 9}|
6| 5
5
9|}
9{}
9{|
6
6| 5
9{| 8
3
9}|
9{}
9{|
6
6| 5
9{| 8
3
69|} 7 {
4;|5 } 7
9
5
7
+i +|> }> {,, , Hessian is 7 69|} { 4;} 7 {5
:5| 5 } 6 { :5|} 6 {5
6
:5| 5 } 6 {
:
:5|} 6 {5 8
43;| 5 } 5 {5
Jacobian
The Mdfreldq is the q q matrix
5 Ci
Ci4 6
Ci4
4
9 C{4 C{5
C{q :
9
:
9 Ci5 Ci5
Ci5 :
9
:
9 C{4 C{5
C{q :
9 .
..
.. :
..
9 .
:
.
9 .
.
. :
7 Ci
Ciq 8
q Ciq
C{4 C{5
C{q
of partial derivatives of the entries in a vector ¿eld
+i4 +{4 > {5 > = = = > {q , > i5 +{4 > {5 > = = = > {q , > = = = > iq +{4 > {5 > = = = > {q ,,
Jacobians resemble Hessians in that the order of the variables in the variable list determines the order of the columns of the matrix, and lexicographic order is usually correct.
The number of variables should be the same as the dimension of the vector if they are
not the same, either a parameter has been included in the variable list, or the vector ¿eld
is independent of one of the variables. In this case, a dialog box asks for the list of variables. In each of the following examples, the variable list is {> |> }. To verify these
examples, choose Jacobian while the insertion point is in the given vector ¿eld.
324
Chapter 9 Vector Calculus
L Vector Calculus + Jacobian
5
6
3 } |
+|}> {}> {|,, Jacobian is 7 } 3 { 8
| { 3
5
5{}
+{5 }> { . }> {} 5 ,, Jacobian is 7 4
}5
5
5{}
5
5
7
3
+{ }> | . f> |} ,, Jacobian is
3
6
3 {5
3
4 8
3 5{}
3
4
}5
6
{5
3 8
5|}
(| is missing)
(f is extra)
Plots of Complex Functions
A complex function I +}, (} and I +}, are both complex) is a challenge to graph, because
the natural graph would require four dimensions.
Conformal Plots
A conformal plot of a complex function I +}, is the image of a two-dimensional rectangular grid of horizontal and vertical line segments. The default is an 44 by 44 grid, with
each of the intervals 3 Re+}, 4 and 3 Im+}, 4 subdivided into 43 equal subintervals. If I +}, is analytic, then it preserves angles at every point at which I 3 +}, 9@ 3
hence, the image is a grid composed of two families of curves that intersect at right
angles.
}4
, put the insertion point in the expression,
To create a conformal plot of I +}, @
}.4
and choose Conformal from the Plot 2D submenu. The number of grid lines and the
view can be changed in the Plot Properties tabbed dialogs.
L Plot 2D + Conformal
}4
}.4
Exercises
325
1
0.8
0.6
0.4
0.2
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
In the following example, Re+}, and Im+}, both range from 6 to 6, the View
Intervals are set at 5 Re+I +},, 7 and 6 Im+I +},, 6, the Grid Size
has been increased to 73 by 73, and Samples per Horizontal Grid Line and Samples
per Vertical Grid Line have both been increased to 93.
L Plot 2D + Conformal
}4
}.4
3
2
1
-2
-1
0
1
2
3
4
-1
-2
-3
Exercises
1. Evaluate the directional derivative of i +{> |> }, @ 6{ 8| . 5} at +5> 5> 4, in the
direction of the outward normal to the sphere {5 . | 5 . } 5 @ <.
s
6@5
2. Find a vector y normal to the surface } @ {5 . | 5 . {5 . | 5
at the point
+{> |> }, 9@ +3> 3> 3, on the surface.
326
Chapter 9 Vector Calculus
pP
3. Let i +{> |> }, @ s
denote Newton’s gravitational potential. Show
5
{ . |5 . } 5
that the gradient is given by
5
6
{
pP
7 | 8
ui +{> |> }, @ +{5 . | 5 . } 5 ,6@5
}
Solutions
1. The directional derivative is given by Gx i +{> |> },, @ ui +{> |> }, x, where x is
a unit vector in the direction of the outward normal to the sphere {5 . | 5 . } 5 @ <.
The vector
5
6
5{
5
u { . | 5 . } 5 @ 7 5| 8
5}
is normal to the sphere {5 . | 5 . } 5 @ <, and at +5> 5> 4, this normal is +7> 7> 5,. A
unit vector in the same direction is given by
5 6 5
6
5 6 5 6
7 7
5@6
7
4
7 8
7 8 7 5@6 8
x@7 7 8
7 @ 9 7 @
5
5
4@6
5
Hence, ui +{> |> }, x evaluated at +5> 5> 4, is 56 .
2. A normal vector is given by
s
6@5
{5 . |5 . {5 . | 5
}
u
$
#
s
s
4
4
5
5
5
5
{ . 6 +{ . | ,{> s
| . 6 +{ . | ,|> 4
@ s
5
5
+{5$. |5 ,
# +{ . | ,
4 . 6{5 . 6|5 4 . 6{5 . 6| 5
@ { s
>| s
> 4
+{5 . | 5 ,
+{5 . | 5 ,
s
4
{ 4 . 6{5 . 6| 5 > | 4 . 6{5 . 6| 5 > +{5 . | 5 ,
@s
+{5 . | 5 ,
Hence, any scalar multiple of
l
k s
{ 4 . 6{5 . 6|5 > | 4 . 6{5 . 6| 5 > +{5 . |5 ,
is also normal to the given surface.
$
#
pP
. Then delete the ¿rst two rows of the
3. Evaluate the expression u s
{5 . | 5 . } 5
Solutions
vector, because p and P are constant parameters.
#
$
pP
u s
{#5 . |5 . } 5
@
@
327
$
p s 5 P5 5 6 {> p s 5 P5 5 6 |> p s 5 P5 5 6 }
+{ .| .} ,
+{ .| .} ,
+{ .| .} ,
pP
6@5
+{ . |5 . } 5 ,
5
+{> |> },
This gives the Newtonian gravitational force between two objects of masses p and
P , with one object at the origin and the other at the point +{> |> },.