ELEMENTARY LINEAR
ALGEBRA AND VECTOR
CALCULUS
MATHEMATICS FOR UNDERGRADUATES
Antony L. Foster
Department of Mathematics (NAC 6/273)
The City College of New York
New York, New York 10031.
Chapter 1.
LINEAR EQUATIONS AND MATRICES
1.1. SYSTEMS OF LINEAR EQUATIONS
One of the most frequently recurring practical problems in almost all fields of study—such as mathematics, physics,
biology, chemistry, economics, all phases of engineering, operations research, the social sciences, and so forth—is that
of solving a system of linear equations. The equation
b = a1 x1 + a2 x2 + · · · + an xn ,
(1)
which expresses b in terms of the variables x1 , x2 , . . . , xn and the constants a1 , a2 , . . . , an , is called a linear
equation. In many applications we are given b and must find numbers x1 , x2 , . . . , xn satisfying (1).
A solution to a linear equation (1) is a sequence of n numbers s1 , s2 , . . . , sn such that (1) is satisfied when
x1 = s1 , x2 = s2 , . . . , xn = sn are substituted in (1). Thus x1 = 2, x2 = 3, and x3 = −4 is a solution to the
linear equation
6x1 − 3x2 + 4x3 = −13,
because
6(2) − 3(3) + 4(−4) = −13.
More generally, a system of m linear equations in n unknowns, or a linear system, is a set of m linear
equations each in n unknowns called x1 , x2 , . . . , xn .
A linear system can be conveniently written as
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
..
..
..
..
.
.
.
.
am1 x1 + am2 x2 + · · · + amn xn = bm .
(2)
Thus the ith equation in the system is
ai1 x1 + ai2 x2 + · · · + ain xn = bi .
In (2) the aij are known constants. Given the values b1 , b2 , . . . , bn , we want to find values of x1 , x2 , . . . , xn that
will satisfy each equation in (2).
A solution to a linear system (2) is a sequence of n numbers s1 , s2 , . . . , sn , which have the property that each
equation in (2) is satisfied when x1 = s1 , x2 = s2 , . . . , xn = sn are substituted.
If the linear system (2) has no solution, it is said to be inconsistent; if it has a solution, it is called consistent. If
b1 = b2 = · · · = bn = 0, then (2) is called a homogeneous system. The solution x1 = x2 = · · · = xn = 0
to a homogeneous system is called the trivial solution. A solution to a homogeneous system in which not all of
x1 , x2 , . . . , xn are zero is called a nontrivial solution.
Consider another system of r linear equations in n unknowns:
c11 x1 + c12 x2 + · · · + c1n xn = d1
c21 x1 + c22 x2 + · · · + c2n xn = d2
..
..
..
..
.
.
.
.
cr1 x1 + cr2 x2 + · · · + crn xn = dr .
We say that (2) and (3) are equivalent if they both have exactly the same solutions.
(3)
Example 1.1.1. The linear system
x1 − 3x2 = −7
2x1 + x2 = 7
(4)
has only the solution x1 = 2 and x2 = 3. The linear system
8x1 − 3x2 = 7
3x1 − 2x2 = 0
10x1 − 2x2 = 14
(5)
also has only the solution x1 = 2 and x2 = 3. Thus (4) and (5) are equivalent.
To find solutions to a linear system we shall use a technique called the method of elimination; that is, we
eliminate some variables by adding a multiple of one equation to another equation. Elimination merely amounts
to the development of a new linear system which is equivalent to the original system but is much simpler to solve.
Readers have probably confined their earlier work in this area to linear systems in which m = n, that is, linear
systems having as many equations as unknowns. In this course we shall broaden our outlook by dealing with systems
in which we have m = n, m < n, and m > n. Indeed, there are numerous applications in which m 6= n.
Example 1.1.2. Consider the linear system
x1 − 3x2 = −3
2x1 + x2 = 8.
(6)
To eliminate x1 , we subtract twice the first equation from the second, obtaining
7x2 = 14,
and equation having no x1 term. Thus we have eliminated the unknown x1 . Then solving for x2 , we have
x2 = 2,
and substituting into the first equation of (6), we obtain
x1 = 3.
Then x1 = 3, x2 = 2 is the only solution to the given linear system.
Example 1.1.3. Consider the linear system
x1 − 3x2 = −7
2x1 + x2 = 7.
(7)
Again, we decide to eliminate x1 . We subtract twice the first equation from the second one, obtaining
0 = 21,
which makes no sense. This means that (7) has no solution; it is inconsistent. We could have come to the same
conclusion from observing that in (7) the left side of the second equation is twice the left side of the first equation,
but the right side of the second equation is not twice the right side of the first equation.
Example 1.1.4. Consider the linear system
x1 + 2x2 + 3x3 = 6
2x1 − 3x2 + 2x3 = 14
3x1 + x2 − x3 = −2.
(8)
To eliminate x1 , we subtract twice the first equation from the second and three times the first equation from the
third, obtaining
−7x2 − 4x3 = 2
(9)
−5x2 − 10x3 = −20.
This is a system of two equations in the unknowns x2 and x3 . We divide the second equation of (9) by -5, obtaining
−7x2 − 4x3 = 2
x2 + 2x3 = 4,
which we write, by interchanging equations as
x2 + 2x3 = 4
−7x2 − 4x3 = 2.
(10)
We now eliminate x2 in (10) by adding 7 times the first equation to the second one, to obtain
10x3 = 30,
or
x3 = 3.
(11)
Substituting this value of x3 into the first equation of (10), we find that x2 = −2. Substituting these values of x2
and x3 into the first equation of (8), we find that x1 = 1. We might observe further that our elimination procedure
has actually produced the linear system
x1 + 2x2 + 3x3 = 6
x2 + 2x3 = 4
(12)
x3 = 3.
obtained by using the first equations of (8) and (10) as well as (11). The importance of the procedure is that although
the linear systems (8) and (12) are equivalent, (12) has the advantage that it is easier to solve.
Example 1.1.5. Consider the linear system
x1 + 2x2 − 3x3 = −4
2x1 + x2 − 3x3 = 4.
(13)
Eliminating x1 , we subtract twice the first equation from the second equation, to obtain
−3x2 + 3x3 = 12.
(14)
We must now solve (14). A solution is
x2 = x3 − 4,
where x3 can be any real number. Then from the first equation of (13),
x1 = −4 − 2x2 + 3x3
= −4 − 2(x3 − 4) + 3x3
= x3 + 4.
Thus a solution to the linear system (13) is
x1 = x3 + 4
x2 = x3 − 4
x3 = any real number.
This means that the linear system (13) has infinitely many solutions. Every time we assign a value to x3 we obtain
another solution to (13). Thus, if x3 = 1, then
x1 = 5,
x2 = −3,
and x3 = 1
is a solution, while if x3 = −2, then
x1 = 2,
x2 = −6,
and x3 = −2
is another solution.
These examples suggest that linear system may have a unique solution, no solution, or infinitely many solutions.
Consider next a linear system of two equations in the unknowns x1 and x2 .
a1 x1 + a2 x2 = c1
(15)
b1 x1 + b2 x2 = c2
The graph of each of these equations is a straight line, which we denote by l1 and l2 respectively. If x1 = s1 , x2 = s2
is a solution to the linear system (15), then the point (s1 , s2 ) lies on both lines l1 and l2 . Conversely, if the point
(s1 , s2 ) lies on both lines l1 and l2 , then x1 = s1 , x2 = s2 is a solution to the linear system (15). Thus we are led
geometrically to the same three possibilities mentioned above in (Figure 1.1).
x2
x2
x2
l2
l2
x1
x1
x1
l1
l2
l1
(a) A unique solution.
(b) No solution.
l1
(c) Infinitely many solutions.
If we examine the method of elimination more closely, we find that it involves three manipulations that can be
performed on a linear system to convert it into an equivalent system. These manipulations are as follows:
1.
We may interchange the ith and jth equations of the system.
2.
We may multiply an equation in the system by a nonzero constant.
3.
We may replace the ith equation by c times the jth equation plus the ith equation,
i =
6 j.
That is, replace
ai1 x1 + ai2 x2 + · · · + ain xn = bi
by
(ai1 + caj1 )x2 + (ai2 + caj2 )x2 + · · · + (ain + cajn )xn = bi + cbj .
It is not difficult to prove that performing these manipulations on a linear system leads to an equivalent system.
Example 1.1.6. Suppose that the ith equation of a linear system such as (2) is multiplied by the nonzero constant
c, obtaining the linear system
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
..
..
..
..
.
.
.
.
cai1 x1 + cai2 x2 + · · · + cain xn = cbi
..
..
..
..
.
.
.
.
am1 x1 + am2 x2 + · · · + amn xn = bm .
(16)
If x1 = s1 , x2 = s2 , · · · xn = sn is a solution to (2), then it is a solution to all the equations in (16) except
possibly for the ith equation. For the ith equation we have
c(ai1 s1 + ai2 s2 + · · · + ain sn ) = cbi
or
cai1 s1 + cai2 s2 + · · · + cain sn = cbi .
Thus the ith equation of (16) is also satisfied. Hence every solution to (2) is also a solution to (16). Conversely,
every solution to (16) also satisfies (2). Hence (2) and (16) are equivalent systems.
In the next Section we develop methods that will enable us to further discuss and solve linear systems of equations.
SUGGESTED EXERCISES FOR SECTION 1
In Exercises 1 through 14, solve the given linear system by the method of elimination. (I suggest that you hold off
on these exercises until later when we discuss the method of Gaussian Elimination.)
Exercise 1.1.1.
x1 + 2x2 = 8
.
3x1 − 4x2 = 4
Exercise 1.1.2.
2x1 − 3x2 + 4x3 = −12
x1 − 2x2 + x3 = −5 .
3x1 + x2 + 2x3 = 1
Exercise 1.1.3.
3x1 + 2x2 + x3 = 2
4x1 + 2x2 + 2x3 = 8 .
x1 − x2 + x3 = 4
Exercise 1.1.4.
x1 + x2 = 5
3x1 + 3x2 = 10.
Exercise 1.1.5.
2x1 + 4x2 + 6x3 = −12
2x1 − 3x2 − 4x3 = 15
3x1 + 4x2 + 5x3 = −8.
Exercise 1.1.6.
x1 + x2 − 2x3 = 5
2x1 + 3x2 + 4x3 = 2.
Exercise 1.1.7.
x1 + 4x2 + 6x3 = 12
3x1 + 8x2 − 2x3 = 4.
Exercise 1.1.8.
3x1 + 4x2 − x3 = 8
6x1 + 8x2 − 2x3 = 3.
Exercise 1.1.9.
x1 + x2 + 3x3 = 12
2x1 + 2x2 + 6x3 = 6.
Exercise 1.1.10.
x1 + x2 = 1
2x1 − x2 = 5
3x1 + 4x2 = 2.
Exercise 1.1.11.
2x1 + 3x2 = 13
x1 − 2x2 = 3
5x1 + 2x2 = 27.
Exercise 1.1.12.
x1 − 5x2 = 6
3x1 + 2x2 = 1 .
5x1 + 2x2 = 1
Exercise 1.1.13.
x1 + 3x2 = −4
2x1 + 5x2 = −8
x1 + 3x2 = −5.
Exercise 1.1.14.
2x1 + 3x2 − x3 = 6
2x1 − x2 + 2x3 = −8
3x1 − x2 + x3 = −7.
Exercise 1.1.15. Show that the linear system obtained by interchanging two equations in (2) is equivalent to (2).
Exercise 1.1.16. Show that the linear system obtained by adding a multiple of an equation in (2) to another
equation is equivalent to (2).
1.2. MATRICES; MATRIX OPERATIONS
Before continuing the study of solving linear systems, we now introduce the notion of a matrix, which will greatly
simplify our notational problems, and develop tools to solve many important applied problems.
If we examine the method of elimination described in Section 1.1, we make the following observation: Only the
numbers in front of the unknowns x1 , x2 , . . . , xn are being changed as we perform the steps in the method of
elimination. Thus we might think of looking for a way of writing a linear system without having to carry along the
unknowns. Matrices enables us to do this—that is, to write linear systems in a compact form that makes it easier
to automate the elimination method on an electronic computer in order to obtain a fast and efficient procedure for
finding solutions. Their use is not, however, merely that of a convenient notation. We now develop operations on
matrices and will work with matrices according to the rules they obey; this will enable us to solve systems of linear
equations and to do other computational problems in a fast and efficient manner. Of course, as any good definition
should do, the notion of a matrix provides not only a new way of looking at old problems but also gives rise to a
great many new questions, some of which we study in these notes.
WHAT ARE MATRICES?
Definition 1.2.1. A matrix (plural matrices) is a rectangular array of objects (usually numbers) denoted by





A = 




a11
a21
..
.
a12
a22
..
.
···
···
a1j
a2j
..
.
ai1
..
.
ai2
..
.
···
aij
..
.
am1
am2
···
amj
···
···
···
···
···
···
a1n
a2n
..
.





.
ain 

.. 
. 
amn
Unless stated otherwise, we assume that all our matrices are composed entirely of real numbers. The ith row of A
is
( ai1 ai2 · · · ain )
(1 6 i 6 m)
while the jth column of A is
a
1j
 a2j
 .
 .
.
amj




(1 6 j 6 n)
If a matrix A has m rows and n columns, we say that A is an m by n (m × n) matrix. If m = n, we say that A is
a square matrix of order n and that the elements a11 , a22 , . . . , ann are on the main diagonal of A. We refer
to aij as the (i, j) entry (entry in ith row and jth column) or (i, j)th element and we often write
A = [aij ] or A = [aij ]m×n
We shall also write Am×n to indicate that A has m rows and n columns. If A is n × n, we merely write An .
Example 1.2.1. The following are matrices:

1 2
A = 4 5
7 8

3
6,
9


2
 −1 
C = 
,
3
4
B = (1
and D =
−2 9 )
0
3
−1 −2
.
In matrix A, the entry a32 = 8; in matrix C, the entry c41 = 4. Here A is a 3 × 3 square matrix, B is a 1 × 3,
matrix C is a 4 × 1 matrix, and matrix D is a 2 × 2 square matrix. In A, the entries a11 = 1, a22 = 5 and a33 = 9
are on the main diagonal.
Whenever a new object is introduced in mathematics, one must determine when two such objects are equal. For
example, in the set of all rational numbers, the numbers 23 and 46 are called equal, although they are not represented
in the same manner. What we have in mind is the definition that a/b equals c/d when ad = bc. Accordingly, we
now have the following definition.
EQUALITY OF MATRICES
Definition 1.2.2. Two m × n matrices A = [aij ] and B = [bij ] are equal if they agree entry by entry, that is, if
aij = bij for i = 1, 2, . . . , m and j = 1, 2, . . . , n.
Example 1.2.2. The matrices



2 −1
1
−3 4  and B =  2
−4 5
y
1
2
0

2 w
x 4
−4 z
are equal if and only if w = −1, x = −3, y = 0, and z = 5.
We next define a number of operations that will produce new matrices out of given matrices; this will enable us to
compute with the matrices and not deal with the equations from which they arise. These operations are also useful
in the applications of matrices.
ADDITION OF MATRICES
Definition 1.2.3. If A = [aij ] and B = [bij ] are both m × n matrices, then their matrix sum A + B is an
m × n matrix C = [cij ] defined by cij = aij + bij , i = 1, 2, . . . , m; j = 1, 2, . . . , n. Thus, to obtain the sum of A
and B, we merely add corresponding (i, j) entries.
Example 1.2.3. Let
A =
Then
A+B =
1
2
−2 3
−1 4
+
1
2
−2 3
−1 4
0
1
2 1
3 −4
and B =
=
1+0
2+1
0
1
2 1
3 −4
−2 + 2
−1 + 3
.
3+1
4−4
=
1
3
0 4
2 0
.
It should be noted that the sum of the matrices A and B is defined only when A and B have the same number
of rows and the same number of columns, that is, only when A and B are of the same size. We now establish the
convention that when A + B is formed, both A and B are of the same size. The basic properties of matrix addition
are considered in the following section and are similar to those satisfied by the real numbers.
SCALAR MULTIPLICATION OF MATRICES
Definition 1.2.4. If A = [aij ] is an m × n matrix and α is a real number, then the scalar multiple of A by α,
is αA, is the m × n matrix C = [cij ], where cij = α aij , i = 1, 2, . . . , m, j = 1, 2, . . . , n; that is, the matrix C is
obtained by multiplying each entry of A by α.
Example 1.2.4. We have
2
4
7
−2 −3
−3 2
=
8 −4 −6
14 −6 4
.
SUBTRACTION: If A and B are m × n matrices, we write A + (−1)B as A − B and call this the difference
between A and B.
We shall sometimes use the summation notation, and we now review this useful and compact notation.
By
n
X
rk ak
k=1
we mean r1 a1 + r2 a2 + · · · + rn an . The letter k is called the index of summation; it is a dummy variable that
can be replaced by another letter. Hence we can write
n
X
k=1
Thus
4
X
i=1
rk ak =
n
X
j=1
rj aj =
n
X
ri ai .
i=1
ri ai = r1 a1 + r2 a2 + r3 a3 + r4 a4 .
The summation notation satisfies the following properties:
1.
n
X
(ri + si )ai =
n
X
α (ri ai ) = α
i=1
2.
(ri ai + si ai ) =
i=1
n
X
i=1
3.
n
X
n
X
ri ai +
i=1
n
X
si ai
i=1
ri ai
i=1
m X
n
X
aij =
j=1 i=1
n X
m
X
aij
i=1 j=1
Property 3 can be interpreted as follows. If we add up the entries in each row of a matrix and then add the resulting
numbers, we obtain the same result as when we add up the entries in each column of the matrix and then add the
resulting numbers.
MULTIPLICATION OF MATRICES
Definition 1.2.5. If A = [aij ] is an m × n matrix and B = [bij ] is an n × p matrix, then the matrix product of
A and B, denoted C = AB = [cij ], is an m × p matrix defined y
n
X
cij =
k=1
aik bkj = ai1 b1j + ai2 b2j + · · · + ain bnj
i = 1, 2, . . . , m
j = 1, 2, . . . , p
Note that the matrix product AB is defined only when the number of columns of A is the same as the number of
rows of B. We also observe that the (i, j) entry of C = AB is obtained by using the ith row of A and the jth
column of B. Thus










a11
a21
..
.
a12
a22
..
.
···
···
a1j
a2j
..
.
ai1
..
.
ai2
..
.
···
aij
..
.
am1
am2
···
amj





= 




c11
c21
..
.
c12
c22
..
.
···
···
c1j
c2j
..
.
ci1
..
.
ci2
..
.
···
cij
..
.
cm1
cm2
···
cmj
···
···
···
···
···
···
···
···
···
···
···
···
a1n
a2n
..
.






ain 

.. 
. 
amn









c1n
c2n
..
.






cin 

.. 
. 
cmn
b11
b21
..
.
b12
b22
..
.
···
···
b1j
b2j
..
.
bi1
..
.
bi2
..
.
···
bij
..
.
bm1
bm2
···
bmj
···
···
···
···
···
···
b1n
b2n
..
.






bin 

.. 
. 
bmn
Example 1.2.5. Let
A =
Then
AB =
1 2
3 1
−1
4


−2 5
and B =  4 −3  .
2
1
(1)(−2) + (2)(4) + (−1)(2) (1)(5) + (2)(−3) + (−1)(1)
(3)(−2) + (1)(4) + (4)(2)
(3)(5) + (1)(−3) + (4)(1)
=
4
6
−2
16
The basic properties of matrix multiplication are considered in the following section. However, we note here that
multiplication of matrices requires much more care than their addition, since the algebraic properties of matrix
multiplication differ from those satisfied by the real numbers. Part of the problem is due to the fact that AB is
defined only when the number of columns of A is the same as the number of rows of B. Thus, if A is an m × n matrix
and B is an n × p matrix, then AB is an m × p matrix. What about BA? Three different situations may occur:
1.
2.
3.
BA may not be defined. This will take place if p 6= m.
If BA is defined, BA will be an n × n matrix and AB will be an m × m matrix, and if m 6= n, AB and BA
are of different sizes.
If BA and AB are of the same size, they may be unequal.
As in the case of addition, we establish the convention that when AB is written, it is defined.
Example 1.2.6. Let A be a 2 × 3 matrix and let B be a 3 × 4 matrix. Then AB is 2 × 4 and BA is not defined.
Example 1.2.7. Let A be 2 × 3 and let B be 3 × 2. Then AB is 2 × 2 and BA is 3 × 3.
Example 1.2.8. Let
A =
Then
AB =
1 2
−1 3
2
−2
3
2
and B =
while
BA =
2 1
0 1
.
1 7
−1 3
.
Thus AB 6= BA.
One might ask why matrix equality and matrix addition are defined in such a natural way while matrix multiplication
appears to be much more complicated. Only a thorough understanding of the composition of functions and the
relationship that exists between matrices and what are called linear transformations would show that the definition
of multiplication given above is the natural one. These topics will be covered later in these notes.
It is sometimes useful to be able to find a column in the matrix product AB without having to multiplying the two
matrices. It is not difficult to show that the jth column of the matrix product AB is equal to the matrix product
ABj , where Bj is the jth column of B.
Example 1.2.9. Let

Then the second column of AB is
1
A =  3
−1

2
2
5

1
AB2 =  3
−1
and B =
−2 3 4
3 2 1
.

 
2 7
3
=  17 
4
2
7
5
We now return to the linear system (2) in Section 1.1 and define the following matrices:

a11
 a21
A = 
 ...
am1
a12
a22
..
.
···
···
..
.
am2
···

 


a1n
x1
b1
a2n 
 x2 
 b2 

, X = 
, B = 
.. 
.. 



 ... 
.
.
amn
xn
bm
We can then write the linear system (2) as AX = B. The matrix A is called the coefficient matrix of the system
and the matrix
a
11
 a21
(A | B) = 
 ..
.
am1
a12
a22
..
.
···
···
..
.
a1n
a2n
..
.
am2
···
amn
|
|
b1
b2
..
.
|
| bm




is called the augmented matrix of the system. The coefficient and augmented matrices of a linear system will play
key roles in our methods of solving linear systems.
Example 1.2.10. Consider the following linear system:
2x1
−2x1
3x1
+ 3x2
+ x2
+ 2x2
− 4x3
+ x3
+ x3
+ x4
+ x4
− 4x4
=
=
=
5
7
3.
We can write this in matrix form as
  x1 
 
3 −4
1
5
x 
0
1
0   2  = 7.
x3
2
0 −4
3
x4

2
 −2
3
The coefficient matrix of this system is
and the augmented matrix is


2 3 −4
1
 −2 0
1
0
3 2
0 −4

2 3 −4
 −2 0
1
3 2
0
Example 1.2.11. The matrix
2 −1 3
3
0 2

1 | 5
0 | 7.
−4 | 3
| 4
| 5
.
is the augmented matrix of the linear system
2x1 − x2 + 3x3 = 4
3x1 + 0x2 + 2x3 = 5.
TRANSPOSE OF MATRICES
Definition 1.2.6. If A = [aij ] is an m × n matrix, then the transpose of A, denoted by AT = [aTij ] is an n × m
matrix defined by aTij = aji . Thus the transpose of A is obtained from A by interchanging the rows and columns of
A.
Example 1.2.12. If
A =
1 2
−3 2
−1
7
,
then AT


1 −3
=  2
2 .
−1 7
TRACE OF A MATRIX
Definition 1.2.7. If A = [aij ] is an n × n matrix, then the trace of A, denoted by Tr(A), is defined as the sum of
all elements on the main diagonal of A (these are elements of A of the form aii ). Thus
Tr(A) =
n
X
aii .
i=1
Example 1.2.13. If

1 2
C = 4 5
7 8

3
6,
9
then
Tr(C) =
3
X
i=1
cii = c11 + c22 + c33 = 1 + 5 + 9 = 15.
SUGGESTED EXERCISES FOR SECTION 1.2
Consider the following matrices for Exercises 1, 2, and 3:




1 0
3 −1 3
1 2 3
A =
, B = 2 1, C = 4
1 5,
2 1 4
3 2
2
1 3
D =
3 −2
2
5
Exercise 1.2.1. If possible, compute:
(a) C + E.
(b)
AB and BA.
(c)
2C − 3E.
(d)
CB + D.
(e)
AB + D2 , where D2 = DD.
(f)
(3)(2A) and 6A.

2
, and E =  0
3

−4 5
1 4.
2 1
Exercise 1.2.2. If possible, compute:
(a) A(BD).
(b)
(AB)D.
(c)
A(C + E).
(d)
AC + AE.
(e)
3A + 2A and 5A.
Exercise 1.2.3. If possible, compute:
(a) AT .
(b)
(AT )T .
(c)
(AB)T .
(d)
B T AT .
(e)
(C + E)T and C T + E T .
(f)
A(2B) and 2(AB).
Exercise 1.2.4.
(a) Let A be an m × n matrix with a row consisting entirely of zeros. Show that if B is an n × p matrix, then the
matrix product AB has a row of zeros.
(b)
Let A be an m × n matrix with a column consisting entirely of zeros and let B be p × m matrix. Prove that
the matrix product BA has a column of zeros.
Exercise 1.2.5. Let A =
1
3
2
2
and B =
2 −1
. Show that AB 6= BA.
−3 4
Exercise 1.2.6. Consider the following linear system
2x1 +
3x1 +
2x1 +
3x2 −
3x2
(a)
Find the coefficient matrix.
(b)
Write the linear system in matrix form.
3x3 +
2x3 +
x4 +
−4x4
x5 =
7
3x5 = −2
=
3
(c)
Find the augmented matrix.
Exercise 1.2.7. Write the linear system whose augmented matrix is

−2 −1 0
2 7
 −3

1
0 0
3
0 1
Exercise 1.2.8. If
a+b
c−d
c+d
a−b
=
4
8
2
3
|
|
|
|

5
3
.
4
6
4 6
, find a, b, c, and d.
10 2
Exercise 1.2.9. Write the following linear system in matrix form.
2x1 + 3x2 = 0
3x2 + x3 = 0
2x1 − x3 = 0.
Exercise 1.2.10. Write the linear system whose augmented matrix is

2
1
3 4
(a)  3 −1
2 0
−2
1 −4 3

| 0
| 3
| 2

2
 3
(b) 
−2
0
1
3 4
−1
2 0
1 −4 3
0
0 0
|
|
|
|

0
3
.
2
0
Exercise 1.2.11. How are the linear systems obtained in Exercise 1.2.10 related?
Exercise 1.2.12. If A = [aij ] is an n × n matrix and α is a real number, then prove that
(a)
Tr(αA) = α Tr(A).
(b)
Tr(A + B) = Tr(A) + Tr(B).
(c)
Tr(AB) = Tr(BA).
Exercise 1.2.13. Compute the trace (see Definition 1.2.7) of each of the following matrices.




2
2
3
1 0 0
1 0
(a)
(b)  2
4
4 .
(c)  0 1 0  .
2 3
3 −2 −5
0 0 1
Exercise 1.2.14. Show that there are no 2 × 2 matrices A and B such that AB − BA =
1 0
.
0 1
Exercise 1.2.15. Show that the jth column of the matrix product AB is equal to the matrix product ABj , where
Bj is the jth column of B.
Exercise 1.2.16. Show that if A X = B has more than one solution, then it has infinitely many solutions. (Hint:
If X1 and X2 are solutions, consider X3 = αX1 + βX2 , where α + β = 1)
1.3. ALGEBRAIC PROPERTIES OF MATRIX OPERATIONS
In this section we consider the algebraic properties of the matrix operations just defined. Many of these properties are
similar to the familiar properties holding for the real numbers. However, there will be striking differences between the
set R of real numbers and the set Mm×n (R) of all m×n matrices in their algebraic behavior under certain operations,
for example, under multiplication (as seen in Section 1.2). Most of the properties will be stated as theorems, whose
proof will be left as exercises.
Theorem 1.3.1. If A and B are in Mm×n (R), then A + B = B + A.
Proof.
Let
A = [aij ], B = [bij ], A + B = C = [cij ], and B + A = D = [dij ].
We must show that cij = dij for all i, j. Now cij = aij + bij and dij = bij + aij for all i, j. Since aij and bij are
real numbers, we have aij + bij = bij + aij , which implies that cij = dij for all i, j. ⊔
⊓
Example 1.3.1. We have
1 2 3
3 2 −1
4 4 2
+
=
−2 1 4
3 −1 2
1 0 6
3 2 −1
1
=
+
3 −1 2
−2
2 3
1 4
.
Theorem 1.3.2. If A, B, and C are in Mm×n (R), then A + (B + C) = (A + B) + C.
Proof.
⊔
⊓
Example 1.3.2. We have
1 2
2
+
3 4
3
1
−2
+
3 1
−2 1
6 4
=
4 3
1 2
2 1
3 1
=
+
+
.
3 4
3 −2
−2 1
Theorem 1.3.3. There exists a unique 0m×n in Mm×n (R) such that A + 0m×n = 0m×n + A = A
Mm×n (R).
for any A ∈
Proof. Let U = [uij ]. Then A + U = A if and only if aij + uij = aij , which holds if and only if uij = 0. Thus
U is the m × n matrix all of whose entries are zero; U is denoted by 0. ⊔
⊓
We call 0m×n the m × n zero matrix. When m = n, we write 0n . When m and n are understood, we shall write
0m×n merely as 0.
Theorem 1.3.4. For any A in Mm×n (R), there exists B in Mm×n (R) such that A + B = 0m×n .
Proof.
⊔
⊓
We can now show that B is unique and that it is (−1)A, which we have already agreed to write as −A, and call it
the negative of A.
Example 1.3.3. If
A =
1 3
−2 4
−2
3
, then − A =
−1
2
−3 2
−4 −3
Theorem 1.3.5. If A is in Mm×n (R) and B is in Mn×p (R), and C is in Mp×q (R), then A(BC) = (AB)C.
Proof.
We shall prove the result for m = 2, n = 3, p = 4, and q = 3. The general proof is completely
analogous.
Let A = [aij ], B = [bij ], C = [cij ], AB = D = [dij ], BC = E = [eij ], (AB)C = F = [fij ], and A(BC) = G =
[gij ]. We must show that fij = gij for all i, j. Now
!
p
p
n
X
X
X
fij =
dik ckj =
air brk ckj
k=1
and
gij =
n
X
air erj =
r=1
Then
fij =
p
X
k=1
= ai1
b1k ckj + ai2
air
r=1
⊔
⊓
p
X
air
r=1
k=1
=
n
X
brk ckj
k=1
!
.
(ai1 b1k + ai2 b2k + · · · + ain bnk )ckj
p
X
n
X
r=1
k=1
p
X
k=1
p
X
brk ckj
k=1
!
b2k ckj + · · · + ain
p
X
bnk ckj
k=1
= gij .
Example 1.3.4. Let
A =
5
2
2 3
−3 4

,
Then
A(BC) =
5
2
2 3
−3 4
and
(AB)C =



−1 1 0
1 0 2
2
2 2  , and C =  2 −3 0  .
0 −1 3
2 1 0
2
B = 0
3

0 3 7
 8 −4 6  = 43
12
9 3 3

19 −1 6 13
16 −8 −8 6

1
2

0
2
0
−3
0
1
16 56
30 8

2
0
43 16 56
=
.

3
12 30 8
0
Recall Example 8 in Section 1.2, which shows that AB need not always equal BA. This is the first significant
difference between multiplication of matrices and multiplication of real numbers.
Theorem 1.3.6. Let A and B be in Mm×n (R) and C be in Mn×p (R), then
(a) (A + B) C = AC + BC.
(b)
If C is in Mm×n (R) and A and B are both in Mn×p (R), then
C (A + B) = CA + CB.
Proof. An easy exercise! ⊔
⊓
Example 1.3.5. Let
A =
2 2 3
3 −1 2
,
B =
0
2
0 1
3 −1
Then
(A + B)C =
2 2
5 2
4
1

,
and


1 0
C = 2 2 .
3 −1

1 0
 2 2  = 18 0
12 3
3 −1
and
AC + BC =
15
7
1
−4
+
3 −1
5 7
=
18 0
12 3
.
Theorem 1.3.7. If α and β are real numbers, A is m × n matrix, and B is an n × p matrix, then
(a) α(β A) = (αβ) A = β(α A).
(b)
A(αB) = α (AB).
Proof. An Easy Exercise. ⊔
⊓
Example 1.3.6. Let
A =
4
2
Then
2(3A) = 2
2 3
−3 4
A(2B) =
4
2
2 3
−3 4

6
= 4
0

3
B = 2
0
and
12 6
9
6 −9 12
We also have
=

−2 1
0 −1  .
1
2
24 12 18
12 −18 24
= 6A.

−4 2
32 −10 16

0 −2 =
= 2(AB).
0
0
26
2
4
Theorem 1.3.8. If α and β are real numbers and A is in Mm×n (R), then
(α + β) A = αA + βA.
Proof. An easy exercise. ⊔
⊓
Theorem 1.3.9. If A and B are both in Mm×n (R) and γ is any real number, then
γ(A + B) = γA + γB.
Proof. An easy exercise. ⊔
⊓
So far we have seen that multiplication and addition of matrices have much in common with multiplication and
addition of real numbers. We now look at some properties of the transpose.
Theorem 1.3.10. If A is in Mm×n (R), then (AT )T = A
Proof. An easy exercise. ⊔
⊓
Theorem 1.3.11. If A and B are both in Mm×n (R) and γ is any real number, then
(a) (γA)T = γAT .
(b)
(A + B)T = AT + B T .
Proof. An easy exercise. ⊔
⊓
Example 1.3.7. Let
A =
Then
AT
Also
1 2
−2 0
3
1
and


1 −2
= 2 0 
3 1
A+B =
4 1
1 2
Now
AT + B T
5
0
B =
BT
and
3
3
−1 2
2 −1
.


3
3
=  −2 2  .
2 −1
(A + B)T
and



4 1
= 1 2.
5 0

1
2  = (A + B)T .
0
4
= 1
5
Theorem 1.3.12. If A is in Mm×n (R) and B is in Mn×p (R), then
(AB)T = B T AT .
Proof. Let A = [aij ] and B = [bij ]; let AB = C = [cij ]. We must prove that cTij is the (i, j) entry in B T AT .
Recall from matrix multiplication that
n
X
cij =
aik bkj , where i = 1, 2, . . . , m and j = 1, 2, . . . , p.
k=1
Thus
cTij = cji =
n
X
ajk bki =
k=1
⊔
⊓
n
X
aTkj bTik =
k=1
n
X
bTik aTkj = the (i, j) entry in B T AT .
k=1
Example 1.3.8. Let
A =
AB =
Then
On the other hand,
AT
and then
1 3
2 −1
1
= 3
2
T


0 1
B = 2 2 .
3 −1
and
and
(AB)T =

2
−1 
3
and
BT =
12 7
5 −3
12 5
7 −3

2
3
T
B A
=
12 7
5 −3
0 2
1 2
3
−1
.
,
= (AB)T .
We also note two other peculiarities of matrix multiplication. If α and β are real numbers, then αβ = 0 can hold
only if α = 0 or β = 0. However, this is not true for matrices. This means it is possible for the matrix product AB
being zero without neither A nor B being zero matrices (very strange indeed!).
1
2
2
4
Example 1.3.9. If A =
and B =
0 0
product AB =
is the zero matrix.
0 0
4 −6
, then neither A nor B is the zero matrix but the matrix
−2 3
The Cancelation Law for Real numbers: If α, β, and γ are real numbers for which αβ = αγ and α 6= 0, it
follows that β = γ. That is, we can cancel out the nonzero factor α. However, the cancelation law does not hold for
matrices, as the following example shows.
Example 1.3.10. If
A =
1 2
2 4
B =
2 1
3 2
then
AB = AC =
but B 6= C.
,
and
8 5
16 10
C =
,
−2 7
5 −1
,
ALGEBRAIC NOTION OF A RING
A Ring is a set R together with two binary operations ⊕ : R × R → R called addition, and ⊙ : R × R → R called
multiplication. This means that we are given some rule on how to add and multiply the elements or objects in R.
The elements or objects in R together with its two binary operations obeys or satisfies the following rules or axioms.
Ring Axioms
1)
A ⊕ B = B ⊕ A for all A and B in R (commutative law of addition).
2)
(A ⊕ B) ⊕ C = A ⊕ (B ⊕ C) for all A, B, and C in R (associative law of addition).
3)
There exist an element 0R ∈ R such that A ⊕ 0R = 0R ⊕ A for all A in R (existence of an additive identity).
4)
For each A in R there corresponds an element −A in R such that A ⊕ (−A) = (−A) ⊕ A = 0R (existence
of additive inverses).
5)
(A ⊙ B) ⊙ C = A ⊙ (B ⊙ C) for all A, B, and C in R (associative law of multiplication).
6)
A ⊙ (B ⊕ C) = (A ⊙ B) ⊕ (A ⊙ C) (left distributive law)
7)
(B ⊕ C) ⊙ A = (B ⊙ A) ⊕ (C ⊙ A) (right distributive law).
8)
There exists an element IR in R (IR 6= 0R ) such that A ⊙ IR = IR ⊙ A = A for all A in R (existence of a
multiplicative identity).
Any set obeying the first 7 axioms is called a Ring. If a ring R happens to obey axiom 8 also, then we say that R
is a ring with unity. Think of unity as an element in the ring R which behaves as if it was the real number 1. If
R is a ring with unity and A ⊙ B = B ⊙ A for all A, B ∈ R, then we say that R is a Commutative Ring with unity.
Otherwise, if there exists A, B ∈ R for which A ⊙ B 6= B ⊙ A, then R is a Non-Commutative Ring with Unity.
Theorem 1.3.13. The set Mm×n (R) of all m × n matrices equipped with addition and multiplication is a noncommutative ring (without unity whenever m 6= n).
Proof. See the theorems and examples of this section. ⊔
⊓
In this section we have developed a number of properties about matrices and their transposes. If a future problem
either asks a question about these ideas or involves these concepts, refer to these properties to help answer the
question. These results can be used to develop many more results.
SUGGESTED EXERCISES FOR SECTION 1.3
Exercise 1.3.1. Prove Theorem 1.3.2.
Exercise 1.3.2. Prove Theorem 1.3.4.
Exercise 1.3.3. Verify Theorem 1.3.5 for the following matrices:
A =
1
3
2 −1
,
B =
−1
3 2
1 −3 4
,


1
0
C =  3 −1  .
1
2
and
Exercise 1.3.4. Prove Theorem 1.3.6.
Exercise 1.3.5. Verify Theorem 1.3.6(b) for the following matrices:
2 −3
2
0 1
2
A =
,
B =
,
and
3 −1 −2
1 3 −2
C =
1 −3
−3
4
.
Exercise 1.3.6. Prove Theorem 1.3.7.
Exercise 1.3.7. Verify Theorem 1.3.7(b) for the following matrices:
1
3
−1
3 2
A =
,
B =
,
2 −1
1 −3 4
and
α = −3.
Exercise 1.3.8. Find a pair of unequal nonzero matrices A and B in M2×2 (R), other than those given in Example 1.3.9, such that AB = O2×2 .
Exercise 1.3.9. Find two different solutions in M2×2 (R) of the matrix equation A2 =
A2 = AA).
0
0
0
0
1 0
0 1
= O2 (Recall
Exercise 1.3.10. Prove Theorem 1.3.8.
Exercise 1.3.11. Verify Theorem 1.3.8 for α = 4, β = −2 and A =
2
4
−3
.
2
2
Exercise 1.3.12. Find two different solutions in M2×2 (R) of the matrix equation A =
Exercise 1.3.13. Prove Theorem 1.3.9.
Exercise 1.3.14. Verify Theorem 1.3.9 for γ

4
A = 1
3
= −3 and

2
−3 
and
2


0 2
B =  4 3.
−2 1
= I2 .
Exercise 1.3.15. Find A, B ∈ M2×2 (R) such that A 6= B and AB =
1
0
0
1
= I2 .
Exercise 1.3.16. Prove Theorem 1.3.10.
Exercise 1.3.17. Find matrices A, B, and C all in M3×3 (R), such that AB = AC with B 6= C and A 6= O2 .
Exercise 1.3.18. Prove Theorem 1.3.11.
Exercise 1.3.19. Verify Theorem 1.3.11 for A =
1 3
2 1
2
−3
Exercise 1.3.20. Verify Theorem 1.3.12 for A =
1 3
2 1
2
−3
, B =
4 2 −1
, and γ = −4.
−2 1
5

3
and B =  2
1

−1
4 .
2
Exercise 1.3.21. Let A be in Mm×n (R) and γ is a real number. Show that if γA = O, then γ = 0 or A = O.
Exercise 1.3.22. Determine all A in M2×2 (R) such that AB = BA for any B in M2×2 (R).
1.4. SPECIAL TYPES OF MATRICES
We have already introduced one special type of matrix, the zero m × n matrix, denoted by Om×n whose entries
consists entirely of zeros. We now consider several other types of matrices whose structure is rather specialized and
for which it will be convenient to have special names
DIAGONAL MATRICES
Definition 1.4.1. An n × n matrix A = [aij ] is called a diagonal matrix if aij = 0 for i 6= j. Thus, for a
diagonal matrix, the entries off the main diagonal are all zero.
SCALAR MATRICES
Definition 1.4.2. A scalar matrix is a diagonal matrix whose diagonal entries are equal.
IDENTITY MATRIX
Definition 1.4.3. The scalar matrix In = [aij ], where aii = 1 and aij = 0 for i 6= j, is called the n × n identity
matrix.
Example 1.4.1. Let

1 0
A = 0 2
0 0

0
0,
3

2
B = 0
0

0 0
2 0,
0 2
and

1 0
I3 =  0 1
0 0

0
0.
1
Then A, B, and I3 are diagonal matrices; B and I3 are scalar matrices; and I3 is the 3 × 3 identity matrix.
It is easy to show that if A is any m × n matrix, then
AIn = A
and
Im A = A.
Also, if A is a scalar matrix, then A = α In for some scalar α.
Definition 1.4.4. Suppose that A is a square matrix, that is, A is in Mn×n (R) = Mn (R) the set of all matrices
with the same number of rows and column with real entries. If p is any positive integer, then we define
If A is in Mn (R), then we also define
Ap = A
| · A ·{z· · · · A} .
p factors of A
A0 = In .
Definition 1.4.5. For nonnegative integers p and q, the familiar law of exponents for the real numbers can also
be proved for matrix multiplication (see Exercise 1.4.25) of a square matrix A:
Ap Aq = Ap+q
and
(Ap )q = Apq .
It should be noted that the rule
(AB)p = Ap B p
does not hold for square matrices unless AB = BA (see Exercise 1.4.26).
UPPER AND LOWER TRIANGULAR MATRICES
Definition 1.4.6. An n × n matrix A = [aij ] is called upper triangular if aij = 0 for i > j. It is called lower
triangular if aij = 0 for i < j.
Example 1.4.2. The matrix


3 3
3 5
0 2

0
3
5
1
A = 0
0
is upper triangular and
1
B = 2
3
is lower triangular.

0
0
2
SYMMETRIC MATRICES
Definition 1.4.7. A matrix A is called symmetric if AT = A.
SKEW SYMMETRIC MATRICES
Definition 1.4.8. A matrix A is called skew symmetric if AT = −A.

1
Example 1.4.3. A =  2
3


2 3
4 5  is a symmetric matrix.
5 6
0 2
Example 1.4.4. B =  −2 0
−3 4

3
−4  is a skew symmetric matrix.
0
We can make a few observations about symmetric and skew symmetric matrices; the proofs of most of these statements
will be left as exercises. It follows from the definitions above that if A is symmetric or skew symmetric, then A is a
square matrix. If A is a symmetric matrix, then the entries of A are symmetric with respect to the main diagonal
of A. Also, A is symmetric if and only if aij = aji and A is skew symmetric if and only if aij = −aji . Moreover, if
A is skew symmetric, then the entries on the main diagonal of A are all zero. An important property of symmetric
and skew symmetric matrices is the following.
Theorem 1.4.1. If A is an n × n matrix, then A = S + K, where S is symmetric and K is skew symmetric.
Moreover, this decomposition is unique.
Proof. Assume that there is such a decomposition A = S + K, where S is symmetric and K is skew symmetric.
We shall determine S and K. Now AT = S T + K T = S − K. Thus we have the expressions
A = S +K
AT = S − K.
Adding these two expressions, we obtain A + AT = 2S, so
1
(A + AT ).
2
S =
Subtracting instead of adding leads to
1
(A − AT ).
2
It is easy to verify that A = S + K, that S is symmetric, and that K is skew symmetric. Thus we have shown that
such a representation is possible and that the expressions for S and K are unique. ⊔
⊓
K =

1
Example 1.4.5. Let A =  4
5

3 −2
6
2 . Then
1
3
S =

1

1

(A + AT ) =  72
2


0


K =  12

7
2
− 21
−2
0
1
2
− 12
0
7
3
2
7
2
3
2
6
3
2
3
2
3



,



,

and A = S + K.
NONSINGULAR (or INVERTIBLE) MATRICES
We now come to a special type of square matrix and formulate the notion corresponding to the reciprocal of a nonzero
real number.
Definition 1.4.9. An n × n matrix A is called nonsingular, or invertible, if there exists an n × n matrix B such
that AB = BA = In . Otherwise, A is called singular, or noninvertible; B is called an inverse of A.
Example 1.4.6. Let A =
inverse of A.
2 3
2 2
and let B =
−1 32
. Since AB = BA = I2 , we conclude that B is an
1 −1
Theorem 1.4.2. The inverse of a matrix, if it exists, is unique.
Proof. Let B and C be inverses of A. Then AB = BA = In and AC = CA = In . We then have B = BIn =
B(AC) = (BA)C = In C = C, which prove that the inverse of a matrix, if it exits, is unique. ⊔
⊓
We now write the inverse of a nonsingular matrix A, as A−1 . Thus
AA−1 = A−1 A = In .
Example 1.4.7. Let
A =
To find A−1 if it exists, we let
A−1 =
Then we must have
so that
a
c
b
d
.
.
b
d
= I2 =
a + 2c b + 2d
3a + 4c 3b + 4d
AA−1 =
1 2
3 4
1 2
3 4
a
c
=
1 0
0 1
1
0
0
1
,
.
Equating corresponding entries of these two matrices, we obtain the linear systems
a + 2c = 1
b + 2d = 0
and
3a + 4c = 0
3b + 4d = 1.
3
2,
The solutions are (you should verify this) a = −1, c =
also satisfies the property that
−2
3
2
a
c
b
d
1
− 21
b = 1, and d = − 21 . Moreover, since the matrix
=
−2
3
2
1
− 21
=
1
0
1
− 12
.
1
3
2
4
−2
0
1
,
we conclude that A is nonsingular and that
A−1 =
Example 1.4.8. Let
A =
To find A−1 if it exists, we let
−1
A
Then we must have
−1
AA
so that
=
a
c
b
d
.
.
= I2 =
a + 2c b + 2d
2a + 4c 2b + 4d
1 2
2 4
1 2
2 4
b
d
=
3
2
a
c
=
1 0
0 1
1
0
0
1
,
.
Equating corresponding entries of these two matrices, we obtain the linear systems
a + 2c = 1
b + 2d = 0
and
2a + 4c = 0
These linear systems have no solutions, so A has no inverse.
We next establish several properties of inverses of matrices.
2b + 4d = 1.
Theorem 1.4.3. If A and B are both nonsingular matrices in Mn (R), then the matrix product AB is nonsingular
and (AB)−1 = B −1 A−1 .
Proof. We have (AB)(B −1 A−1 ) = A(BB −1 )A−1 = (AIn )A−1 = AA−1 = In . Similarly, (B −1 A−1 )(AB) = In .
Therefore, AB is nonsingular. Since the inverse of a matrix is unique, we conclude that (AB)−1 = B −1 A−1 . ⊔
⊓
Corollary 1.4.1. If A1 , A2 , . . . , Ar are nonsingular matrices in Mn (R), then the matrix product A1 A2 · · · An is
−1
−1
nonsingular and (A1 A2 · · · Ar )−1 = A−1
r Ar−1 · · · A1 .
Proof. An easy exercise. ⊔
⊓
Theorem 1.4.4. If A is a nonsingular matrix in Mn (R), then A−1 is nonsingular and (A−1 )−1 = A.
Proof. An easy exercise. ⊔
⊓
Theorem 1.4.5. If A is a nonsingular matrix in Mn (R), then AT is nonsingular and (AT )−1 = (A−1 )T .
Proof. We have AA−1 = In . Taking transposes of both sides, we obtain (A−1 )T AT = InT = In . Taking
transposes of both sides of the equation A−1 A = In , we find, similarly, that (AT )(A−1 )T = In . These equations
imply that (A−1 )T = (AT )−1 . ⊔
⊓
Example 1.4.9. If
A =
then from Example 1.4.7
A−1 =
−2
3
2
1
− 21
and
1 2
3 4
,
(A−1 )T =
3
−2
2
1 − 21
.
Also (you should verify this!)
T
A
=
1 3
2 4
and
T −1
(A )
=
3
−2
2
1 − 21
.
It follows from Theorem 1.4.5 that if A is a symmetric nonsingular matrix, then A−1 is symmetric (see Exercise 1.4.14).
Suppose that A is nonsingular. Then AB = AC implies that B = C (see Exercise 1.4.8) and AB = On implies
that B = On (see Exercise 1.4.11).
LINEAR SYSTEMS AND INVERSES
If A is an n × n matrix, then the linear system A X = B is a system of n equations in n unknowns. Suppose that
A is nonsingular. Then A−1 exists and we can multiply A X = B by A−1 on both sides, obtaining
A−1 (A X) = A−1 B,
or
In X = A−1 B.
Moreover, X = A−1 B is clearly a solution to the given linear system. Thus, if A is nonsingular, we have a unique
solution.
Applications. This observation is useful in industrial problems. Many physical models are described by linear
systems. This means that if n values are used as inputs (which can be arranged as the n × 1 matrix X), then m
values are obtained as outputs (which can be arranged as the m × 1 matrix B) by the rule A X = B. The matrix
A is inherently tied to the process. Thus suppose that a chemical process has a certain matrix A associated with it.
Any change in the process may result in a new matrix. In fact, we speak of a black box, meaning that the internal
structure of the process does not interest us. The problem frequently encountered in systems analysis is that of
determining the input to be used to obtain a desired output. That is, we want to solve the linear system A X = B
for X as we vary B. If A is a nonsingular square matrix, an efficient way of handling this is as follows. Compute
A−1 once; then whenever we change B, we find the corresponding solution X by forming A−1 B.
Example 1.4.10. Consider an industrial process whose matrix is the matrix A of Example 1.4.7. If B is the output
matrix
8
,
6
then the input matrix X is the solution to the linear system A X = B. Using the result from Example 1.4.7, we
have
−2
1
8
−10
−1
X = A B =
=
.
3
− 21
6
9
2
On the other hand, if B is the output matrix
then
X = A−1
10
20
10
20
,
=
0
.
5
SUGGESTED EXERCISES FOR SECTION 1.4
Exercise 1.4.1.
(a) Show that if A is in Mm×n (R), then Im A = A and AIn = A.
(b)
Show that if A is any scalar matrix in Mn (R), then A = γ In for some real number γ.
Exercise 1.4.2. Prove that the sum, product, and scalar multiple of diagonal, scalar, and upper (lower) triangular
matrices is diagonal, scalar, and upper (lower) triangular, respectively.
Exercise 1.4.3.
(a) Show that A is symmetric if and only if aij = aji for all i, j.
(b)
Show that A is skew symmetric if and only if aij = − aji for all i, j.
(c)
Show that if A is skew symmetric, then the elements on the main diagonal of A are all zero.
Exercise 1.4.4. Show that if A is a symmetric matrix, then AT is symmetric.
Exercise 1.4.5. Show that if A is any n × n matrix, then
(a) AAT and AT A are symmetric.
(b)
A + AT is symmetric.
(c)
A − AT is skew symmetric.
Exercise 1.4.6. Let A and B be symmetric matrices.
(a) Show that A + B is symmetric.
(b)
Show that AB is symmetric if and only if AB = BA.


3 −2 1
Exercise 1.4.7. Write the matrix A =  5
2 3  as a sum of a symmetric and a skew symmetric ma−1 6 2
trix.endexercise
Exercise 1.4.8. Show that if AB = AC and A is nonsingular, then B = C.
Exercise 1.4.9. Find the inverse of A =
1 3
.
5 2
Exercise 1.4.10.
Exercise 1.4.11. Show that if A is nonsingular and AB = On for an n × n matrix B, then B = On .
Exercise 1.4.12. Let A =
a
c
b
. Show that A is nonsingular if and only if ad − bc 6= 0
d
Exercise 1.4.13. Consider the linear system A X = B, where A is the matrix defined in Exercise 1.4.9.
(a)
(b)
3
Find a solution if B =
.
4
5
Find a solution if B =
.
6
Exercise 1.4.14. Prove that if A is symmetric and nonsingular, then A−1 is symmetric.
Exercise 1.4.15. Consider the homogeneous system A X = O, where A is n × n. If A is nonsingular, show that
the only solution is the trivial one, X = O.
Exercise 1.4.16. Prove that if one row (column) of the n × n matrix A consists entirely of zeros, then A is singular.
(Hint: Assume that A is nonsingular, that is, there exists an n × n matrix B such that AB = BA = In . Establish
a contradiction.)
Exercise 1.4.17.
Exercise 1.4.18. Prove Corollary 1.4.1.
Exercise 1.4.19. Prove Theorem 1.4.4
Exercise 1.4.20. Show that the matrix A =
Exercise 1.4.21. Let
2 3
4 6


3 2 −1
A =  0 −4 3 
0 0
0
is singular.
and
Verify that A + B and AB are upper triangular.

6 −3
B = 0 2
0 0

2
4.
3
Exercise 1.4.22. Find two 2 × 2 singular matrices whose sum is nonsingular.
Exercise 1.4.23. Find two 2 × 2 nonsingular matrices whose sum is singular.
Exercise 1.4.24. If A is a nonsingular matrix whose inverse is
2
4
1
, find A.
1
Exercise 1.4.25. Let p and q be nonnegative integers and let A be a square matrix. Show that
Ap Aq = Ap+q
and
(Ap )q = Apq .
Exercise 1.4.26. If AB = BA, and p is a nonnegative integer, show that (AB)p = Ap B p .
Exercise 1.4.27. If p is a nonnegative integer and α is a scalar, show that
(α A)p = αp Ap .
Exercise 1.4.28. Consider an industrial process with associated linear system A X = B, where A is n×n. Suppose
that
1 2
A−1 =
.
1 3
4
8
Find the input matrix for each of the following output matrices: (a)
,
(b)
.
6
15
Exercise 1.4.29. If


4
0 0
D =  0 −2 0  .
0
0 3
Find D−1 .
Exercise 1.4.30. If
−1
A
=
3
1
2
3
and
B
−1
=
2
3
5
−2
,
find (AB)−1 .
Exercise 1.4.31. Describe all skew symmetric scalar matrices.
Exercise 1.4.32. Describe all matrices that are both upper and lower triangular.
1.5. ECHELON FORM OF A MATRIX
In this section we take the elimination method for solving linear systems, learned in high school, and systemize it
by introducing the language of matrices. This will result in two methods for solving a system of m linear equations
in n unknowns. These methods take the augmented matrix of the linear system, perform certain operations on it,
and obtain a new matrix that represents an equivalent linear system (that is, has the same solutions as the original
linear system). The important point here is that the latter linear system can be solved very easily.
For example, if

1
0
0
2 0
1 1
0 1

|
3
|
2
| −1
represents the augmented matrix of a linear system, then the solution is easily found from the corresponding equations
x1
+
2x2
x2
+
x3
x3
=
3
=
2
= −1.
The task of this section is to manipulate the augmented matrix representing a given linear system into a form from
which the solution can easily be found.
Definition 1.5.1. An m × n matrix A is said to be in reduced row echelon form (r.r.e.f) if it satisfies the
following properties:
(a)
All rows consisting entirely of zeros, if any, are at the bottom of the matrix.
(b)
The first nonzero entry (called the leading entry) in each row not consisting entirely of zeros must
be a 1, called a leading one.
(c)
If row i is any row not consisting entirely of zeros, then the leading one of row i is to the right of the
leading ones occurring in any rows k where k < i.
(d)
If a column contains a leading entry of some row, then all other entries in that column are zero.
If A satisfies properties (a), (b), and (c), it is said to be in row echelon form (r.e.f). In Definition 1.5.1, there may
be no rows that consist entirely of zeros.
A similar definition can be formulated in the obvious manner for reduced column echelon form (r.c.e.f) and
column echelon form (c.e.f).
Example 1.5.1. The following are matrices in row echelon form:

1
0

A = 0

0
0
and
5
1
0
0
0
0
0
0
0
0
2
3
1
0
0

−2
4

1
4
8

0
7 −2  , B = 
0

0
0
0
0
0

0
0
C = 
0
0
0
0
0
0
1
0
0
0
3
0
0
0
5
1
0
0
0
1
0
0
0
0
1
0

0
0
,
0
1

9
3

1
0
7
−2
0
0
Example 1.5.2. The following are matrices in reduced row echelon form:

1
0
B = 
0
0
0
1
0
0
0
0
1
0

0
0
,
0
1

1
0

D = 0

0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0

−2
4
4
8

7 −2 

0
0
0
0
and

The following matrices are not in reduced

1 2
E = 0 0
0 0

1
0
H = 
0
0
1 2
E = 0 0
0 0
0
1
0

0 1
2 3.
0 0
row echelon form. (Why not?)



0
4
1 0
3 4
0
0 ,
G =  0 2 −2 5  ,
1 −3
0 0
1 2

0
3 4
1 −2 5 
,
1
2 2
0
0 0

1
0
J = 
0
0

2
3 4
1 −2 5 

0
1 2
0
0 0
We shall now show that every matrix can be put into row (column) echelon form, or into reduced row (column)
echelon form, by means of certain row (column) operations.
Definition 1.5.2. An elementary row (column) operation on a matrix A is any one of the following operations:
(a)
Interchange rows (columns) i and j of A.
(b)
Multiply row (column) i of A by α 6= 0.
(c)
Add α times row (column) i of A to row (column) j of A, i 6= j.
Observe that when a matrix is viewed as the augmented matrix of a linear system, the elementary row operations are
equivalent, respectively, to interchanging tow equations, multiplying an equation by a nonzero constant, and adding
a multiple of one equation to another equation.
Definition 1.5.3. An m × n matrix A is said to be row (column) equivalent to an m × n matrix B if B can be
obtained by applying a finite sequence of elementary row (column) operations to A.
Example 1.5.3. The matrix
is row equivalent to

2 4
1 3
−1 2

3
2
3

4 8
−1 2
−1 7

6
3,
8
1
A = 2
1
2
D = 1
4
because if we add twice row 3 of A to row 2 of A, we obtain

1
2 4
B =  4 −1 7
1 −1 2
Interchanging rows 2 and 3 of B, we obtain

Multiplying row 1 of C by 2, we obtain D.
We can easily show (see Exercise 1.5.1) that:
1
2 4
C =  1 −1 2
4 −1 7

3
8.
3

3
3,
8
(a)
Every matrix is row equivalent to itself;
(b)
if A is row equivalent to B, then B is row equivalent to A; and
(c)
if A is row equivalent to B and B is row equivalent to C, then A is row equivalent to C.
In view of (b), both statements “A is row equivalent to B” and “B is row equivalent to A” can be replaced by “A
and B are row equivalent.” A similar statement holds for column equivalence.
Theorem 1.5.1. Every nonzero m × n matrix A = [aij ] is row (column) equivalent to a matrix in row (column)
echelon form.
Proof. We shall prove that A is row equivalent to a matrix in row echelon form, that is, by using only elementary
row operations we can transform A into a matrix in row echelon form. A completely analogous proof using elementary
column operations establishes the result for column equivalence.
Step 1: We look in matrix A for the first column with a nonzero entry; say this is column j and say
that this first nonzero entry in column j occurs in row i; that is, entry aij 6= 0. Now interchange (if
necessary), the rows 1 and i, thus obtaining matrix B = [bij ] where entry b1j 6= 0.
Step 2: Multiply all entries in row 1 of B by 1/b1j , obtaining C = [cij ] whose entry c1j = 1.
Step 3: Now if chj 6= 0 where 2 6 h 6 m, then to row h of C we add −chj times row 1; all elements
in column j, rows 2, 3, . . . , m are zero. Denote the resulting matrix by D. Note that we have used only
elementary row operations.
Step 4: Next, consider the (m − 1) × n submatrix A1 of D obtained by mentally deleting the first
row of D.
We now repeat the four step procedure above with submatrix A1 instead of matrix A. Continuing this way, we
obtain a matrix H in row echelon form which is row equivalent to A. ⊔
⊓
Example 1.5.4. Let

0
0
A = 
2
2
2
0
2
0

3 −4 1
2
3 4
.
−5
2 4
−6
9 7
Find a matrix H in echelon form which is row equivalent to A.
Solution. Step 1: Column 1 is the first (counting from left to right) column in A with a nonzero entry. The first
(counting from top to bottom) nonzero entry in the first column occurs in the third row, that is, a31 . We interchange
the first and third rows of A to produce a matrix B given as

2
0
B = 
0
2
Step 2: Multiply the first row of B by
1
b11
=
1
2,
2
0
2
0

−5
2 4
2
3 4
.
3 −4 1
−6
9 7
to produce a matrix C given as

1
0
C = 
0
2
1
0
2
0
− 25
2
3
−6
1
3
−4
9
Step 3: Add −2 times the first row of C to the fourth row of C to
entry in the first column is d11 = 1.

1
1 − 25
1
0
2
3
0
D = 
0
2
3 −4
0 −2 −1
7

2
4

1
7
produce a matrix D in which the only nonzero

2
4

1
3
Step 4: Identify A1 as the submatrix A1 obtained by mentally deleting the first row of D; do not physically erase
the first row of D.
Repeat the four steps above using the submatrix A1 which we write below with the first row of D separated from
its submatrix A1 .
1
− 25
1

1

3 4
−4 1  Interchange the first and Second rows of A1 to obain
7 3
0
0
2
A1 =  0
2
3
0 −2 −1
1
− 25
1

1
1
− 25
1
3
2
0
1
C1 =  0
0
2
0 −2 −1
1

1
0 1
D1 =  0 0
0 0
− 25
3
2
2
2
2

−4 1
3 4  Multiply the first row of B1 by 1/2 to obain
7 3
0
2
3
B1 =  0
0
2
0 −2 −1

2
1
2
−2
3
7
1
2
1
−2
3
3
4  Add two times the first row of C1 to the third row to obain
3
2
1
2


4
4
Identify A2 as the submatrix A2 obtained by mentally deleting the first row of D1 ; do not physically erase the first
row of D1 .
Repeat the four steps above using the submatrix A2 which we write below with the first row of D1 separated from
its submatrix A2 . No row of A2 needs to be interchanged; So then we have B2 = A2 .
1
A2 =
1
−2
0 0
0 0
2
2
3
3
1
C2 =
1
2
0 0
0 0
1
2
3
2
2
4
1 − 52
= B2 Multiply the first row of B2 by 1/2 to obtain
1 2
−2
3
Finally, add -2 times the first row of C2 to its second row to obtain
1 2
1
3
2
−2
1
2
0 0
0 0
1
0
3
2
2
0
0
4
4
1
1
D2 =
1 − 52
1
2
3
2
0
1 2
3
2
0
1 − 52
0
.
The matrix

1


0

H = 

0

0
is in row echelon form and is row equivalent to A. ⊔
⊓
1 − 52
1
1
3
2
−2
0
1
3
2
0
0
0
2







2

1
2
0
When doing hand computations, it is sometimes possible to avoid messy fractions by suitably modifying the steps
in the procedure.
Theorem 1.5.2. Every nonzero m × n matrix A = [aij ] is row (column) equivalent to a matrix in reduced row
(column) echelon form.
Proof. We proceed as in Theorem 1.5.1, obtaining a matrix H in row echelon form which is row equivalent to A.
In H, if row i contains a nonzero entry, then its first (counting from left to right) nonzero entry is a leading one.
Let’s suppose this leading one is in column j which we denote by cj . Then c1 < c2 < · · · < cj < . . . < cr where
r (1 6 r 6 m) is the number of nonzero rows in H. Add suitable multiples of row i of H to all the rows of H
preceding row i to make all entries in column cj and rows i − 1, i − 2, . . . , 1 of H equal to zero; except the leading
ones. The result is a matrix K in reduced row echelon form which has been obtained from H by elementary row
operations only and is thus now row equivalent to H. Since A is row equivalent to H and H is row equivalent to K,
then A is row equivalent to K. An analogous proof can be given to show that A is column equivalent to a matrix in
reduced column echelon form. ⊔
⊓
It can be shown, with some difficulty, that:
For a given nonzero m × n matrix A,there is only one matrix B in reduced row (column) echelon form
that is row (column) equivalent to A.
The proof of this statement is omitted.
Example 1.5.5. Suppose that we wish to find a matrix in reduced row echelon form that is row equivalent to the
matrix A of Example 1.5.4. Starting with the matrix H obtained there, we add −1 times the second row to the first
row, obtaining

3 
1 0 −4 3
2
3
1
 0 1 2 −2 2 

.
3
0 0 1
2
2
0 0 0
0 0
In this matrix we add -3/2 times the third row to its second row and 4 times the third row to its first row. This
yields

19 
1 0 0
9
2
17
 0 1 0 − 4 − 52 

,
3
0 0 1
2
2
0 0 0
0
0
which is in reduced row echelon form and is row equivalent to A.
We now apply these results to the solution of linear systems.
Theorem 1.5.3. Let A X = B and C X = D be two linear systems each of m equations and n unknowns. If the
augmented matrices (A | B) and (C | D) are row equivalent, then the linear systems are equivalent; that is, they
have exactly the same solutions.
Proof. This follows from the definition of row equivalence and from the fact that the three elementary row
operations on the augmented matrix are the three manipulations on linear systems, discussed in Section 1.1, which
yield equivalent linear systems. We also note that if one system has no solution, then the other system has no
solution. ⊔
⊓
Corollary 1.5.1. If A and B are row equivalent m × n matrices, then the homogeneous systems A X = O and
B X = O are equivalent.
Proof. An easy exercise. ⊔
⊓
We now pause to observe that we have developed the essential features of two very straightforward methods for
solving linear systems. The idea consists of starting with the linear system A X = B, then obtaining a partitioned
matrix (C | D) in either row echelon form or reduced row echelon form that is row equivalent to the augmented
matrix (A | B). Now (C | D) represents the linear system C X = D, which is quite simple to solve because of the
structure of (C | D), and the set of solutions to this system gives precisely the set of solutions to A X = B. The
method where (C | D) is in row echelon form is called Gaussian elimination; the method where (C | D) is in
reduced row echelon form is called Gauss-Jordan reduction. These methods are used often and computer codes
of their implementation are widely available.
We thus consider the linear system C X = D, where C is m × n, and (C | D) is in row echelon form. Then for
example, (C | D) is of the following form:

1
0

 ..
.

0

0

0

.
 ..
0
c12
0
..
.
c13
1
...
c24
0
...
...
...
0
c1n
c2n
...
1
0
c(k−1)n
1
0
..
.
...
0
|
|

d1
d2
..
.





dk−1 
.
dk 

dk+1 

.. 
. 
dm
|
|
|
|
|
|
This augmented matrix represents the linear system
x1
+ c12 x2
+
c13 x3
x3
+
+
...
c24 x4
..
.
+
...
+
+
c1n xn
c2n xn
=
=
d1
d2
xn−1
+
=
=
=
dk−1
dk
dk+1
=
dm .
0x1
..
.
+
...
+
c(k−1)n xn
xn
0xn
..
.
0x1
+
...
+
0xn
First, if dk+1 = 1, then the linear system C X = D has no solution, since at least one equation is not satisfied.
If dk+1 = 0, which implies that dk+2 = · · · = dm = 0, we then obtain xn = dk , xn−1 = dk−1 − c(k−1)n xn =
dk−1 − c(k−1)n dk , and continue using backward substitution to find the remaining unknowns corresponding to the
leading entry in each row. Of course, in the solution, some of the unknowns may be expressed in terms of others
that can take on any values whatsoever. This merely indicates that C X = D has infinitely may solutions. On the
other hand, every unknown may have a determined value, indicating that the solution is unique.
Example 1.5.6. Consider the linear system C X = D whose augmented matrix (C | D) is in row echelon form as

1
..
0
(C . D) = 
0
0
Then
2
1
0
0
3
2
1
0
4 5
3 −1
2 3
1 2
|
|
|
|

6
7
.
7
9
x4 = 9 − 2x5
x3 = 7 − x3 − 2x4 − 3x5 = 7 − 2(9 − 2x5 ) − 3x5 = −11 + x5
x2 = 7 − 2x3 − 3x4 + x5 = 2 + 5x5
x1 = 6 − 2x2 − 3x3 − 4x4 − 5x5 = −1 − 10x5
x5 = any real number.
Thus all solutions X are of the form
x1 = −1 − 10r
x2 = 2 + 5r
x3 = −11 + r
x4 = 9 − 2r
x5 = r, any real number.
or




−1
−10
 2 
 5 




X =  −11  +  1  r, where r is any real number.




9
−2
0
1
Since r can be assigned any real number, the given linear system as infinitely many solutions.
Example 1.5.7. If

1
(C | D) =  0
0
2 3
1 2
0 0

5
6
1
4 |
3 |
0 |
is the augmented matrix in row echelon form of the linear system A X = B,then linear system C X = D has no
solution, for the last equation is
0x1 + 0x2 + 0x3 + 0x4 = 1,
which can never be satisfied. Thus A X = B has no solution.
Example 1.5.8. If
then

1
0
(C | D) = 
0
0
2
1
0
0
3
2
1
0
4
3
2
1
|
|
|
|

5
6
,
7
8
x1 = 0
x2 = 0
x3 = −9
x4 = 8.


0
 0
X = 
.
−9
8
or
The solution to C X = D is unique.
If (C | D) is in reduced row echelon form, then we can solve C X = D without backward substitution, but, of
course, it takes more effort to put a matrix in reduced row echelon form than in row echelon form.
Example 1.5.9. If
then
Example 1.5.10. If
then

1
0
(C | D) = 
0
0
0
1
0
0
0
0
1
0
0
0
0
1
|
|
|
|

5
6
,
7
8
 
5
6
X =  .
7
8

1 1
(C | D) =  0 0
0 0
2 0
0 1
0 0
− 52
1
2
0
|
|
|
3
2
1
2
0

,
1
1
− x5
2
2
2
5
x1 =
− x2 − 2x3 + x5 ,
3
2
x4 =
where x2 , x3 , an x5 can take on any real numbers. Thus a solution is of the form
x1 =
x2 =
x3 =
x4 =
x5 =
2
5
− r − 2s + t
3
2
r
s
1
1
− t
2
2
t,
or in matrix form as
2



 5
−1
−2
2
0
 1
 0
 0
 






X = 0 +  0 r +  1 s +  0  t
1




 1
−2
0
0
2
0
0
0
1
3

where r, s, and t are any real numbers
We now solve a linear system both by Gaussian elimination and by Gauss-Jordan reduction.
Example 1.5.11. Find solutions (if any) of the linear system
x1
2x1
3x1
+
−
+
2x2
3x2
x2
+
+
−
3x3
2x3
x3
=
=
=
6
14
−2.
Solution. We form the augmented matrix of the system as

1
2
3 |
 2 −3
2 |
3
1 −1 |

6
14  .
−2
Next, we start the Gaussian elimination method by applying the elementary row operations to the above augmented
matrix.

1
2
3
2
3 |
−3
2 |
1 −1 |

1
0
3

2
3 |
6
−7 −4 |
2
1 −1 | −2

1
0
0
2
3
|
−7 −4 |
−5 −10 |

1
0
0

2
3 | 6
1
2 | 4
−7 −4 | 2

1
0
0

2 3 | 6
1 2 | 4 
0 10 | 30

2 3
1 2
0 1
1
0
0

6
14 
−2

| 6
| 4.
| 3
Add -2 times the first row to the second row to obtain
Add -3 times the first row to the third row to obtain

6
2  Multiply the third row by −1/5 and interchange the second and third rows to obtain
−20
Add 7 times the second row to the third row to obtain
Multiply the third row by 1/10 to obtain
The above matrix is in row echelon form. This means that x3 = 3 and from the second row
x2 + 2x3 = 4,
so
x2 = 4 − 2(3) = −2.
From the first row
x1 + 2x2 + 3x3 = 6,
which implies that
x1 = 6 − 2x2 − 3x3 = 6 − 2(−2) − 3(3) = 1.
Thus x1 = 1, x2 = −2, and x3 = 3 is the solution to the system. This gives the solution by Gaussian elimination.
⊔
⊓
If, instead, we wish to use Gauss-Jordan reduction, we would transform the last matrix into reduced row echelon
form by the following steps:
Add -2 times the second

1 0
0 1
0 0
row to the first row:

−1 | −2
2 |
4
Add -2 times the third row to the second row to obtain
1 |
3

1
0
0
0 −1
1
0
0
1

0 0
1 0
0 1
1
0
0

| −2
| −2 
|
3
Add the third row to the first row to obtain

|
1
| −2  .
|
3
The solution is x1 = 1, x2 = −2, and x3 = 3, as before.
NOW WE CONSIDER A HOMOGENEOUS SYSTEM A X = O CONSISTING OF m EQUATIONS
IN n UNKNOWNS.
a
11
 a21
(A | O) = 
 ..
.
am1
a12
a22
..
.
···
···
..
.
a1n
a2n
..
.
am2
···
amn
| 0
| 0
.

| ..
| 0
Example 1.5.12. Consider the homogeneous system whose augmented matrix is

1
0

0
0
0
0
0
0
0
1
0
0
0
0
1
0
2
3
4
0
|
|
|
|

0
0
.
0
0
Since the augmented matrix is in reduced row echelon form, the solution is easily seen as
x1 = −2r
x2 = s
x3 = −3r
x4 = −4r
x5 = r,
or


 
−2
0
 0
1


 
X =  −3  r +  0  s, where r and s are any real numbers.


 
−4
0
1
0
In Example 1.5.12 we solved a homogeneous system of m (= 4) linear equations in n(= 5) unknowns, where m < n
and the augmented matrix A was in reduced row echelon form. We can ignore any row of the augmented matrix that
consists entirely of zeros. Thus let rows 1, 2, . . . , r of A be the nonzero rows, and let the 1 in row i occur in column
ci . We are then solving a homogeneous system of r equations in n unknowns, r < n, and in this special case (A is
in reduced row echelon form) we can solve for xc1 , xc2 , . . . , xcr in terms of the remaining n − r unknowns. Since the
latter can take on any real values, there are infinitely many solutions to the system A X = O; in particular, there
is a nontrivial solution. We now show that this situation holds whenever we have m < n; A does not have to be in
reduced row echelon form.
Theorem 1.5.4. Any homogeneous system A X = O consisting of m linear equations in n unknowns always has a
nontrivial solution if m < n, that is, if the number of unknowns exceeds the number of equations.
Proof. Let B be a matrix in reduced row echelon form that is row equivalent to A. Then the homogeneous systems
A X = O and B X = O are equivalent. If we let r be the number of nonzero rows of B, then r 6 m. If m < n,
we conclude that r < n. We are then solving r equations in n unknowns and can solve for r unknowns in terms of
the remaining n − r unknowns, the latter being free to take on any values we please. Thus B X = O, and hence
A X = O has a nontrivial solution. ⊔
⊓
We shall soon use this result in the following equivalent form: If A is m × n and A X = O has only the trivial
solution, then m > n.
Example 1.5.13. Consider the homogeneous system
x1
x1
x1
+
x2
+ x3
+
2x2
+ x3
The augmented matrix

1
A = 1
1
is row equivalent to
Hence the solution is

1
0
0
x1 = −r
x2 = r
x3 = −r
x4 = r
or
1
0
2
0 0
1 0
0 1
+
1 1
0 1
1 0
x4
x4
= 0
= 0
= 0.

| 0
| 0
| 0

1 | 0
−1 | 0  .
1 | 0


−1
 1
X = 
 r, any real number.
−1
1
A useful property of matrices in reduced row echelon form (see Exercise 1.5.3) is that if A is an n × n matrix in
reduced row echelon form 6= In , then A has a row consisting entirely of zeros.
SUGGESTED EXERCISES FOR SECTION 1.5
Exercise 1.5.1. Prove the following statements:
(a)
Every matrix is row equivalent to itself.
(b)
If A is row equivalent to B, then B is row equivalent A.
(c)
If A is row equivalent to B and B is row equivalent to C, then A is row equivalent to C.
Exercise 1.5.2. Let

0
0
A = 
0
0
0
2
1
3

−1 2 3
3
4 5
.
3 −1 2
2
4 1
(a)
Find a matrix B in row echelon form that is row equivalent to A.
(b)
Find a matrix C in reduced row echelon form that is row equivalent to A.
Exercise 1.5.3. Let A be an n × n matrix in reduced row echelon form. Prove that if A 6= In , then A has a row
consisting entirely of zeros.
Exercise 1.5.4. Let


1 −2 0 2
 2 −3 −1 5 
A = 
.
1 3
2 5
1 1
0 2
(a)
Find a matrix B in row echelon form that is row equivalent to A.
(b)
Find a matrix C in reduced row echelon form that is row equivalent to A.
Exercise 1.5.5. Consider the linear system
x1
x1
3x1
+
−
+
x2
2x2
x2
+ 2x3
+ x3
+ x3
= −1
= −5
= 3.
(a)
Find all solutions, if any exists, by using the Gaussian elimination method.
(b)
Find all solutions, if any exists, by using the Gauss-Jordan reduction method.
Exercise 1.5.6. Repeat Exercise 1.5.5 for each of the following linear systems.
(a)
x1
x1
3x1
+
−
+
x2
2x2
x2
+
+
+
2x3
x3
x3
+
+
−
3x4
x4
x4
= 13
= 8
= 1.
(c)
2x1
3x1
x1
6x1
5x1
+
−
+
x2
2x2
x2
−
x2
+
+
−
+
+
x3
x3
x3
x3
2x3
−
−
−
−
−
2x4
6x4
x4
9x4
8x4
= 1
= −2
= −1
= −2
= 3.
Exercise 1.5.7.

1
1
(a)
0
(c)

In Exercises 7, 8, and 9, solve the


1 1 | 0
1
1
1 0 | 3
(b)
1 1 | 1
1
1 2
1 1
5 7

3 | 0
1 | 0
9 | 0
Exercise 1.5.8. (a)
Exercise 1.5.9. (a)

(d)
1 2
1 3
1 0
3 1
0 1
2 1

1
1
2
3
4
1
2

1

1
2
2
0
0
2
1
Exercise 1.5.11. Show that A =
1 2
1 2

| 8
| 7
| 3
and X =
a b
c d
3 | 0
1 | 0
+ x3
− x3
+ x3
=
=
=
1
3
2.

1
0
1
(b)
(b)
+ x2
+ x2
+ x2
linear system, if it is consistent, with given augmented matrix.

2 3 | 0
1 1 | 0
1 2 | 0

| 7
| 4 

| 5 

| 11
| 11
a b
Exercise 1.5.10. Let A =
c d
solution if and only if ad − bc 6= 0.
x1
x1
2x1
(b)

1
2

0
2

−3 | 0
−3 | 3 
−1 | −1
1 3
2 1
0 2
2
3
1
1
1
0
2
4
|
|
|
|

0
0

0
0
x1
. Show that the linear system A X = O has only the trivial
x2
is row equivalent to I2 if and only if ad − bc 6= 0.
Exercise 1.5.12. In the following linear system, determine all values of λ for which the resulting linear system has:
(1) No solution.
(b)
A unique solution.
(c)
Infinitely may solutions.
x1
x1
x1
+ x2
+ 2x2
+ x2
−
+
+
x3
x3
(λ2 − 5)x3
= 2
= 3
= λ.
Exercise 1.5.13. Repeat Exercise 1.5.12 for the linear system
x1
2x1
2x1
+ x2
+ 3x2
+ 3x2
+
+
+
x3
2x3
(λ2 − 1)x3
=
2
=
5
= λ + 1.
Exercise 1.5.14.
(a) Formulate the definitions of column echelon form and reduced column echelon form of a matrix.
(b)
Prove that every m × n nonzero matrix is column equivalent to a matrix in column echelon form.
Exercise 1.5.15. Prove that every m × n nonzero matrix is column equivalent to a matrix in reduced column
echelon form.
Exercise 1.5.16. Repeat Exercise 1.5.12 for the linear system
x1
x1
+
+
x2
(λ2 − 8)x2
=
=
3
λ.
Exercise 1.5.17. Let A be the matrix in Exercise 1.5.2.
(a) Find a matrix in column echelon form that is column equivalent to A.
(b)
Find a matrix in reduced column echelon form that is column equivalent to A.
Exercise 1.5.18. Repeat Exercise 1.5.12 for the linear system
x1
x1
x1
+
+
+
x2
2x2
x2
+
+
+
x3
x3
(λ2 − 5)x3
=
=
=
2
3
λ.
Exercise 1.5.19. Repeat Exercise 1.5.17 for the matrix

1
2
3
2
1
1

3 4 5
3 −1 2  .
2 4 1
Exercise 1.5.20. Show that if the homogeneous system
(a − r)x1
cx1
+
+
dx2
(b − r)x2
= 0
= 0
has a nontrivial solution, then r satisfies the equation (a − r)(b − r) − cd = 0.
Exercise 1.5.21. Let A X = B, B 6= O be a consistent linear system.
(a) Show that if X1 is a solution to the linear system A X = B and Y1 is a solution to the associated homogeneous
system A X = O, then X1 + Y1 is a solution to the system A X = B.
(b)
Show that every solution X to A X = B can be written as X1 + Y1 , where X1 is a particular solution to
A X = B and Y1 is a solution to A X = O. [Hint: Let X = X1 + (X − X1 ).]
Note: It is suggested that you use the methods of this section to solve the exercises in Section 1.1.
1.6. ELEMENTARY MATRICES; FINDING A−1
In this section we develop a method for finding the inverse, A−1 of a square matrix A if it exists. This method is
such that we do not have to first find out whether A−1 exists. Give the square matrix A, we start to find A−1 , if
in the course of the computation we hit a certain condition, then we know that A−1 does not exist. Otherwise, we
proceed to the end and obtain A−1 . This method requires that the three elementary row operations (a), (b), and (c)
defined in the previous section be performed on A. We clarify these notions by starting with the following definition.
Definition 1.6.1. An n × n matrix E is called an elementary matrix if it was obtained by performing elementary
row or elementary column operations on In .
Example 1.6.1. The following are elementary matrices:


1
0,
0


0
0,
1
0 0
E1 =  0 1
1 0
1 2
E3 =  0 1
0 0

1
E2 =  0
0
and

1 0
E4 =  0 1
0 0

0 0
−2 0  ,
0 1

3
0.
1
Matrix E1 was obtain by interchanging the first and third rows of I3 ; E2 was obtained by multiplying the second
row of I3 by -2; and E4 was obtained by adding 3 times the first column of I3 to its third column.
Theorem 1.6.1. Let A be an m × n matrix and B is the matrix obtained by performing a single elementary row
(column) operation on A. Let E be the elementary matrix obtained from Im (In ) by performing the same elementary
row (column) operation as was performed on A. Then B = EA (B = AE).
Proof. An easy exercise. ⊔
⊓
Theorem 1.6.1 says that an elementary row operation on A can be achieved by pre-multiplying A (multiplying A
on the left) by the corresponding elementary matrix E; an elementary column operation on A can be obtained by
post-multiplying A (multiplying A on the right) by the corresponding elementary matrix.
Example 1.6.2. Let A =

−5
first row. Thus B =  −1
3
-2 times the third row of I3


3 2 1
2 3 4  and let B result from A by adding -2 times the third row of A to its
0 1 2
−3
4 . Now let E be the elementary matrix that is obtained from I3 by adding
2


1 0 −2
to the its first row. Thus E =  0 1
0 . It is easy to verify that B = EA.
0 0
1
1
 −1
3
3 0
2 3
0 1
Theorem 1.6.2. If A and B are nonzero m × n matrices, then A is row (column) equivalent to B if and only if
B = Ek Ek−1 · · · , E2 E1 A (B = AE1 E2 · · · Ek−1 Ek ), where E1 , E2 , · · · , Ek−1 , Ek are elementary matrices.
Proof. We prove only the theorem for row equivalence. If A is row equivalent to B, then B results from A by a
sequence of elementary row operations. This implies that there exist elementary matrices E1 , E2 , . . . , Ek such that
B = Ek Ek−1 · · · E2 E1 A. Conversely, if B = Ek Ek−1 · · · , E2 E1 A, where the Ei′ s are elementary matrices, then B
results from A by a sequence of elementary row operations, which implies that A is row equivalent to B. ⊔
⊓
Theorem 1.6.3. An elementary matrix E is nonsingular and its inverse, E −1 is an elementary matrix of the same
type.
Proof. An easy exercise. ⊔
⊓
Thus an elementary row operation can be “undone” by another elementary row operation of the same type.
We now obtain an algorithm for finding the inverse, A−1 (if it exists) of a given square matrix A; first, we prove the
following lemma.
Lemma 1.6.1. Let A be an n × n matrix and let the homogeneous system A X = O have only the trivial solution
X = O. Then A is row equivalent to In .
Proof. Let B be a matrix in reduced row echelon form which is row equivalent to A. Then the homogeneous system
A X = O and B X = O are equivalent, and thus B X = O also has only the trivial solution. It is clear that if r is
the number of nonzero rows of B, then the homogeneous system B X = O is equivalent to the homogeneous system
whose coefficient matrix consists of the nonzero rows of B and is therefore r × n. Since this last homogeneous system
only has the trivial solution, we conclude from Theorem 1.5.4 that r > n. Since B is n × n, r ≤ n. Hence r = n,
which means that B has no zero rows. Thus B = In . ⊔
⊓
Theorem 1.6.4. An n × n matrix A is nonsingular if and only if A is a product of elementary matrices (a finite
number).
Proof. If A is a product of finitely many elementary matrices E1 , E2 , . . . , Ek , then A = E1 E2 · · · , Ek−1 Ek .
Now each elementary matrix is nonsingular, and the product of nonsingular matrices is again nonsingular; therefore,
A is nonsingular. Conversely, if A is nonsingular, then A X = O implies that A−1 (A X) = A−1 (O) = O, so
In X = O or X = O. Thus A X = O has only the trivial solution. Lemma 1.6.1 then implies that A is row
equivalent to In . This means that there exists elementary matrices E1 , E2 , . . . , Ek such that
In = Ek Ek−1 · · · E2 E1 A.
−1
It then follows that A = (Ek Ek−1 · · · E2 E1 )−1 = E1−1 E2−1 · Ek−1
Ek−1 . Since the inverse of an elementary matrix
is an elementary matrix, we have established the result. ⊔
⊓
Corollary 1.6.1. An n × n matrix A is nonsingular if and only if A is row equivalent to In .
Proof. If A is row equivalent to In , then In = Ek Ek−1 · · · E2 E1 A, where the Ei are all elementary matrices.
−1
Therefore, it follows that A = E1−1 E2−1 · · · Ek−1
Ek−1 . Now the inverse of an elementary matrix is an elementary
matrix, and so by Theorem 1.6.4 A is nonsingular.
Conversely, if A is nonsingular, then A is a product of elementary matrices, A = Ẽ (a product of finitely many
elementary matrices). Now A = A In = ẼIn which implies that A is row equivalent to In . ⊔
⊓
Theorem 1.6.5. The homogeneous system of n linear equations in n unknowns A X = O has a nontrivial solution
if and only if A is singular.
Proof. We can see that Lemma 1.6.1 and Corollary 1.6.1 imply that if the homogeneous system A X = O, where
A is an n × n matrix, has only the trivial solution X = O, then A is nonsingular. Conversely, consider A X = O,
where A is n × n, and let A be nonsingular. Then A−1 exists and we form A−1 (A X) = A−1 O = O. Thus X = O,
which means that the homogeneous system has only the trivial solution. ⊔
⊓
1 2
2 4
be the matrix defined in Example 1.4.8, which is singular; i.e., A−1 does not
1 2
x1
0
exist. Consider the homogeneous system AX = 0; that is,
=
. The reduced row echelon form
2
4
x
0
2
1 2 | 0
of the augmented matrix is
, and so a solution is
0 0 | 0
Example 1.6.3. Let A =
x1 = −2t

x2 = t or X = 
−2
1

 t,
where t is any real number. Thus the homogeneous system has a nontrivial solution.
At the end of the proof of Theorem 1.6.4, we had
−1
A = E1−1 E2−1 · · · Ek−1
Ek−1 ,
from which it follows that
−1
A−1 = (E1−1 E2−1 · · · Ek−1
Ek−1 )−1 = Ek Ek−1 · · · E2 E1 .
This now provides an algorithm for finding A−1 . Thus we perform elementary row operations on A until we get In ;
the product of the elementary matrices Ek Ek−1 · · · E2 E1 then gives A−1 .
An Algorithm for find A−1 : A convenient way of organizing the computing process is to write down the augmented
matrix (A | In ). Then
(Ek Ek−1 · · · E2 E1 )(A | In ) = (Ek Ek−1 · · · E2 E1 A | Ek Ek−1 · · · E2 E1 ) = (In | A−1 ).
In other words, for a given n × n matrix A the algorithm for finding A−1 as follows:
Step 1. Form the augmented n × 2n matrix (A | In ) where In is the n × n identity matrix.
Step 2. Bring the matrix (A | In ) to reduced row echelon form using only elementary row operations.
Step 3. Check if the reduced row echelon form have a leading one in each of the first n columns
(counting from left to right), then the reduced row echelon form of (A | In ) is the matrix (In | A−1 )
and A−1 is the n × n submatrix which remains after deleting In from (In | A−1 ).
Otherwise, there is either a row consisting entirely of zeros or else there is a leading one in a column
beyond the nth column (Remember the matrix (A | In ) has 2n columns!). In this case A is considered
to be singular.

1 1
Example 1.6.4. For the 3 × 3 matrix A =  0 2
5 5
Solution.

1
3 . Find A−1 if exists.
1
Step 1. Form the 3 × 6 augmented matrix (A | I3 )

1
(A | I3 ) =  0
5
1
2
5
1 | 1
3 | 0
1 | 0

0 0
1 0.
0 1
Step 2. Bring the augmented matrix above to reduced row echelon form. Here are the elementary row operations
that were used:
1.
−5R1 + R4 .
2.
1
2 R2 .
3.
−R2 + R1 .
4.
5.
6.
− 41 R3 .
− 32 R3 + R2 .
1
2 R3
+ R1 .
The reduced row echelon form of (A | I3 ) is

1


(I3 | A−1 ) =  0

0
Step 3. Hence

0 0
|
13
8
− 21
− 18
1 0
| − 15
8
1
2
3
8
0 1
|
5
4
0
− 14
− 12
− 81
1
2
3
8
0
− 41
13
8


A−1 =  − 15
 8
5
4
It is easy to verify that AA−1 = A−1 A = I3 .



.




.

The question that arise at this point is how can the algorithm tell us if A is singular? that is, when the above
algorithm for finding A−1 fails. In step 3 of the algorithm we said if the reduced row echelon form of (A | In ) has
at least one row consisting entirely of zeros or else there is a column beyond the nth column (counting from left to
right) containing a leading one. This really means that A is row equivalent to some matrix B containing a row of
zeros. Thus the answer is that A is singular if and only if A is row equivalent to a matrix B having at least one row
that consists entirely of zeros. We now prove this result.
Theorem 1.6.6. An n × n matrix A is singular if and only if A is row equivalent to a matrix B that has a row of
zeros.
Proof. First, let A be row equivalent to a matrix B that has a row consisting entirely of zeros. From Exercise 1.4.16
in section 4 it follows that B is singular. Now B = Ek Ek−1 · · · E2 E1 A, where E1 , E2 , . . . , Ek−1 , Ek are elementary
matrices. If A is nonsingular, then B is nonsingular, a clear contradiction.
Conversely, if A is singular, then A is not row equivalent to In , by Corollary 1.6.1. Thus A is row equivalent to a
matrix B 6= In , which is in reduced row echelon form. From Exercise 1.5.3 of section 5 it follows that B must have
a row of zeros. ⊔
⊓
This means that in order to find A−1 , we do not have to determine, in advance, whether or not it exists. We merely
start to calculate A−1 ; if at any point in the computation we find a matrix B that is row equivalent to A and has a
row of zeros, then A−1 does not exist.

1
Example 1.6.5. Consider the matrix A =  1
5

2 −3
−2
1 . Determine if A is singular or not.
−2 −3
Solution. Apply the algorithm for calculating A−1 . After applying the three elementary row operations
−R1 
+

1
2 −3
R2 , −5R1 + R3 , −3R2 + R3 to (A | I3 ) we find that A is row equivalent to the matrix B =  0 −4
4 .
0
0
0
Since B has a row of zeros, we stop and conclude that A is a singular matrix. ⊔
⊓
In section 4 we defined an n × n matrix B to be the inverse of the n × n matrix A if AB = In and BA = In . We
now show that one of these equations follows from the other.
Theorem 1.6.7. If A and B are n × n matrices such that AB = In , then BA = In . Thus B = A−1 .
Proof. We first show that if AB = In , then A is nonsingular. Suppose that A is singular. Then A is row equivalent
to a matrix C with a row of zeros. Now C = Ek Ek−1 · · · E2 E1 A, where all the Ei are elementary matrices. Then
CB = Ek Ek−1 · · · E2 E1 AB, so AB is row equivalent to CB. Since CB has a row of zeros, we conclude from
Theorem 1.6.6 that AB is singular. Then AB = In is impossible, because In is nonsingular. This contradiction
shows that A is nonsingular, and so A−1 exists. Multiplying both sides of the equation AB = In by A−1 on the left,
we then obtain A−1 (AB) = (A−1 A)B = In B = B = A−1 In = A−1 . Thus B = A−1 and BA = A−1 A = In
holds. ⊔
⊓
This theorem tells us that when using multiplication to verify that matrix B is the inverse of A it is enough to verify
only one of the two equations AB = In or BA = In but not both.
SUGGESTED EXERCISES FOR SECTION 1.6
Exercise 1.6.1. Prove Theorem 1.6.1.
Exercise 1.6.2. Let A be a 4 × 3 matrix. Find the elementary matrix E, which as a premultiplier of A, that is as
EA, performs the following elementary row operations on A:
(a) Multiplies the second row of A by -2.
(b) Adds 3 times the third row of A to the fourth row of A.
(c) Interchanges the first and third rows of A.
Exercise 1.6.3. Let A be a 3 × 4 matrix. Find the elementary matrix F , which as a postmultiplier of A, that is as
AF , performs the following elementary column operations on A:
(a) Adds -4 times the first column of A to the second column of A.
(b) Interchanges the second and third columns of A.
(c) Multiplies the third column of A by 4..
Exercise 1.6.4. Prove Theorem 1.6.3 (Hint: Find the inverse of the elementary matrices represented by the various
elementary row operations.)
Exercise 1.6.5. Find the inverse, if it exists, of


1 1
2 3
1 1

1 1
3 1
2 −1
9 1


2 2
3 2
1 3
1
(a)  1
0
1
1
(c) 
1
5
1
(e)  1
1


1
1 1
2 −1 2 

−1 2 1
3
3 2
1
1
(b) 
1
1

1
2

1
6

1 2
(d)  1 3
1 0
a b
be a 2 × 2 matrix and denote by det(A) the real number ad − bc. Then the
c d
statement A is nonsingular if and only if det(A) 6= 0 holds. Show that
Exercise 1.6.6. Let A =

1
2
1
A−1 =
provided that the statement holds.
1
det(A)
d
−c
−b
a
.


2 3 −1
3
 1 0
Exercise 1.6.7. Let A = 
. Find the elementary matrix that as a postmultiplier of A performs the
0 2 −3
−2 1
3
following elementary column operations on A:
(a) Multiplies the third column of A by -3.
(b) Interchanges the second column and third columns of A.
(c) Adds -5 times the first column of A to the third column of A.

1 2
Exercise 1.6.8. Prove that A =  0 1
1 0
(Hint: First, write the inverse as a product

3
2  is nonsingular and write it as a product of elementary matrices.
3
of elementary matrices then use Theorem 1.6.3.)
Exercise 1.6.9. Which of the following homogeneous systems have a nontrivial solution?
x1
+
a)
x1
+
3x1
(c) −2x1
2x1
2x2
2x2
2x2
+
+
+
+ x2
+ 2x2
− 3x2
3x3
2x3
3x3
=
=
=
0
0
0.
+ 3x3
− 4x3
+ 5x3
=
=
=
(b)
2x1
x1
−3x1
Exercise 1.6.11. Find the inverse of A =

1
2
the hint in Exercise 1.6.8.)
Exercise 1.6.14. Find the inverse

1
 −1
(a) 
2
3

1
3
2
4

3
2
0
x3
3x3
2x3
= 0
= 0
= 0.
For the nonsingular ones find the inverse.



3
1 2 3
2
(d)  1 1 2  .
2
0 1 1

3
3 .
4
is nonsingular and write it as a product of elementary matrices (See
(if possible) of the following matrices:


2 −3
1
3 1
3 −3 −2 
(b)  2 1

0
1
5
1 2
1 −2
5
1 2
(c)  1 1
1 1
−
−
+
3
.
4
1 2
Exercise 1.6.12. Find the inverse of A =  0 2
1 2
Exercise 1.6.13. Show that A =
x2
2x2
x2
0
0
0.
Exercise 1.6.10. Find out which of the following matrices are singular.

1 2
1 3
1 3
(a)
(b)
(c)  1 1
2 6
−2 6
0 1
+
−
−

2
(d)  0
1

2
2
2

1 3
1 2.
0 3

1
1
 0 −2
Exercise 1.6.15. Find the inverse of A = 
1
2
0
3

2
1
0
0

1 −2
2
1
Exercise 1.6.16. If A is a nonsingular matrix whose inverse is
4
1
2
, find A.
1
Exercise 1.6.17. Prove that two m× n matrices A and B are row equivalent if and only if there exists a nonsingular
matrix P such that B = P A (Hint: Use Theorem 1.6.2 and Theorem 1.6.4.)
Exercise 1.6.18. Let A and B be row equivalent n × n matrices. Prove that A is nonsingular if and only if B is
nonsingular.
Exercise 1.6.19. Let A and B be n × n matrices. Show that if AB is nonsingular, then A and B must be
nonsingular. (Hint: Use Theorem 1.6.5.)
Exercise 1.6.20. Let A be an m × n matrix. Show that A is row equivalent to Om×n if and only if A = Om×n .
Exercise 1.6.21. Let A and B be m × n matrices. Show that A is row equivalent to B if and only if AT is column
equivalent to B T .
Exercise 1.6.22. Show that a matrix which has a row or a column consisting entirely of zeros must be singular.
Exercise 1.6.23. Find value(s) of λ for which the inverse of

1 1
A = 1 0
1 2
exists. What is A−1 ?

0
0
λ
Exercise 1.6.24.
(a) Is (A + B)−1 = A−1 + B −1 ?
(b)
Is (γ A)−1 =
1
γ
A−1 ?
Exercise 1.6.25. For what value(s) of λ does the homogeneous system
(λ − 1)x1
2x1
have a nontrivial solution?
+
x2
+ (λ − 1)x2
= 0
= 0
1.7. EQUIVALENT MATRICES
– WE WILL NOT DISCUSS IT IN CLASS –
Chapter 2.
REAL VECTOR SPACES
VECTORS IN THE PLANE—A REVIEW
In many applications we deal with measurable quantities, such as pressure, mass, and speed, which can be completely
described by giving their magnitude. They are called scalars and will be denoted by lowercase Latin letters. There
are many other measurable quantities, such as velocity, force, and acceleration, which require for their description
not only magnitude, but also a sense of direction. These are called vectors and their study comprises this chapter.
Vectors will be denoted by bold lowercase letters. You perhaps have already encountered vectors in elementary
physics and in the calculus.
We quickly review the definition of a vector in the plane.
We draw a pair of perpendicular lines intersecting at a point O, called the origin. One of the lines, the x-axis
is usually taken in a horizontal position. The other line, the y-axis, is then taken in a vertical position. We now
choose a point on the x-axis to the right of O and a point on the y-axis above O to fix the units of length and
positive directions on the x- and y-axes. Frequently, but not always, these points are chosen so that they are both
equidistant from O, that is, so that the same unit of length is used for both axes. The x and y axes together are
called coordinate axes and they form a rectangular coordinate system or a Cartesian coordinate system.
x2
x1
each point P in the plane we associate an ordered pair (x, y) of real numbers, its coordinates. Conversely, we can
associate a point in the plane with each ordered pair of real numbers. The point P with coordinates (x, y) is denoted
P (x, y), or simply as (x, y). The set of all points in the plane is denoted by R2 , it is called 2-space.
Consider the 2 × 1 matrix
α =
x
,
y
where x and y are real numbers (note the use of square brackets instead of round ones used in the previous chapter).
With α we associate the directed line segment with the initial point at the origin and the terminal point at P (x, y).
−→
The directed line segment from O to P is denoted by OP ; O called its tail and P its head. We distinguish the tail
and the head by placing an arrow at the head. A directed line segment has a direction, indicated by the arrow at its
head. The magnitude of a directed line segment is its length. Thus a directed line segment can be used to describe
−→
force, velocity, or acceleration. Conversely, with a directed line segment OP with tail O(0, 0) and head P (x, y) we
can associate the 2 × 1 matrix
x
.
y
x
A vector in the plane is a 2×1 matrix α =
, where x and y are real numbers, which are called the components
y
of α. We refer to a vector in the plane merely as a vector.
Thus with every vector we can associated a directed line segment and conversely, with every directed line segment
we can associate a vector. Frequently, the notions of directed line segment and vector are used interchangeably, and
a directed line segment is called a vector. Since a vector is a matrix, the vectors
α1 =
x1
y1
and α2 =
x2
y2
are said to be equal if x1 = x2and y1 = y2 . That is, two vectors are equal if their respective components are
x
equal. With each vector α =
we can associate the unique point P (x, y); conversely, with each point P (x, y) we
y
x
associate the unique vector
. Thus we also write the vector α as (x, y). Of course, this association is carried out
y
−→
by means of the directed line segment OP , where O is the origin and P is the point with coordinates (x, y). Thus
the plane may be viewed both as the set of all points or as the set of all vectors. For this reason, and depending
upon the context,
we sometimes take R2 as the set of all ordered pairs (x, y) and sometimes as the set of all 2 × 1
x
matrices
(or directed line segments).
y
Let
α1 =
x1
y1
and β =
x2
y2
be two vectors in the plane. The sum of the vectors α and β is the vector
α+β =
x1 + y1
.
x2 + y2
We can interpret vector addition geometrically, as follows. In the figure below, the directed line segment γ is parallel
to β, it has the same length as β, and its tail is the head (x1 , y1 ) of α so its head is (x1 + y1 , y1 + y2 ). Thus the
vector with tail at 0 and head at (x1 + y1 , x2 + y2 ) is α + β. We can also describe α + β as the diagonal of the
parallelogram defined by α and β, as shown in the diagram.
–Linear Algebra To be Continued—
INTENTIONALLY LEFT BLANK
Chapter 3
INTRODUCTON TO INTEGRATION IN VECTOR FIELDS
CURVES, SURFACES, LINE INTEGRALS, VECTOR FIELDS, GREEN’S THEOREM,
SURFACE INTEGRALS AND STOKE’S THEOREM AND THE DIVERGENCE THEOREM
CURVES IN THE PLANE AND IN SPACE
In this section we discuss two mathematical formulations of the intuitive notion of a curve. The precise relation
between them turns out to be quite subtle, so we shall begin by giving some examples of curves of each type and
practical ways of passing between them.
3.1 WHAT IS A CURVE?
If asked to give an example of a curve, you might give a straight line, say 𝑦𝑦 − 2𝑥𝑥 = 1 (even though this is not
'curved'!), or a circle, say, 𝑥𝑥 2 + 𝑦𝑦 2 = 1, or perhaps a parabola, say 𝑦𝑦 − 𝑥𝑥 2 = 0.
𝑦𝑦 − 2𝑥𝑥 = 1
𝑦𝑦 − 𝑥𝑥 2 = 0
All of these curves are described by means of their cartesian equation
𝑥𝑥 2 + 𝑦𝑦 2 = 1
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑐𝑐
where, 𝑓𝑓 is a function of two variable 𝑥𝑥 and 𝑦𝑦 and 𝑐𝑐 is a constant. From this point of view, a curve is a set of
points, namely
∁ = {(𝑥𝑥, 𝑦𝑦) ∈ ℝ2 : 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑐𝑐}.
(1)
These examples are all curves in the plane ℝ2 , but we can also consider curves in ℝ3 (3-D space)--- for
example, the x-axis in ℝ3 is the straight line given by
{(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) ∈ ℝ3 : 𝑦𝑦 = 0 & 𝑧𝑧 = 0}
and more generally a curve in ℝ3 might be defined by a pair of equations
𝑓𝑓1 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑐𝑐1 ,
𝑓𝑓2 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑐𝑐2
Curves of this kind are called level curves, the idea being that the curve in Eq (1), for example, is the set of
points (𝑥𝑥, 𝑦𝑦) in the plane at which the quantity 𝑓𝑓(𝑥𝑥, 𝑦𝑦) reaches the 'level' 𝑐𝑐.
However, there is another way to think about curves which turns out to be more useful in many situations. For
this, a curve is viewed as the path traced out by a moving point. Thus, if 𝛄𝛄(t) denotes the position vector
of the point at time 𝑡𝑡, the curve is described by a function 𝛄𝛄 of a scalar parameter 𝑡𝑡 with vector values (in ℝ2
for a plane curve, in ℝ3 for a space curve i.e., for a curve in 3-D space). We use this idea to give our first
formal definition of a curve in ℝ𝑛𝑛 (we shall be interested only in the cases when 𝑛𝑛 = 2 𝑎𝑎𝑎𝑎𝑎𝑎 𝑛𝑛 = 3, but it is
convenient to treat both cases simultaneously):
Definition 3.1.1 A parametrized curve in ℝ𝑛𝑛 is a continuous function or map given by
𝛄𝛄 ∶ 𝑰𝑰 → ℝn , 𝜸𝜸 = 𝛄𝛄(t) = 〈γ1 (t), γ2 (t), … , γn (t)〉 for 𝑡𝑡 in some open interval 𝑰𝑰 ⊆ ℝ
where the continuous functions (usually called maps) 𝛾𝛾1 (𝑡𝑡), 𝛾𝛾2 (𝑡𝑡), … , 𝛾𝛾𝑛𝑛 (𝑡𝑡) are called the component
functions of 𝜸𝜸 and 𝑡𝑡 is called the parameter.
The symbol 𝑰𝑰 denotes the parameter interval for 𝜸𝜸 and it is usually (but not always) an open interval of the
form 𝑰𝑰 = {𝑡𝑡 ∈ ℝ ∶ 𝛼𝛼 < 𝑡𝑡 < 𝛽𝛽 𝑓𝑓𝑓𝑓𝑓𝑓 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝛼𝛼, 𝛽𝛽 ∈ ℝ}. A parametrized curve whose image, 𝜸𝜸(𝑰𝑰), is contained
in a level curve ∁ is called a parametrization of (part of) ∁.
The following examples illustrate how to pass from level curves to parametrized curves and back again in
practice.
Example 3.1.1 Let us find parametrization 𝛄𝛄(t) of the plane curve given as ∁= {(𝑥𝑥, 𝑦𝑦) ∈ ℝ2 : 𝑦𝑦 − 𝑥𝑥 2 = 0} (called
a parabola). You are perhaps more familiar with it written in functional form as: 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2 , 𝑥𝑥 ∊ ℝ
Suppose 𝛄𝛄(𝑡𝑡) = 〈𝛾𝛾1 (𝑡𝑡), 𝛾𝛾2 (𝑡𝑡)〉, then the component functions 𝛾𝛾1 and 𝛾𝛾2 of 𝛄𝛄 must satisfy the relation
𝛾𝛾2 (𝑡𝑡) = 𝛾𝛾2 (𝑡𝑡)2
(2)
for all values of 𝑡𝑡 ∈ 𝑰𝑰 where 𝛄𝛄 is defined (yet to be decided), and ideally every point (𝑥𝑥, 𝑦𝑦) on the parabola ∁ should be
equal to 〈𝛾𝛾1 (𝑡𝑡), 𝛾𝛾2 (𝑡𝑡)〉 for some value of 𝑡𝑡 ∈ 𝑰𝑰. Of course, there is an obvious solution to Eq (2): Take 𝛾𝛾1 (𝑡𝑡) =
𝑡𝑡, 𝛾𝛾2 (𝑡𝑡) = 𝑡𝑡 2 . To get every point on the parabola we must allow 𝑡𝑡 to take every real number value (since the 𝑥𝑥coordinate of 𝛄𝛄(𝑡𝑡) is just 𝑡𝑡, and the 𝑥𝑥-coordinate of a point on the parabola can be any real number), so we must take 𝐼𝐼
to be ℝ. Thus, the parametrization we seek is:
𝛄𝛄 ∶ ℝ ⟶ ℝ2 , 𝛄𝛄(𝑡𝑡) = 〈𝑡𝑡, 𝑡𝑡 2 〉.
You might ask: Is the parametrization given above for the parabola the only one ? The answer is no! Another
choice for parametrization of the parabola is 𝛄𝛄 ∶ ℝ ⟶ ℝ2 , 𝛄𝛄(𝑡𝑡) = 〈𝑡𝑡 3 , 𝑡𝑡 6 〉. Yet another is 𝛄𝛄 ∶ ℝ ⟶ ℝ2 , 𝛄𝛄(𝑡𝑡) =
〈2𝑡𝑡, 4𝑡𝑡 2 〉, and of course there are (infinitely many) others (see if you can think of some others!). So the parametrization
of a given level curve 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑐𝑐 is not unique.
Example 3.1.2 Now we try to find a parametrization of the plane curve as ∁= {(𝑥𝑥, 𝑦𝑦) ∈ ℝ2 : 𝑥𝑥 2 + 𝑦𝑦 2 = 1} (called
the unit circle with center at the origin (0,0)). It is tempting to take 𝛾𝛾1 (𝑡𝑡) = 𝑡𝑡 , 𝛾𝛾2 (𝑡𝑡) = √1 − 𝑡𝑡 2 as a choice for
parametrization (perhaps motivated by the previous example!). But this parametrization is only a parametrizes to upper
half of the circle, because √1 − 𝑡𝑡 2 is always nonnegative. Similarly, if we had taken 𝛾𝛾1 (𝑡𝑡) = 𝑡𝑡, 𝛾𝛾2 (𝑡𝑡) = −√1 − 𝑡𝑡 2 , we
would only have a parametrization of the lower half of the circle.
If we want to a parametrization of the whole circle, we must (try again!). We need functions 𝛾𝛾1 (𝑡𝑡) and 𝛾𝛾2 (𝑡𝑡) such that
𝛾𝛾1 (𝑡𝑡)2 + 𝛾𝛾2 (𝑡𝑡)2 = 1
(3)
for all 𝑡𝑡 ∈ 𝐼𝐼, such that every point (𝑥𝑥, 𝑦𝑦) on the circle is equal to 〈𝛾𝛾1 (𝑡𝑡), 𝛾𝛾2 (𝑡𝑡)〉 for some 𝑡𝑡 ∈ 𝐼𝐼. There is an
obvious solution to Eq (3): Take 𝛾𝛾1 (𝑡𝑡) = cos(𝑡𝑡) 𝑎𝑎𝑎𝑎𝑎𝑎 𝛾𝛾2 (𝑡𝑡) = 𝑠𝑠𝑠𝑠𝑠𝑠(𝑡𝑡) (since cos2 (𝑡𝑡) + sin2 (𝑡𝑡) = 1 for all
real values of 𝑡𝑡). We can take 𝐼𝐼 = ℝ, although this is overkill: Any open interval 𝐼𝐼 whose length greater than
2𝜋𝜋 will suffice. Thus a desired parametrization of the unit circle is
𝛄𝛄 ∶ [0, 2𝜋𝜋] ⟶ ℝ2 , 𝛄𝛄(𝑡𝑡) = 〈cos 𝑡𝑡 , sin 𝑡𝑡〉
Note: In our parametrization of the unit circle chose an interval 𝐼𝐼 which is not an open interval (rather it is
closed! contrary to Definition 2.1). The reason for this choice is purely convenient and geometrical is see
should be clear in section 4.
TRY THIS EXERCISE NOW! Find a parametrization 𝛄𝛄 of the circle with radius 𝑟𝑟 > 0 centered at a point (ℎ, 𝑘𝑘) ∈
ℝ2 (Do not choose all of ℝ as your parameter interval). Choose your parameter interval so that distinct values of your
parameter variable corresponds to distinct points on the circle.
The next example shows how to pass from parametrized curves back to level curves.
Example 3.1.3 Consider the parametrized curve (called an astroid) given by
𝛄𝛄(𝑡𝑡) = 〈cos3 𝑡𝑡, sin3 𝑡𝑡〉.
Since cos2 𝑡𝑡 + sin2 𝑡𝑡 = 1 for all 𝑡𝑡, the coordinates 𝑥𝑥 = cos 3 𝑡𝑡, 𝑦𝑦 = sin3 𝑡𝑡 of the point 𝛄𝛄(𝑡𝑡) satisfy
2
2
𝑥𝑥 3 + 𝑦𝑦 3 = 1.
This level curve coincides with the image of the map 𝛄𝛄.
In this section, we shall be studying curves (and later, surfaces) using methods of the calculus.
To
differentiate a vector-valued function such as 𝛄𝛄(𝑡𝑡) (as in Definition 3.1.1), we differentiate component-wise:
If
𝜸𝜸(𝑡𝑡) = 〈𝛾𝛾1 (𝑡𝑡), 𝛾𝛾2 (𝑡𝑡), … , 𝛾𝛾𝑛𝑛 (𝑡𝑡)〉,
then
𝑑𝑑𝛄𝛄
𝑑𝑑𝛾𝛾1 𝑑𝑑𝛾𝛾2
𝑑𝑑𝛾𝛾𝑛𝑛
〉,
=〈
,
,… ,
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑 2 𝛄𝛄
𝑑𝑑 𝑡𝑡 2
=〈
𝑑𝑑 2 𝛾𝛾1 𝑑𝑑 2 𝛾𝛾2
𝑑𝑑𝑡𝑡 2
,
𝑑𝑑𝑡𝑡 2
,… ,
𝑑𝑑 2 𝛾𝛾𝑛𝑛
𝑑𝑑 𝑡𝑡 2
〉, etc.
We say that a parametrized curve 𝛄𝛄 is smooth if each of its component functions 𝛾𝛾1 , 𝛾𝛾2 , … , 𝛾𝛾𝑛𝑛 is smooth i.e., if
all derivative
𝑑𝑑𝜸𝜸𝑖𝑖 𝑑𝑑 2 𝜸𝜸𝑖𝑖 𝑑𝑑 3 𝜸𝜸𝑖𝑖
𝑑𝑑𝑑𝑑
,
𝑑𝑑 𝑡𝑡 2
,
𝑑𝑑𝑡𝑡 3
, … exists, for 𝑖𝑖 = 1,2, … , 𝑛𝑛. In most cases, all parametrized curves studied here
in this section will be assumed to be smooth.
Definition 3.1.2
If 𝛄𝛄 is a parametrized curve, then at some 𝑡𝑡 ∈ 𝑰𝑰 its first derivative 𝑑𝑑𝛄𝛄/𝑑𝑑𝑑𝑑 is called the
tangent vector of 𝛄𝛄 at the point 𝛄𝛄(𝑡𝑡).
To see the reason for this terminology, note that the vector
1
[𝛄𝛄(𝑡𝑡 + Δ𝑡𝑡) − 𝛄𝛄(𝑡𝑡)]
Δ𝑡𝑡
is parallel to the chord joining the two points 𝛄𝛄(𝑡𝑡) 𝑎𝑎𝑎𝑎𝑎𝑎 𝛄𝛄(𝑡𝑡 + Δ𝑡𝑡) of the image ∁ of 𝛄𝛄.
𝛄𝛄(𝑡𝑡 + Δ𝑡𝑡)
𝛄𝛄(𝑡𝑡)
We expect, as Δ𝑡𝑡 ⟶ 0 (tends to 0), the chord parallel to the tangent to ∁ at 𝛄𝛄(t). Hence, the tangent should be
parallel to
lim
Δ𝑡𝑡→0
1
𝑑𝑑𝛄𝛄
[𝛄𝛄(𝑡𝑡 + Δ𝑡𝑡) − 𝛄𝛄(𝑡𝑡)] =
= 𝛄𝛄′ (𝑡𝑡).
Δ𝑡𝑡
𝑑𝑑𝑑𝑑
The following result is intuitively clear:
Theorem 3.1.1 If the tangent vector of a parametrized curve is constant, then the image of the curve is (part of) a
straight line.
proof If
𝑑𝑑𝛄𝛄
𝑑𝑑𝑑𝑑
= a for all 𝑡𝑡 ∈ 𝐼𝐼, where 𝐚𝐚 is a constant vector, we have, integrating component-wise,
𝑡𝑡
𝑡𝑡
𝑑𝑑𝛄𝛄
𝑑𝑑𝑑𝑑 = � 𝐚𝐚 𝑑𝑑𝑑𝑑 = 𝑡𝑡𝐚𝐚 + 𝐛𝐛,
𝑡𝑡 0 𝑑𝑑𝑑𝑑
𝑡𝑡 0
𝛾𝛾(𝑡𝑡) = �
where b is another constant vector. If 𝐚𝐚 ≠ 𝟎𝟎, this is the parametric equation of the straight line parallel to 𝐚𝐚 and passing
through the point with position vector 𝒃𝒃 = 𝜸𝜸(𝒕𝒕𝟎𝟎 ):
𝑡𝑡𝒂𝒂
𝜸𝜸(𝒕𝒕)
𝒃𝒃
𝒂𝒂
If 𝐚𝐚 = 𝟎𝟎, the image of 𝛾𝛾 is a single point (namely, the point with position vector 𝐛𝐛 = 𝛄𝛄(𝑡𝑡0 ) ).
Q.E.D
Problems
1. Is the curve 𝛄𝛄(𝑡𝑡) = 〈𝑡𝑡 2 , 𝑡𝑡 4 〉 a parametrization of the parabola = 𝑥𝑥 2 ? (Explain).
2. Find parametrizations of the following level curves
a. 𝑦𝑦 2 − 𝑥𝑥 2 = 1 ;
b.
𝑥𝑥 2
4
+
𝑦𝑦 2
9
= 1.
3. Find the cartesian equations of the following parametrized curves.
a. 𝜸𝜸(𝑡𝑡) = 〈cos2 𝑡𝑡, sin2 𝑡𝑡〉;
b. 𝛄𝛄(𝑡𝑡) = 〈𝑒𝑒 𝑡𝑡 , 𝑡𝑡 2 〉.
4. Calculate the tangent vectors of the curves in Problem 3.
5. Calculate the tangent vector at each point of the astroid sketched in Example 3.1.3. At which points is the
tangent vector zero?
1
6. Show that 𝛄𝛄(𝑡𝑡) = 〈cos2 𝑡𝑡 − 2 , sin(𝑡𝑡) cos(𝑡𝑡) , sin⁡
(𝑡𝑡)〉 is a parametrization of the curve of intersection of
1
1
the circular cylinder of radius 2 and axis the 𝑧𝑧-axis with the sphere of radius 1 and center �− 2 , 0, 0�. (This
is called Viviani's Curve)
3.2 ARC-LENGTH
If 𝒗𝒗 = 〈𝑣𝑣1 , 𝑣𝑣2 , … , 𝑣𝑣𝑛𝑛 〉 is a vector in ℝ𝑛𝑛 , its length or magnitude is
∥ 𝒗𝒗 ∥= �𝑣𝑣12 + 𝑣𝑣22 + ⋯ + 𝑣𝑣𝑛𝑛2 .
If 𝒖𝒖 is another vector in ℝ𝑛𝑛 , ∥ 𝒖𝒖 − 𝒗𝒗 ∥ is the length of the straight line segment joining the points in ℝ𝑛𝑛 with
position vectors 𝒖𝒖 and 𝒗𝒗.
To find a formula for the length of any parametrized curve, say, 𝜸𝜸, note that, if Δ𝑡𝑡 is sufficiently small, the part
of the image ∁ of 𝜸𝜸 between 𝛾𝛾(𝑡𝑡) and 𝛾𝛾(𝑡𝑡 + Δ𝑡𝑡) is nearly a straight line. so its length is approximately
∥ 𝜸𝜸(𝑡𝑡 + Δ𝑡𝑡) − 𝜸𝜸(𝑡𝑡) ∥.
Again, since Δ𝑡𝑡 is sufficiently small, the vector
approximately
1
Δ𝑡𝑡
[𝜸𝜸(𝑡𝑡 + Δ𝑡𝑡) − 𝜸𝜸(𝑡𝑡)] ≈ 𝜸𝜸′ (𝑡𝑡), so the length L is
∥ 𝜸𝜸′ (𝑡𝑡) ∥ Δ𝑡𝑡
(4)
If we want to calculate the length of (a not necessarily small) part of ∁, we can divide it up into segments, each
of which corresponds to a small increment Δ𝑡𝑡 in 𝑡𝑡, calculate the length of each segment using Eq (4), and add
up the results.
Letting Δ𝑡𝑡 → 0 should then give the exact length.
This motivates the following definition:
Definition 3.2.1
by
The arc-length of a parametrized curve 𝜸𝜸 starting at the point 𝜸𝜸(𝑡𝑡0 ) is the function 𝑠𝑠(𝑡𝑡) given
𝑡𝑡
𝑡𝑡
𝑑𝑑𝜸𝜸
∥ 𝑑𝑑𝑑𝑑 = � ∥ 𝜸𝜸′ (𝜏𝜏) ∥ 𝑑𝑑𝑑𝑑
𝑠𝑠(𝑡𝑡) = � ∥
𝑑𝑑𝑑𝑑
𝑡𝑡 0
𝑡𝑡 0
Thus, 𝑠𝑠(𝑡𝑡0 ) = 0 and 𝑠𝑠(𝑡𝑡) is positive or negative according to whether 𝑡𝑡 is larger or smaller than 𝑡𝑡0 . If we choose a
𝜆𝜆
different starting point, say, 𝜸𝜸(𝜆𝜆), the resulting arc-length 𝑠𝑠̃ differs from 𝑠𝑠 by the constant ∫𝑡𝑡 ∥ 𝜸𝜸′ (𝑢𝑢) ∥ 𝑑𝑑𝑑𝑑 .
0
Example 3.2.1 For a logarithmic spiral
𝜸𝜸(𝑡𝑡) = 〈𝑒𝑒 0.2𝑡𝑡 cos(𝑡𝑡) , 𝑒𝑒 0.2𝑡𝑡 sin(𝑡𝑡)〉,
80
60
40
20
0
-20
-40
-60
-80
-100
-120
-100
0
-50
150
100
50
we have
𝛄𝛄′ = 〈𝑒𝑒 0.2𝑡𝑡 (0.2cos(𝑡𝑡) − sin(𝑡𝑡)), 𝑒𝑒 0.2𝑡𝑡 (cos(𝑡𝑡) + 0.2 sin(𝑡𝑡))〉
∴ ∥ 𝜸𝜸′ ∥2 = 𝑒𝑒 0.4𝑡𝑡 (0.2 cos(𝑡𝑡) − sin(𝑡𝑡))2 + 𝑒𝑒 0.4𝑡𝑡 (0.2 sin(𝑡𝑡) + cos(𝑡𝑡))2 = 1.04𝑒𝑒 0.4𝑡𝑡 .
Hence, the arc-length of gamma starting at 𝜸𝜸(0) = 〈1,0〉 (which corresponds to the point (1,0) ) is
𝑡𝑡
𝑠𝑠 = 𝑠𝑠(𝑡𝑡) = ∫0 √1.04𝑒𝑒 0.4𝜏𝜏 𝑑𝑑𝑑𝑑 =
√1.04
0.2
(𝑒𝑒 0.2𝑡𝑡 − 1) .
If s is the arc-length of a curve 𝜸𝜸 starting at the point 𝛄𝛄(𝑡𝑡0 ), we have
𝑠𝑠 ′ (𝑡𝑡) =
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
=
𝑑𝑑
𝑡𝑡
∫ ∥ 𝜸𝜸′ (𝜏𝜏) ∥ 𝑑𝑑𝑑𝑑 =∥ 𝜸𝜸′ (𝑡𝑡) ∥.
𝑑𝑑𝑑𝑑 𝑡𝑡
0
Thinking of 𝛄𝛄(𝑡𝑡) as the position of a moving point at time 𝑡𝑡,
change of distance along the curve).
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
(5)
is the speed of the point (the rate of
For this reason, we make the following definition
Definition 3.2.2 If 𝛄𝛄: 𝑰𝑰 ⟶ ℝ𝑛𝑛 is a parametrized curve, its speed at the point 𝛄𝛄(𝑡𝑡) is ∥ 𝛄𝛄′ (𝑡𝑡) ∥, and the curve 𝛄𝛄 is
called a unit-speed curve if the tangent vector, 𝛄𝛄′ (𝑡𝑡) , is a unit vector (i.e., ∥ 𝛄𝛄′ (𝑡𝑡) ∥ = 1) for all 𝑡𝑡 ∈ 𝑰𝑰.
We shall see many examples of formulas and results relating to curves that take on a much simpler form when the curve
is unit speed. The reason for this simplification is given in the next theorem. Although this admittedly looks
uninteresting at first sight, it will be extremely useful for what follows.
Theorem 3.2.1 Let 𝐧𝐧(𝑡𝑡) be a unit vector that is a smooth function of a parameter 𝑡𝑡. Then, the dot product
𝐧𝐧′ (𝑡𝑡) ∙ 𝐧𝐧(𝑡𝑡) = 0
for all 𝑡𝑡, i.e., 𝐧𝐧′ (𝑡𝑡) = 0 or else 𝐧𝐧′ (t) is perpendicular to 𝐧𝐧(𝑡𝑡) for all 𝑡𝑡. In particular, If 𝜸𝜸 is a unit-speed curve, then
is either 0 or perpendicular to
proof
𝑑𝑑𝜸𝜸
𝑑𝑑𝑑𝑑
.
𝑑𝑑 2 𝜸𝜸
𝑑𝑑 𝑡𝑡 2
We use the 'dot product' formula' for differentiating dot products of vector-valued functions 𝐚𝐚(𝑡𝑡) and 𝐛𝐛(𝑡𝑡):
𝑑𝑑
𝑑𝑑𝑑𝑑
(𝐚𝐚 ∙ 𝐛𝐛) =
𝑑𝑑𝐚𝐚
𝑑𝑑𝑑𝑑
∙ 𝐛𝐛 + 𝐚𝐚 ∙
𝑑𝑑𝐛𝐛
𝑑𝑑𝑑𝑑
.
Using this to differentiate both sides of the equation 𝐧𝐧 ∙ 𝐧𝐧 = 1 with respect to 𝑡𝑡 gives
𝐧𝐧′ ∙ 𝐧𝐧 + 𝐧𝐧 ∙ 𝐧𝐧′ = 0,
so 2𝐧𝐧′ ∙ 𝐧𝐧 = 0. The last part follows by taking 𝐧𝐧 = 𝛄𝛄′
Q.E.D
Problems
1. Calculate the arc-length of the catenary 𝛄𝛄(𝑡𝑡) = 〈𝑡𝑡, cosh 𝑡𝑡〉 starting at the point (0,1).
2. Show that the following curves a unit-speed curves
(i)
(ii)
1
3
1
3
𝛄𝛄(𝑡𝑡) = 〈 3 (1 + 𝑡𝑡)2 , 3 (1 − 𝑡𝑡)2 ,
4
3
𝑡𝑡
√2
〉;
𝛄𝛄(𝑡𝑡) = 〈5 cos 𝑡𝑡 , 1 − sin 𝑡𝑡 , − 5 cos 𝑡𝑡〉.
3. A cycloid is the plane curve traced out by a point on the circumference of a circle as it rolls without
slipping along a straight line. It can be shown that if the straight line is the x-axis and the circle has radius
𝛼𝛼 > 0, the cycloid can be parametrized as
𝜸𝜸: ℝ ⟶ ℝ2 ,
𝜸𝜸(𝑡𝑡) = 𝛼𝛼〈𝑡𝑡 − sin 𝑡𝑡 , 1 − cos 𝑡𝑡〉
Calculate the arc-length along the cycloid corresponding to one complete revolution of the circle.
4. The circular helix is the space curve given by
𝜸𝜸: 𝑰𝑰 ⟶ ℝ3 , 𝜸𝜸 = 𝜸𝜸(𝑡𝑡) = 〈cos αt , sin αt, βt〉,
where 𝛼𝛼 and 𝛽𝛽 are constants. Calculate the arc-length function 𝑠𝑠 along the circular helix starting at the point 𝜸𝜸(0)
3.3 REPARAMETRIZATION
We saw in Examples 3.1.1 and 3.1.2 of section 3 that a given level curve can have many parametrizations, and
it is important to understand the relationship between them.
Definition 3.3.1 A parametrized curve 𝝋𝝋: 𝑱𝑱 ⟶ ℝ𝑛𝑛 is called a reparametrization of 𝜸𝜸: 𝑰𝑰 ⟶ ℝ𝑛𝑛 if there is a
smooth bijective map (i.e. a one-to-one and onto map) 𝜏𝜏: 𝑱𝑱 ⟶ 𝑰𝑰, 𝑡𝑡 = 𝜏𝜏(𝑢𝑢) such that the inverse map 𝜏𝜏 −1 : 𝑰𝑰 ⟶ 𝑱𝑱, 𝑢𝑢 =
𝜏𝜏 −1 (𝑡𝑡) is also smooth and
𝝋𝝋: 𝑱𝑱 ⟶ ℝ𝑛𝑛 , 𝝋𝝋(𝑢𝑢) = (𝜸𝜸 ∘ 𝜏𝜏)(𝑢𝑢) = 𝜸𝜸(𝜏𝜏(𝑢𝑢))
Note that, since 𝜏𝜏has a smooth inverse, 𝜸𝜸 is a reparametrization of 𝝋𝝋:
𝝋𝝋�𝜏𝜏 −1 (𝑡𝑡)� = 𝜸𝜸 �𝜏𝜏�𝜏𝜏 −1 (𝑡𝑡)�� = 𝜸𝜸(𝑡𝑡) for all 𝑡𝑡 ∈ 𝐼𝐼.
Two curves that are reparametrizations of each other have the same image, so they should have the same geometric
properties.
Example 3.3.1 In Example 3.1.2, we gave the parametrization 𝜸𝜸(𝑡𝑡) = 〈cos 𝑡𝑡, sin 𝑡𝑡〉 for the circle 𝑥𝑥 2 + 𝑦𝑦 2 = 1.
Another parametrization is
𝝋𝝋(𝑢𝑢) = 〈sin 𝑢𝑢, cos 𝑢𝑢〉
(since sin2 𝑠𝑠 + cos2 𝑠𝑠 = 1 for all s). To see that 𝝋𝝋 is a reparametrization of 𝜸𝜸, we have to find a reparametrization map
𝜏𝜏 such that
𝜋𝜋
2
〈cos⁡
(𝜏𝜏(𝑢𝑢)), sin⁡
(𝜏𝜏(𝑢𝑢))〉 = 〈sin 𝑢𝑢, cos 𝑢𝑢〉
𝜋𝜋
2
One solution is 𝑡𝑡 = 𝜏𝜏(𝑢𝑢) = − 𝑢𝑢 (observe that this is a smooth bijective map! whose inverse is 𝑢𝑢 = 𝜏𝜏 −1 (𝑡𝑡) = − 𝑡𝑡
which is also a smooth bijective map. Thus 𝜸𝜸 is also a reparametrization of 𝝋𝝋 too!)
As we remarked in the previous section, the analysis of a curve is simplified when it is known to be unit-speed. It is
therefore important to know exactly which curve have unit-speed reparametrizations.
Definition 3.3.2 A point 𝜸𝜸(𝑡𝑡) of a parametrized curve 𝜸𝜸 is called a regular point if the tangent vector at 𝜸𝜸(𝑡𝑡)
does not vanish, i.e., if 𝜸𝜸′ (𝑡𝑡) ≠ 𝟎𝟎 (please keep in mind that 𝟎𝟎 = 〈0,0, … ,0〉 in ℝ𝑛𝑛 and 0 is a real number); otherwise,
𝜸𝜸(𝑡𝑡) is a singular point. A curve 𝜸𝜸 is regular if all its points are regular points (equivalently, A curve is regular if it
has no singular points).
Before we show the relationship between regularity and unit-speed reparametrization, we note two simple properties of
regular curves. Although these results are not particularly appealing, they will be very important for what is to follow.
Theorem 3.3.1 If 𝛾𝛾 is a regular curve, then any reparametrization 𝜑𝜑 of 𝛾𝛾 is again regular.
proof
We aim to show that the curve 𝜑𝜑 is a regular curve (i.e.,
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
is never zero) Since 𝜑𝜑 is a reparametrization of 𝛾𝛾, we
have smooth bijective reparametrization maps 𝑡𝑡 = 𝜏𝜏(𝑢𝑢) and 𝑢𝑢 = 𝜏𝜏 −1 (𝑡𝑡) having the property 𝜏𝜏�𝑡𝑡 −1 (𝑡𝑡)� = 𝑡𝑡. Now
differentiating on both sides with respect to 𝑡𝑡 and using the chain rule yields
𝑑𝑑𝑑𝑑 𝑑𝑑𝜏𝜏 −1
=1
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
This shows that 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 is never zero. Since 𝝋𝝋(𝑢𝑢) = 𝜸𝜸�𝜏𝜏(𝑢𝑢)�, and 𝜸𝜸 is a regular curve, another application of the chain
rule gives
𝑑𝑑𝝋𝝋
=
𝑑𝑑𝑑𝑑
𝑑𝑑𝜸𝜸 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
≠ 𝟎𝟎 (do you see why?)
Q.E.D
Theorem 3.3.2 If 𝜸𝜸 is a regular curve, its arc-length, 𝑠𝑠 (see Definition 3.1.3), starting at any point 𝜸𝜸(𝑡𝑡) of 𝜸𝜸, is a
smooth function of 𝑡𝑡.
proof We have already seen that (whether or not 𝜸𝜸 is regular) 𝑠𝑠 is a differentiable function of 𝑡𝑡 and that
𝑑𝑑𝑑𝑑
= ∥ 𝜸𝜸′ (𝑡𝑡) ∥.
𝑑𝑑𝑑𝑑
To simplify the notation, assume form now on that 𝛾𝛾 is a plane curve, say
𝜸𝜸(𝑡𝑡) = 〈𝑥𝑥(𝑡𝑡), 𝑦𝑦(𝑡𝑡)〉,
where 𝑥𝑥 and 𝑦𝑦 are smooth functions of 𝑡𝑡. Define 𝑓𝑓: ℝ2 ⟶ ℝ by
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = �𝑥𝑥 2 + 𝑦𝑦 2 ,
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
, �
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
= 𝑓𝑓 �
𝑑𝑑𝑑𝑑 2
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 2
𝑑𝑑𝑑𝑑
= �� � + � �
(6)
The crucial point is that 𝑓𝑓 is smooth on ℝ2 without the origin, which means that all the partial derivatives of 𝑓𝑓 of all
orders exists and are continuous functions except at the origin (0,0). For example,
𝜕𝜕𝜕𝜕
𝑥𝑥
(𝑥𝑥, 𝑦𝑦) =
,
𝜕𝜕𝜕𝜕
�𝑥𝑥 2 + 𝑦𝑦 2
𝜕𝜕𝜕𝜕
𝑦𝑦
(𝑥𝑥, 𝑦𝑦) =
,
𝜕𝜕𝜕𝜕
�𝑥𝑥 2 + 𝑦𝑦 2
are well defined and continuous except where 𝑥𝑥 = 𝑦𝑦 = 0, and similarly for higher derivatives. Since 𝜸𝜸 is regular 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑
and 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 are never both zero, so the chain rule and Eq (6) shows that 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 is smooth. For example,
and similarly for the higher derivatives of 𝑠𝑠.
𝑑𝑑2 𝑠𝑠 𝜕𝜕𝜕𝜕 𝑑𝑑2 𝑥𝑥 𝜕𝜕𝜕𝜕 𝑑𝑑2 𝑦𝑦
=
+
,
𝑑𝑑𝑡𝑡 2 𝜕𝜕𝜕𝜕 𝑑𝑑𝑡𝑡 2 𝜕𝜕𝜕𝜕 𝑑𝑑𝑡𝑡 2
Q.E.D
The main result we want is
Theorem 3.3.4 A parametrized curve 𝜸𝜸: 𝑰𝑰 ⟶ ℝ𝑛𝑛 has a unit-speed reparametrization if and only if it is
regular.
Suppose the 𝜸𝜸: 𝑰𝑰 ⟶ ℝ𝑛𝑛 is a parametrized curve with a unit-speed reparametrization 𝝋𝝋: 𝑱𝑱 ⟶ ℝ𝑛𝑛 . Let
𝜏𝜏: 𝑱𝑱 ⟶ 𝑰𝑰, 𝑡𝑡 = 𝜏𝜏(𝑢𝑢) be the reparametrization map. Then we have
proof
𝝋𝝋(𝑢𝑢) = 𝜸𝜸�𝝉𝝉(𝑢𝑢)� = 𝜸𝜸(𝑡𝑡)
Since 𝝋𝝋 is unit-speed, ∥
𝑑𝑑𝝋𝝋
𝑑𝑑𝑑𝑑
∴∥
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝝋𝝋 𝑑𝑑𝜸𝜸 𝑑𝑑𝑑𝑑
=
,
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑑𝑑𝜸𝜸 𝑑𝑑𝑑𝑑
𝑑𝑑𝝋𝝋
∥=∥
∥ � �.
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
∥ = 1, so clearly 𝑑𝑑𝜸𝜸/𝑑𝑑𝑑𝑑 cannot be zero (i.e. 𝜸𝜸 is a regular curve).
𝑑𝑑𝜸𝜸
Conversely, suppose that the tangent vector
know that
∴
𝑑𝑑𝑑𝑑
≠ 𝟎𝟎 (i.e., 𝛾𝛾 is a regular parametrized curve). By Eq (5), we
> 0 for all 𝑡𝑡, where 𝑠𝑠 is the arc-length of 𝛾𝛾 starting at any point of the curve, and by Theorem
3.3.2 we have that 𝑠𝑠 is a smooth function of 𝑡𝑡. It follows from the inverse function theorem of multivariable
calculus that 𝑠𝑠: 𝐼𝐼 ⟶ ℝ is injective (i.e. one-to-one), that its image is an open interval 𝐽𝐽, and that the inverse
map 𝑠𝑠 −1 : 𝐽𝐽 ⟶ 𝐼𝐼 is also smooth. (If you are not familiar with the inverse function theorem you should accept
these statements has true for now; until you become familiar with it). We take 𝜏𝜏 = 𝑠𝑠 −1 and let 𝜑𝜑 be the
corresponding reparametrization of 𝛾𝛾, so that
𝝋𝝋(𝑠𝑠) = 𝜸𝜸(𝑡𝑡).
Then,
∴ ∥
𝑑𝑑𝝋𝝋 𝑑𝑑𝑑𝑑 𝑑𝑑𝜸𝜸
= ,
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑑𝑑𝝋𝝋
𝑑𝑑𝑑𝑑
∴ ∥
∥
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝝋𝝋
𝑑𝑑𝑑𝑑
=∥
𝑑𝑑𝜸𝜸
∥ = 1.
𝑑𝑑𝑑𝑑
∥=
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
(by Eq (5))
Q.E.D
The proof of Theorem 3.3.2 shows that the arc-length is essentially the only unit-speed parameter on a regular
curve.
Corollary 3.3.1 Let 𝜸𝜸 ∶ 𝑰𝑰 ⟶ ℝ𝑛𝑛 be a regular curve and let 𝝋𝝋 ∶ 𝑱𝑱 ⟶ ℝ𝑛𝑛 be a unit-speed reparametrization of 𝜸𝜸:
𝝋𝝋�𝑢𝑢(𝑡𝑡)� = 𝜸𝜸(𝑡𝑡) for all 𝑡𝑡,
where 𝑢𝑢 is a smooth function of 𝑡𝑡. Then, if 𝑠𝑠 is the arc-length of 𝜸𝜸 (starting at any point), we have
𝑢𝑢 = ±𝑠𝑠 + 𝑐𝑐,
(7)
where 𝑐𝑐 is a constant. Conversely, if 𝑢𝑢 is given by Eq (7) for some value of 𝑐𝑐 and with either sign, then 𝝋𝝋 is a unitspeed reparametrization of 𝜸𝜸
proof The calculation in the first part of the proof of Theorem 3.3.4 shows that 𝑢𝑢 gives a unit-speed reparametrization
of 𝜸𝜸 if and only if
𝑑𝑑𝑑𝑑
Hence, 𝑢𝑢 = ±𝑠𝑠 + 𝑐𝑐 for some constant 𝑐𝑐.
𝑑𝑑𝑑𝑑
= ±∥
𝑑𝑑𝜸𝜸
𝑑𝑑𝑑𝑑
∥= ±
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
,
(by Eq (5))
Q.E.D
Although every regular curve has a unit-speed reparametrization, this may be very complicated, or even impossible to
write down 'explicitly', as the following examples show.
Example 3.3.1 Consider the logarithmic spiral
we found in Example 3.1.4 that
𝜸𝜸(𝑡𝑡) = 〈𝑒𝑒 0.2𝑡𝑡 cos 𝑡𝑡, 𝑒𝑒 0.2𝑡𝑡 sin 𝑡𝑡〉,
∥ 𝜸𝜸′ (𝑡𝑡) ∥2 = 1.04𝑒𝑒 0.04𝑡𝑡 .
This is never zero, so 𝜸𝜸 is regular. The arc-length of 𝛾𝛾 starting at (1,0) was found to be 𝑠𝑠 = √1.04(𝑒𝑒 0.2𝑡𝑡 − 1). Hence,
𝑡𝑡 =
1
0.2
ln �
s
√1.04
+ 1�, so a unit-speed reparametrization of 𝜸𝜸 is given by the rather unwieldly formula
𝝋𝝋(𝑠𝑠) = 〈 �
1
𝑠𝑠
𝑠𝑠
1
𝑠𝑠
+ 1� cos � ln �
+ 1�� , �
+ 1� sin � 𝑙𝑙𝑙𝑙 �
+ 1�� 〉
0.2
0.2
√1.04
√1.04
√1.04
√1.04
𝑠𝑠
Example 3.3.2 The twisted cubic is the space curve given by
𝜸𝜸(𝑡𝑡) = 〈𝑡𝑡, 𝑡𝑡 2 , 𝑡𝑡 3 〉,
−∞ < 𝑡𝑡 < ∞
We have
𝜸𝜸′ (𝑡𝑡) = 〈1, 2𝑡𝑡, 3𝑡𝑡 2 〉,
∴ ∥ 𝜸𝜸′ (𝑡𝑡) ∥ = �1 + 4𝑡𝑡 2 + 9𝑡𝑡 4 .
This is never zero, so 𝛾𝛾 is regular. The arc-length starting at 𝜸𝜸(0) = 𝟎𝟎 = 〈0,0,0〉 (is the position vector associated with
origin (0,0,0) ) is
𝑡𝑡
𝑡𝑡
𝑠𝑠 = 𝑠𝑠(𝑡𝑡) = ∫0 ‖𝜸𝜸′ (𝑢𝑢)‖ 𝑑𝑑𝑑𝑑 = ∫0 √1 + 4𝑢𝑢2 + 9𝑢𝑢4 𝑑𝑑𝑑𝑑.
This integral cannot be evaluated in terms of familiar elementary functions (i.e,, logarithms, exponentials,
trigonometric, polynomials, etc). The integral above is an example of an elliptic integral.
Our final example shows that a given level curve can have both regular and non-regular parametrizations.
=
Example 3.3.3 For the parametrization 𝜸𝜸 = 𝜸𝜸(𝑡𝑡) = 〈𝑡𝑡, 𝑡𝑡 2 〉 of the level curve (or the parabola)
𝑥𝑥 2 − 𝑦𝑦 = 0, we have
𝑑𝑑𝜸𝜸
𝑑𝑑𝑑𝑑
= 𝜸𝜸′ (𝑡𝑡) = 〈1, 2𝑡𝑡〉 ≠ 𝟎𝟎 for all 𝑡𝑡 , so 𝜸𝜸 is a regular curve. But the curve 𝜑𝜑 given by
𝜑𝜑(𝑡𝑡) = 𝛾𝛾(𝑢𝑢(𝑡𝑡)) = 𝛾𝛾(𝑡𝑡 3 ) = 〈𝑡𝑡 3 , 𝑡𝑡 6 〉
is a reparametrization of the parabola 𝜸𝜸 (where 𝑢𝑢 = 𝑢𝑢(𝑡𝑡) = 𝑡𝑡 3 is the reparametrization map) . This time
𝝋𝝋′ (𝑡𝑡) = 𝑢𝑢′ (𝑡𝑡)𝜸𝜸′ �𝑢𝑢(𝑡𝑡)� = 3𝑡𝑡 2 〈1, 2𝑡𝑡 3 〉 = 〈3𝑡𝑡 2 , 6𝑡𝑡 5 〉, and this is zero when 𝑡𝑡 = 0, so the reparametrization 𝝋𝝋 is not
regular. Thus showing that a regular curve may have both regular and non-regular parametrizations.
Problems
1. Which of the following curve are regular?
(i)
𝛾𝛾(𝑡𝑡) = 〈cos 2 𝑡𝑡, sin2 𝑡𝑡〉 for −∞ < 𝑡𝑡 < ∞;
𝜋𝜋
(ii)
the same is curve as in (i), but with 0 < 𝑡𝑡 < 2 ;
(iii)
𝛾𝛾(𝑡𝑡) = 〈𝑡𝑡, cosh 𝑡𝑡〉 for −∞ < 𝑡𝑡 < ∞.
2. The cissoid of Diocles (shown above) is the plane curve whose equation in terms of polar coordinates
(𝑟𝑟, 𝜃𝜃) (you should what polar coordinates are look it up if you don't!) is
𝜋𝜋
𝜋𝜋
𝑟𝑟 = 𝑓𝑓(𝜃𝜃) = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠,
− < 𝜃𝜃 < .
2
2
write down a parametrization of the cissoid using 𝜃𝜃 as your parameter and show that the curve given by
𝑡𝑡 3
2
〈𝑡𝑡
〉,
𝜸𝜸(𝑡𝑡) =
,
−1 < 𝑡𝑡 < 1,
√1 − 𝑡𝑡 2
is a reparametrization of it (of the cissoid that is).
3.4 LEVEL CURVES vs. PARAMETRIZED CURVES
We shall now try to clarify the precise relationship between the two types of curve we have considered in
previous sections.
Level curves in the generality we have defined them are not always the kind of objects we would want to call
curves. For example, the level 'curve' 𝑥𝑥 2 + 𝑦𝑦 2 = 0 is a single point. The correct conditions to impose on a
function 𝑓𝑓(𝑥𝑥, 𝑦𝑦) in order that 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑐𝑐, where 𝑐𝑐 is a constant, will be an acceptable level curve in the plane
are contained in the following theorem, which shows that such level curves can be parametrized. Note that we
might as well assume that 𝑐𝑐 = 0 (since we could replace 𝑓𝑓 by 𝑓𝑓 − 𝑐𝑐).
Theorem 3.4.1 Let 𝑓𝑓(𝑥𝑥, 𝑦𝑦) be a smooth function of two variables x and y. Assume that ∇𝑓𝑓(𝑥𝑥, 𝑦𝑦) ≠ 𝟎𝟎 for all
points (𝑥𝑥, 𝑦𝑦) of the level curve given by
∁= {(𝑥𝑥, 𝑦𝑦) ∈ ℝ2 : 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 0}.
If 𝑃𝑃(𝑥𝑥0 , 𝑦𝑦0 ) is a point of ∁, there is a regular parametrized curved 𝜸𝜸 = 𝜸𝜸(𝑡𝑡), defined on an open interval 𝑰𝑰 containing 0,
such that the curve 𝜸𝜸 passes through 𝑃𝑃(𝑥𝑥0 , 𝑦𝑦0 ) when 𝑡𝑡 = 0 and 𝜸𝜸(𝑡𝑡) is contained in ∁ for all 𝑡𝑡.
I will not supply a proof of the theorem because you may not have the machinery necessary to understand it (knowledge
of the Inverse function Theorem for Multivariable calculus and some point set topology are required!). However, I will
provide enough reasons as to why you should believe it is true, simply because you have studied multivariable calculus
of functions of two and three variables. Before I proceed, a few explanations are in order. To say that a function of
two variables is a 'smooth function' means that all the partial derivatives of all orders exists and all are continuous
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
function on the domain of the function. The assumption that 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔(𝑓𝑓) = ∇ 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 〈
not both
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
(𝑥𝑥, 𝑦𝑦) = 0 and
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
(𝑥𝑥, 𝑦𝑦),
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
(𝑥𝑥, 𝑦𝑦)〉 ≠ 𝟎𝟎 means
(𝑥𝑥, 𝑦𝑦) = 0 holds (usually one says the 'gradient of 𝑓𝑓 never vanishes').
To understand the significance of the conditions placed on f(x, y) in the theorem, we suppose for a moment that
𝑄𝑄(𝑥𝑥0 + Δ𝑥𝑥, 𝑦𝑦0 + Δ𝑦𝑦) is a point of the level curve near the point P(x_0,y_0) which is on the level curve as well. Then
we know that 𝑓𝑓(𝑥𝑥0 + Δ 𝑥𝑥, 𝑦𝑦0 + Δ 𝑦𝑦) = 0. We al so know that
0 = 𝑓𝑓(𝑥𝑥0 + Δ𝑥𝑥, 𝑦𝑦0 + Δ𝑦𝑦) − 𝑓𝑓(𝑥𝑥0 , 𝑦𝑦0 ) ≈
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
(𝑥𝑥0 , 𝑦𝑦0 )Δ𝑥𝑥 +
If follows that for sufficiently small Δ𝑥𝑥 and Δ𝑦𝑦 we get that
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
(𝑥𝑥0 , 𝑦𝑦0 )Δ𝑥𝑥 +
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
(𝑥𝑥0 , 𝑦𝑦0 )Δ𝑦𝑦 (this you should know!).
(𝑥𝑥0 , 𝑦𝑦0 )Δ𝑦𝑦 ≈ 0,
(8)
which says that the vector 〈Δ𝑥𝑥, Δ𝑦𝑦〉 is nearly (and equal if we let both Δ𝑥𝑥 and Δ𝑦𝑦 approach 0) tangent to the level curve
at 𝑃𝑃 and that the vector ∇𝑓𝑓 is perpendicular to the level curve at 𝑃𝑃 (this statement should not surprise you because you
have seen it many times in Calculus III ).
𝑦𝑦
∁
∇𝑓𝑓
〈Δ𝑥𝑥, Δ𝑦𝑦〉
P
𝑥𝑥
The hypothesis in the theorem tells us that ∇𝑓𝑓 never vanishes at every point P of the level curve. Suppose
𝜕𝜕𝜕𝜕
now for example, that 𝜕𝜕𝜕𝜕 ≠ 0 at 𝑃𝑃. Then,∇𝑓𝑓 is not parallel to the 𝑥𝑥-axis at 𝑃𝑃, so the tangent vector to the
tangent line to the level curve at P is not parallel to the 𝑦𝑦-axis.
∁
Tangent line at 𝑃𝑃
𝑃𝑃(𝑥𝑥0 , 𝑦𝑦0 )
Rectangle on which functon exists
𝑥𝑥 = 𝑎𝑎2
𝑥𝑥 = 𝑎𝑎_1 𝑥𝑥 = 𝑥𝑥0
This implies that vertical lines of the form 𝑥𝑥 = 𝑎𝑎 near the line 𝑥𝑥 = 𝑥𝑥0 all intersect the level curve in a unique point
(𝑥𝑥, 𝑦𝑦) near 𝑃𝑃. In other words, the equation
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 0
(9)
has a unique solution 𝑦𝑦 near 𝑦𝑦0 for every 𝑥𝑥 near 𝑥𝑥0 . Note that this may fail to be the case if the tangent to the level
curve at 𝑃𝑃 is parallel to the 𝑦𝑦-axis:
𝑃𝑃
In this example, lines 𝑥𝑥 = 𝑎𝑎 for 𝑎𝑎 < 𝑥𝑥0 do not meet the level curve near 𝑃𝑃, while those vertical lines just to the right of
𝑥𝑥 = 𝑥𝑥0 and near P meets the level curve in more than one point.
The statement in red about 𝑓𝑓 in the last paragraph means that there is a function 𝑔𝑔(𝑥𝑥), defined for 𝑥𝑥 near 𝑥𝑥0 , such that
𝑦𝑦 = 𝑔𝑔(𝑥𝑥) is the unique solution of Eq (9) near 𝑥𝑥0 . We can now define a parametrization \gamma of the part of the level
curve near 𝑃𝑃 by
𝜸𝜸(𝑡𝑡) = 〈𝑡𝑡, 𝑔𝑔(𝑡𝑡)〉.
If we accept that 𝑔𝑔 is smooth (which follows from the inverse function theorem), then 𝜸𝜸 is certainly regular since
is obviously never zero. This 'proves' the theorem.
𝛾𝛾 ′ (𝑡𝑡) = 〈1, 𝑔𝑔′ (𝑡𝑡)〉
Q.E.D
3.5. LINE INTEGRALS, VECTOR FIELDS, GREEN’S THEOREM, SURFACE
INTEGRALS, STOKE’S THEOREM
LINE INTEGRALS OF SCALAR FIELDS IN THE PLANE AND IN SPACE
Definition 3.5.1.
Let f (x, y, z) be any continuous scalar function of three variables defined on a Domain E
(This is a usually a connected open region of R3 . Suppose that γ be any smooth parametrized space curve given by
γ(s) = hx(s), y(s), z(s)i) with s as the arc-length parameter and satisfying the inequality a 6 s 6 b.
We denote the line integral of f along the path γ with respect to arc length from a to b to be:
In space
Z
f (x, y, z) ds =
lim
kPk→0
γ
or
Z
In the Plane
f (x, y) ds =
γ
n
X
f (x∗i , yi∗ , zi∗ ) ∆si (provided this limit exists)
i=1
lim
kPk→0
n
X
f (x∗i , yi∗ ) ∆si (provided this limit exists)
i=1
Theorem 3.5.1. [Evaluation Theorem for Line Integrals with respect to Arc-length] If the plane
curve γ is given parametrically by x = x(s), y = y(s), a 6 s 6 b or the space curve γ by x = x(s), y =
y(s), z = z(s), a 6 s 6 b, then
(In the plane)
Z
Z
f (x, y) ds =
γ
or
(In Space)
Z
f (x, y, z) ds =
γ
b
f (x(s), y(s)) ds
a
Z
b
f (x(s), y(s), z(s)) ds
a
In practice, the arc-length parameter s is usually difficult to obtain, therefore, the above Evaluation Theorem becomes
impractical. A more practical Evaluation Theorem for line integrals with respect to arc-length is given in the next
theorem.
Theorem 3.5.2. [Evaluation Theorem for Line Integrals with respect to Arc-length] If the curve γ
is given parametrically by
x = x(t), y = y(t), a 6 t 6 b or by x = x(t), y = y(t), z = z(t), a 6 t 6 b,
then
Z
f (x, y) ds =
γ
or
Z
γ
f (x, y, z) ds =
Z
b
f (x(t), y(t))
a
Z
a
b
f (x(t), y(t), z(t))
p
[x′ (t)]2 + [y ′ (t)]2 dt =
Z
a
b
f (x(t), y(t)) k γ ′ (t) k dt
Z
p
[x′ (t)]2 + [y ′ (t)]2 + [z ′ (t)]2 dt =
a
b
f (x(t), y(t), z(t)) k γ ′ (t) k dt
Example 3.5.1. How to find the mass of a Helical Spring: Find the mass of a spring in the shape of the
Helix defined parametrically by γ(t) = h2 cos t, t, 2 sin ti, for 0 6 t 6 6π, with density ρ(x, y, z) = 2y.
Pn
Solution.
First
i=1 ρ(xi , yi , zi )∆si so that
R Mass, m is approximated as follows: m ≈
Pn we assume that the
m = limkpk→0 i=1 ρ(xi , yi , zi )∆si = γ ρ(x, y, z) ds. Since the density of the spring at point (x, y, z) on the spring
(the curve γ) is given as ρ(x, y, z) = 2y = 2t and the arc-length element ds is given by
ds = k γ ′ (t) k dt =
Thus,
m = Mass =
Z
p
√
(−2 sin t)2 + (1)2 + (2 cos t)2 dt = 5 dt
ρ(x, y, z) ds =
γ
Z
6π
√
√
2 5 t dt = 36π 2 5 mass-units
0
⊔
⊓
PROBLEMS YOU SHOULD TO DO
R
x2 y ds, where oriented curve γ is determined by the parametric equations
p
x = 3 cos(t), y = 3 sin(t), 0 6 t 6 π/2. Also show that the parametrization x = 9 − y 2 , y = y, 0 6 y 6 3,
gives the same value. Draw a picture of γ too! and if the above integral is impossible to evaluate by hand, then
don’t worry about it!
Exercise 3.5.1. Evaluate (if possible)
γ
R
Exercise 3.5.2.
Evaluate (if possible) γ (2x2 − 3yz) ds, where the parametrized curve γ is given by γ(t) =
hcos t, sin t, cos ti, for 0 6 t 6 2π. Do not Draw a picture of γ. If the above line integral is impossible to evaluate
by hand, then don’t worry about it!
Theorem 3.5.3. Let f (x, y, z) be a continuous function defined on some region D containing the parametrized
curve γ. Then, if γ is a piece-wise smooth curve, with γ = γ1 ∪ γ2 ∪ · · · ∪ γn , where γ1 , γ2 , . . . , γn are all smooth
and where the terminal point of γi is the same as the initial point of γi+1 , for i = 1, 2, . . . , n − 1, we have
(i)
(ii)
R
−γ
R
γ
R
f (x, y, z) ds =
f (x, y, z) ds =
γ
f (x, y, z) ds and
Pn
R
i=1 γi
f (x, y, z) ds.
Interpretation of Line Integrals of f (x, y) with respect to Arc-Length
Rb
As before, if y = f (x) ≥ 0 for all x ∈ [a, b] then a f (x) dx measures the area of the plane region bounded by the
vertical lines x = a, y = Rb and the interval [a, b] and the graph of y = f (x). In a similar way, if z = f (x, y) ≥ 0 for
all (x, y) ∈ D ⊆ R2 , then γ f (x, y) ds measures the surface area of the vertical cylinder (usually called a “curtain”)
bounded above by the graph of z = f (x, y) and the vertical lines (in space parallel to the z-axis) passing through
the initial and terminal points of the curve γ (with the arc-length parameter s) and also bounded below by the curve
γ itself.
Example
3.5.2. Evaluating a line integral over a Piece-Wise smooth Curve γ: Evaluate the line integral
R
γ (3x − y) ds, where the plane cure γ is the line segment from (1, 2) to (3, 3), followed by the portion of the circle
x2 + y 2 = 18 traversed from (3, 3) clockwise around to (3, −3).
Solution. Since γ is piece-wise smooth, we must find smooth parametrization of the smooth portions of γ which
are γ1 (the line segment from (1, 2) to (3, 3)) and γ2 the circular part of the circle x2 + y 2 = 18 traversed clockwise
from (3, 3) to (3, −3).
(A) For the line Segment portion: Use the parametrization formula for any line segment given as:
ϕ(t) = (1 − t)ϕ0 + tϕ1 for 0 6 t 6 1.
Now treat the initial point of the line segment as position vector γ10 = ϕ0 = h1, 2i and the terminal point (3, 3) as
the position vector γ11 = ϕ1 = h3, 3i so that
γ1 (t) = h1 + 2t, 2 + ti for 0 6 t 6 1.
√
Thus ds = k γ1′ (t) k dt = k h2, 1i k dt = 5 dt and f (γ1 (t)) = f (1 + 2t, 2 + t) = 3(1 + 2t) − (2 + t) = 1 + 5t and
so we have
Z
Z
Z 1
Z 1
p
√
7√
2
2
f (x, y) ds =
(3x − y) ds =
[3(1 + 2t) − (2 + t)] 2 + 1 dt =
(5t + 1) 5 dt =
5
2
γ1
γ1
0
0
(B) For the curved portion of γ: This is not a line segment so our approach has to be different from that in
part (A). The usual parametrization of a circle with counterclockwise orientation is given as:
γ(t) = hR cos t, R sin ti for α 6 t 6 β.
√
We want a parametrization with clockwise orientation of a circle of radius R = 3 2. So we want to start when
−t = π/4 and end when −t = −π/4. This means replacing t in the counterclockwise parametrization of a circle
with −t will gives the clockwise orientation we seek: Thus using the trigonometric identities cos(−t) = cos t and
sin(−t) = − sin t we arrive at
√
√
γ2 (t) = h3 2 cos t, −3 2 sin ti for − π/4 6 t 6 π/4.
√
√
√
√
√
Now f (γ2 (t)) = f (3 2 cos t, −3 2 sin t) = 3(3 2 cos t) − (−3 2 sin t) = 3 2(3 cos t + sin t) and
ds = k
and
γ2′ (t)
q
√
√
√
√
√
k dt = k h−3 2 sin t, −3 2 cos ti k dt = (−3 2 sin t)2 + (−3 2 cos t)2 dt = 3 2 dt
Z
γ2
f (x, y) ds =
Z
π/4
−π/4
√
(3 cos t + sin t) 18 dt = 54 2.
Finally, combining the results of (A) and (B) we have that
Z
f (x, y) ds =
γ
Z
f (x, y) ds +
γ1
Z
f (x, y) ds =
γ2
Z
γ
(3x − y) ds =
√
7√
5 + 54 2.
2
⊔
⊓
R
Exercise 3.5.3. Evaluate (if possible) γ (x2 + y 2 ) ds, where γ is the piecewise smooth plane curve given by the
circle x2 + y 2 = 4 traversed clockwise from (0, 2) to (0, −2) and the line segment from (0, −2) to (−2, −2) and the
line segment from (−2, −2) to (−2, 2) and finally the line segment from (−2, 2) to (0, 2). Draw a picture of γ. If the
above line integral is impossible to evaluate by hand, then don’t worry about it!
Definition 3.5.2.
The line integral of f (x, y, z) with respect to the parameter x along the smooth
parametrized space-curve γ(t) = hx(t), y(t), z(t)i for α 6 t 6 β is written as:
Z
f (x, y, z) dx =
γ
lim
k P k→0
n
X
f (xi , y1 , zi ) ∆xi (provided that is limit exists )
i=1
and independent of how we choose the points (xi , yi , zi ) on the curve.
Likewise, we define the line integral of f (x, y, z) with respect to the parameter y along the smooth parametrized
space-curve γ(t) = hx(t), y(t), z(t)i for α 6 t 6 β as:
Z
f (x, y, z) dy =
γ
lim
k P k→0
n
X
f (xi , y1 , zi ) ∆yi
i=1
and the line integral of f (x, y, z) with respect to the parameter z along the smooth parametrized space-curve
γ(t) = hx(t), y(t), z(t)i for α 6 t 6 β as:
Z
f (x, y, z) dz =
γ
lim
k P k→0
n
X
f (xi , y1 , zi ) ∆zi .
i=1
In each case, the line integral is defined whenever the corresponding limit exists and is independent of how we choose
the points (xi , yi , zi ) on the curve.
Theorem 3.5.4. [Evaluation Theorem of Line Integrals with respect to the coordinate axes] Let
P (x, y, z), Q(x, y, z) and R(x, y, z) be a continuous functions defined on a Path-connected region E ⊆ R3 containing
the smooth parametrized space-curve γ(t) = hx(t), y(t), z(t)i for α 6 t 6 β. Then
Z
P (x, y, z) dx =
Z
Q(x, y, z) dy =
Z
R(x, y, z) dz =
γ
Z
β
Z
β
P (x(t), y(t), z(t)) x′ (t) dt
Q(x(t), y(t), z(t)) y ′ (t) dt
α
γ
IN THE PLANE
β
α
γ
Notation:
Z
R(x(t), y(t), z(t)) z ′ (t) dt.
α
Z
P (x, y) dx + Q(x, y) dy =
γ
IN SPACE
Z
γ
P (x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz =
Z
P (x, y) dx +
γ
Z
γ
P (x, y, z) dx +
Z
Q(x, y) dy
γ
Z
γ
Q(x, y, z) dy +
Z
R(x, y, z) dz.
γ
PROBLEM 4 Evaluating A Line Integral in Space: You Try these Now!
R
Evaluate the line integral γ (4xz + 2y) dx, where the piecewise smooth space-curve γ is made up of the line segment
(a) from (2, 1, 0) to (4, 0, 2) and the line segment (b) from (4, 0, 2) to (2, 1, 0).
R
Calculate γ 4x dx + 2y dz, where γ is the curve consisting of the line segment from (5, 1, 0) to (0, 1, 1) followed by
the line segment from (0, 1, 1) to (3, 5, 1) and followed by the line segment from (3, 5, 1) to (0, 0, 0).
Exercise 3.5.4.
Evaluate
Z
P (x, y) dx + Q(x, y) dy =
γ
Z
γ
x2
y
−x
dx + 2
dy,
2
+y
x + y2
where γ(t) = hcos3 t, sin3 ti; 0 6 t 6 π/2. [This hint might be helpful for evaluating the definite integral: Let
u = tan3 t]
Theorem 3.5.5. If f (x, y, z) is a continuous function defined on some path-connected region E of R3 and containing
the parametrized curve γ, then
(i)
If γ is a piecewise smooth curve, then
Z
−γ
(ii)
f (x, y, z) dx = −
Z
f (x, y, z) dx.
γ
If γ = γ1 ∪ γ1 ∪ · · · ∪ γn where all the γ1 , γ2 , . . . , γn are all smooth and the terminal point of γi is the same
as the initial point of γi+1 for all i = 1, 2, . . . , n − 1, then
Z
f (x, y, z) dx =
γ
n Z
X
i=1
f (x, y, z) dx.
γi
Theorem 3.5.6. Let x = (x1 , x2 , . . . , xn ) be a point in a path-connection region D of Rn containing a smooth
parametrized curve γ, for α 6 t 6 β and if f (x) is continuous on D. Then
Z
γ
f (x) ds =
Z
β
α
g(t) dt = G(β) − G(α)
where G(t) is an anti-derivative of g(t) = f (γ(t))k γ ′ (t) k
Problem 5 Consider the function f (x, y) = x2 + y 2 which is continuous on D = R2 . Let γ(t) = hR cos t, R sin ti
be a circle of radius
R R > 0 in D with counterclockwise orientation i.e., 0 6 t 6 2π. Provide an interpretation of
the line integral γ f (x, y) ds. Can you provide the answer to this integral without actually evaluating the integral
directly? (I had evaluated it in class; but try providing the result without looking it up or evaluating it directly)
2-D and 3-D VECTOR FIELDS
Consider this practical example: Suppose that as a Structural Engineer, you were sent by your employer (who wants
to build a bridge that takes you across the Hudson River in New York City) to take velocity readings (of the river)
in a 1 mile length of the river (a straight part of the Hudson). You would have to setup a rectangular coordinate
grid similar to R2 . At special points in your grid you would probably draw arrows (some long and some short point
in different directions) to represent the velocity of the river at point (x, y). In fact, from a mathematical point of
view, you actually have a graph called a vector field. Note this type of graph is different from the graphs that
you’ve seen thus far. To each point (x, y) your region (The river) a unique arrow (representing velocity of the river)
was assigned. Clearly, it seems that we have a vector function F : D → R2 whose range consists only of vectors in
R2 (viewed as a 2-D vector space) and whose domain D is a subset of R2 . The components of the function F could
depend on other factors at the point (x, y) chosen. Thus we make the following definition.
Definition 3.5.3. A vector field in a region D of R2 (the plane) is a function F : D → R2 (a mapping from the
plane to a vector space) defined by
F(x, y) = hP (x, y), Q(x, y)i for (x, y) ∈ D
where the component functions P (x, y) and Q(x, y) are scalar functions.
A vector field in a region E of R3 (in space) is a function F : E → R3 (a mapping from the space to a 3-D vector
space) defined by
F(x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, z)i for (x, y, z) ∈ E
where the component functions P (x, y, z), Q(x, y, z) and R(x, y, z) are scalar functions.
Example 3.5.3.
LINE INTEGRALS OF VECTOR FIELDS IN THE PLANE AND IN SPACE
WHAT IS A WORK INTEGRAL?
Suppose that the vector field F = F(x, y, z) = P (x, y, z) i + Q(x, y, z) j + R(x, y, z) k represents a force throughout
a region in space (it might be the force of gravity or an electromagnetic force of some kind) and that
γ(t) = hg(t), h(t), k(t)i, a 6 t ≤ b,
is a smooth curve in the region. Then the integral of F · T, the scalar component of F in the direction of the curve’s
unit tangent vector, over the curve is called the work done by F over the curve from a to b.
Definition 3.5.4. [Work Over a Smooth Curve] The work done by a force F = P (x, y, z) i + Q(x, y, z) j +
R(x, y, z) k over a smooth curve γ from t = a to t = b is
Z t=b
W =
F · T ds.
(1)
t=a
where the scalar coordinate functions (or component functions) P (x, y, z), Q(x, y, z) and R(x, y, z) are continuous (and possibly have continuous first partial derivatives) on an open and path-connected domain E ⊆ R3 .
We wish to find the work done by the vector field F in moving a particle along a piece-wise smooth (or smooth)
parametrized unit-speed curve (or path) γ with arc-length parametrization contained in E .
In order to accomplish this (calculate work done), we let γ(s) = x(s) i + y(s) j + z(s) k (or γ(s) = hx(s), y(s), z(s)i)
be the position vector for a point P0 (x, y, z) on the curve γ. If T = T(s) = x′ (s) i + y ′ (s) j + z ′ (s) k is the unit
tangent vector at P0 , then FT = F · T (a scalar function) is the tangential component of the vector field F
at the point P0 . The work done by the vector field F in moving the particle from point P0 a short distance ∆s
along the curve is approximately FT ∆s = F · T ∆s, and consequently the work done in moving the particle from
arbitrary point A to B along γ is defined to be the work integral given by
Z
Z
W =
FT ds =
F · T ds
γ
γ
Different ways of writing the work integral
Work =
Z
β
α
(F · T) ds =
Z
FT ds
(The Definition of the Work Integral)
γ
Z d~γ
=
F·
ds
ds
γ
=
Z
γ
F · d~γ
Expanded to include ds; emphasizes the
arc-length parameter s and the velocity vector d~γ /ds
(Compact differential form)
=
Z d~γ
dt
F·
dt
γ
=
Z dx
dy
dz
P
+Q
+R
dt (Abbreviates the components of γ)
dt
dt
dt
γ
=
Z
P dx + Q dy + R dz
dy
ds
j+
Expanded to include dt; emphasizes the parameter t
and the velocity vector d~γ /dt
(dt′ s canceled; the most commonly used form).
γ
and T =
dx
ds
i+
dz
ds
k = h dx
ds ,
dy dz
ds , ds i
=
d~
γ
ds
and kTk = 1
HOW TO EVALUATE A WORK INTEGRAL
To evaluate the work integral, take these steps.
Step 1. Find a Parametrization of the curve γ with parameter t (if this has not been given)
Step 2. Evaluate F on the curve γ as a function of the parameter t, i.e., Calculate g(t) = F(γ(t)).
Step 3. Find γ~′ (t) = d~γ /dt i.e., Take the first derivative with respect to the parameter t of each component of
the parameterize curve γ.
Step 4. Dot the two vector functions F(γ(t)) with γ~′ (t), i.e., calculate F(~γ (t)) · γ~′ (t)
Step 5. Integrate the function of a single variable computed in Step 4 with respect to the parameter t from
t = α to t = β (the parameter interval you have from Step 1); i.e., Evaluate the definite integral
Z
t=β
t=α
F(~γ (t)) · γ~′ (t) dt.
Here’s a Quick Example
Example 3.5.4. [Finding work done by a Variable Force Over a Space Curve] Find the work done by
the variable force F given by
F(x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, z)i = hy − x2 , z − y 2 , x − z 2 i on E = R3
in moving a particle along the path (or smooth curve) given by γ(t) = ht, t2 , t3 i; 0 6 t 6 1 from the point
(0, 0, 0) to the (1, 1, 1) (Note: This is not by any means a straight path).
R
Solution. To Calculate the work integral γ F · d~γ using the 5 Steps above to accomplish this Task.
Step 1. A parametrization of the path or curve γ has been given as
γ = γ(t) = ht, t2 , t3 i; α = 0 6 t 6 1 = β.
Step 2. Calculate the force at each point on the path, i.e., F(γ(t)).
F(γ(t)) = F(t, t2 , t3 ) = ht2 − t2 , t3 − t4 , t − t6 i = h0, t3 − t4 , t − t6 i; 0 6 t 6 1.
Step 3. Find γ~′ (t).
γ~′ (t) = hx′ (t), y ′ (t), z ′ (t)i = h1, 2t, 3t2 i; 0 6 t 6 1.
Step 4. Dot the two vector functions F(γ(t)) and γ~′ (t).
F(γ(t)) ·
d~γ
= h0, t3 − t4 , t − t6 i · h1, 2t, 3t2 i
dt
= (t3 − t4 )(2t) + (t − t6 )(3t2 )
= 2t4 − 2t5 + 3t3 − 3t8 ; 0 6 t 6 1.
Step 5. Integrate the expression 2t4 − 2t5 + 3t3 − 3t8 over the parameter interval 0 6 t 6 1.
Z 1
WORK =
(2t4 − 2t5 + 3t3 − 3t8 ) dt
0
= 2/5 t5 − 1/3 t6 + 3/4 t4 − 1/3 t9
⊔
⊓
t=1
t=0
=
29
(Work Units).
60
Exercise 3.5.5. Find the work done by the Gravitational vector field
x
y
z
F = F(x, y, z) = −GmM
,
,
(x2 + y 2 + z 2 )3/2 (x2 + y 2 + z 2 )3/2 (x2 + y 2 + z 2 )3/2
in moving a particle along the straight-line curve γ in space from A(0, 3, 0) to B(4, 3, 0). Here F is measure in
Newtons and length in meters
Exercise 3.5.6. Write out the details for the work integral along a piecewise smooth curve γ defined in an open
and connected region D of R2 for some vector field F(x, y) where (x, y) is always in D.
WHAT ARE FLOW AND CIRCULATION INTEGRALS?
Suppose we have a vector field F(x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, z)i and suppose that instead of calling F
a force field, we now view F as the velocity field of a fluid (liquid, air, molten lava, etc) flowing through a region in
space (a tidal basin or the turbine chamber of a hydroelectric generator, for example). Under these circumstances,
the integral of the quantity F · T along a curve γ in the region gives us the fluid’s flow along γ.
Definition 3.5.5. If γ(t) is a smooth parametrized curve in the domain of a continuous velocity field F, the flow
along the curve from t = α to t = β is
Z
Z
Flow of F along γ =
FT ds =
F · T ds
(2).
γ
γ
The integral in this case is called a flow integral.
If the curve γ is a simple closed curve, the flow integral is called the circulation of F around the curve γ.
I
I
Circulation of F around the closed path γ =
FT ds =
F · T ds.
γ
γ
(Circulation is just the flow of the fluid around a simple closed curve.)
Note: We do not calculate flow and circulation integrals any differently than we calculate work integrals. Our
interpretation of the vector field F (as either velocity or force) completely determines whether we are calculating
(flow, circulation) or work respectively. The 5 step procedure given above is still the same.
Here are two quick examples
Example 3.5.5. [Finding Flow along a Helix] A fluid’s velocity field F(x, y, z) = hx, z, yi. Find the flow
along the helix
γ(t) = hcos(t), sin(t), ti, 0 6 t ≤ π/2.
R
Solution.
To Calculate the Flow integral γ F · d~γ using the same 5-step procedure outlined above for work
integrals.
Step 1. A parametrization of the path or curve γ has been given as
γ = γ(t) = hcos(t), sin(t), ti; α = 0 6 t 6 π/2 = β.
Step 2. Calculate the velocity at each point alonge the path, i.e., F(γ(t)).
F(γ(t)) = F(cos(t), sin(t), t) = hcos(t) t, sin(t)i, 0 6 t 6 π/2.
Step 3. Find γ~′ (t).
γ~′ (t) = hx′ (t), y ′ (t), z ′ (t)i = h− sin(t), cos(t), 1i; 0 6 t 6 π/2.
Step 4. Dot the two vector functions F(γ(t)) and γ~′ (t).
F(γ(t)) · γ~′ (t) = hcos(t), t, sin(t)i · h− sin(t), cos(t), 1i
= − sin(t) cos(t) + t cos(t) + sin(t).
Step 5. Integrate the expression − sin(t) cos(t) + t cos(t) + sin(t). over the given parameter interval 0 6 t 6
π/2.
Z π/2
Flow =
(− sin(t) cos(t) + t cos(t) + sin(t)) dt
0
=
⊔
⊓
t=π/2
cos2 (t)
π
1
+ t sin(t)
=
− .
2
2
2
t=0
Example 3.5.6. [Finding Circulation Around a circle] Find the circulation of the field F(x, y) = hx − y, xi
around the unit circle
γ(t) = hcos t, sin ti, 0 ≤ t ≤ 2π.
H
Solution. To Calculate the Circulation integral γ F · d~γ using the same 5-step procedure for work integrals
Step 1. A parametrization of the unit circle γ has been given as
γ = γ(t) = hcos(t), sin(t)i; 0 = α 6 t 6 β = 2π.
Step 2. Calculate the velocity at each point along the circle:
F(γ(t)) = F(cos(t), sin(t)) = hcos(t) − sin(t), cos(t)i, 0 6 t 6 2π.
Step 3. Find γ~′ (t).
γ~′ (t) = hx′ (t), y ′ (t)i = h− sin(t), cos(t)i; 0 6 t 6 2π.
Step 4. Dot the two vector functions F(γ(t)) and γ~′ (t).
F(γ(t)) · γ~′ (t) = hcos(t) − sin(t), cos(t)i · h− sin(t), cos(t)i
= − sin(t) cos(t) + sin2 t + cos2 (t)
= 1 − sin t cos t.
Step 5. Integrate the expression 1 − sin(t) cos(t) over the given parameter interval 0 6 t 6 2π.
Circulation =
Z
2π
0
(1 − sin(t) cos(t)) dt
t=2π
sin2 t
= t−
= 2π.
2 t=0
⊔
⊓
FLUX ACROSS A CURVE IN THE PLANE
To find the rate at which a fluid is entering or leaving a region enclosed by a smooth curve γ in the xy-plane, we
calculate the line integral over γ of F · n, the scalar component of the fluid’s velocity field in the direction of the
curve’s outward pointing normal vector. The value of this integral is the flux of F across γ. Flux is Latin for flow,
but many flux calculations involve no motion at all. If F were an electric field or a magnetic field, for instance, the
integral of F · n would still be called the flux of the field across γ.
Definition 3.5.6. [Flux Across a Closed Curve in the Plane] If γ is a smooth closed curve in the domain
of a continuous vector field F(x, y) = hP (x, y), Q(x, y)i in the plane and if n is the outward-pointing unit normal
vector on γ, the flux of F across γ is
I
Flux of F across γ =
F · n ds
(3)
γ
PATH INDEPENDENCE, POTENTIAL FUNCTIONS, AND CONSERVATIVE VECTOR FIELDS
In gravitational and electric fields, the amount of work it takes to move a mass or a charge from one point to
another depends only on the object’s initial and final positions and not on the path taken in between. Now we
discuss the notion of path independence of work integrals and describe some properties of vector fields in which the
work integrals are path independent.
PATH INDEPENDENCE AND CONSERVATIVE VECTOR FIELDS
R
If A and B are two points in an open region E in space, the work F · d~γ done in moving a particle from A to B
by a field F defined on E usually depends on the path taken. For some special fields, however, the integral’s
value
R
is the same for all paths from A to B. If this is true for all points A and B in E , we say that the integral F · d~γ is
path independent in E and that the field F is conservative on E .
Definition 3.5.7. [Path Independence and Conservative Field] Let F be a field defined on an open region
RB
E in space and suppose that for any two points A and B in E the workR A F · d~γ done in moving a particle from A
to B is the same value over all paths γ from A to B. Then the integral F · d~γ is said to be path independent in
E and the vector field F is said to be conservative on E .
Under conditions normally met in practice, a vector field F is conservative if and only if it is the gradient field of a
scalar function ϕ; that is, if and only if F = ∇ϕ for some ϕ. The function ϕ is then called a potential function
for F.
Definition 3.5.8. [Potential Function] If F is a vector field defined on E and F = ∇ϕ for some scalar function
ϕ defined on an open region E in space, then ϕ is called a potential function for F on E .
An electric potential is a scalar function whose gradient field is an electric field. A gravitational potential is a scalar
function whose gradient field is a gravitational field, and so on. As we will see, once we have found a potential
function ϕ for a vector field F, we can evaluate all the work integrals in the domain of F with the formula
Z
B
A
F · d~γ =
Z
B
A
∇ϕ · d~γ = ϕ(B) − ϕ(A).
If you think of ∇F for functions F of several variables as being something like the derivative f ′ for functions of the
single variable, then you see that the above equation is the vector calculus analogue of the Fundamental Theorem of
Calculus formula
Z
b
a
f ′ (x) dx = f (b) − f (a).
Conservative vector fields have other remarkable properties we as we go along. For example, saying that a vector
field F is conservative on E is equivalent to saying that the integral of F around every closed path (or closed curve)
in E is zero. Naturally, we need to impose conditions on the curves, vector fields, and domains to make the above
equation and its implications hold.
Assumptions we make from Now on: Connectivity
1.
All curves we consider are piecewise smooth, recall that it means the curve is made up of finitely many
smooth pieces connected end to end, as discussed in the section about Curves given earlier.
2.
The component functions of the vector field F have continuous first partial derivatives on some region E
in space. When F = ∇ϕ, this continuity requirement guarantees that the mixed second derivatives of the
potential function ϕ are equal (Clairaut’s Theorem), a result we will find revealing in studying conservative
vector fields.
3.
E is an open region in space. This means essentially that every point in E is the center of a sphere that lies
entirely in E .
4.
E is connected (all in one piece) which means in an open region every point of E can be connected to every
other point of E by a smooth curve that lies entirely in E .
Theorem 3.5.7. [Independence of Path Theorem] Let F be a continuous vector field on an open and connected
set D (subset of Rn ). Then the line integral
Z
γ
F · d~γ
is independent of path in D if and only if F = ∇ϕ for some scalar function ϕ : Rn → R (ϕ is called a potential
function for F); that is, if and only if F is a conservative vector field on D.
Notice that this theorem is just a generalized restatement of the definition given above.
Theorem 3.5.8. [Equivalent Conditions for line integrals] Let F be a continuous vector field on an open
connected subset D of Rn . Then the following conditions are equivalent:
A) The vector field F is conservative on D (i.e., F = ∇ϕ for some scalar function ϕ : Rn → R)
B) The line integral
C)
R
γ
R
γ
F · d~γ is independent of path in D.
F · d~γ = 0 for every closed path γ in D.
Theorem 3.5.9. [Test for Path Independence in Space] Let F(x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, z)i,
where the scalar functions P (x, y, z), Q(x, y, z) and R(x, y, z) are continuous together with their first-order partial
derivatives in an open and connected subset E of R3 . Then the vector field F is conservative on E if and only if
∇ × F = 0, that is, if and only if
∂P
∂Q
=
,
∂y
∂x
∂P
∂R
=
,
∂z
∂x
∂Q
∂R
=
∂z
∂y
In the two variable case, where F(x, y) = hP (x, y), Q(x, y)i. F is conservative on D if and only if
∂P
∂Q
=
∂y
∂x
Theorem 3.5.10. [The Fundamental Theorem for Line Integrals] Let γ be a piecewise smooth curve given
parametrically by γ = γ(t), a 6 t 6 b, which begins at a = γ(a) and ends at b = γ(b). If the scalar function ϕ is
continuously differentiable on an open set containing the curve γ, then
Z
∇ϕ(γ) · d~γ = ϕ(b) − ϕ(a)
γ
Exercise 3.5.7. For each point (x, y, z) in R3 /{(0, 0, 0)}, let F(x, y, z) be a vector pointed toward the origin (0, 0, 0)
with magnitude inversely proportional to the distance from the origin; that is, let
F = F(x, y, z) = −κ
γ
= −κ
k γ k2
x
y
z
,
,
x2 + y 2 + z 2 x2 + y 2 + z 2 x2 + y 2 + z 2
where γ = hx, y, zi. Show that vector field F is conservative on it’s domain E of definition by finding a potential
function for F.
3.6. GREEN’S THEOREM IN THE PLANE
In the preceding section, we learned how to evaluate flow integrals for conservative fields. We found a potential
function for the field, evaluated it at the path endpoints, and calculated the integral as the appropriate difference of
those values.
In this section, we see how to evaluate flow and flux integrals across closed plane curves when the vector field is not
conservative. The means for doing so is a theorem known as Green’s Theorem which converts line integrals to double
integrals.
Green’s Theorem is one of the great theorems of calculus. It is deep and surprising, and has far-reaching consequences.
In pure mathematics, it ranks in importance with the Fundamental Theorem of Calculus. In applied mathematics,
the generalizations of Green’s Theorem in three dimensions provide the foundation for theorems about electricity,
magnetism, and fluid flow.
We talk in terms of velocity fields of fluid flows because fluid flows are easy to picture. Be aware, however, that
Green’s Theorem applies to any vector field satisfying certain mathematical conditions. It does not depend for its
validity on the field’s having a particular physical interpretation.
Flux Density at a Point: Divergence
We need two new ideas for Green’s Theorem. The first is the idea of the flux density of a vector field at a point,
which in mathematics is called the divergence of the vector field. We obtain it in the following way:
Suppose we are given a vector field F(x, y) = P (x, y) i + Q(x, y) j which we interpret as the velocity
field of a fluid flow in the plane and that the first partial derivatives of P and Q are continuous at each
point of a region D. We let (x, y) be a point in the region and let R denote a small rectangle with
one of its vertices at (x, y) that, along with its interior, lies entirely in the region R. The sides of the
rectangle, parallel to the coordinate axes, have lengths ∆x and ∆y. The rate at which fluid leaves the
rectangle across the bottom edge is approximately
F(x, y) · (−j)∆x = −Q(x, y) ∆x.
This is the scalar component of the velocity at (x, y) in the direction of the outward normal times the
length of the segment. If the velocity is in meeters per second, for example, the exit rate will be in
meters per second times meters or square meters per second. The rates at which the fluid crosses the
other three sides in the directions of their outward normals can be estimated in a similar way. All told,
we have:
Exit Rates:
Top:
F(x, y + ∆y) · j ∆x = Q(x, y + ∆) ∆x
Bottom: F(x, y) · (−j) ∆x = −Q(x, y) ∆x
Right: F(x + ∆x, y) · i ∆y = P (x + ∆x) ∆y
Left:
F(x, y) · (−i) ∆y = −P (x, y) ∆y.
Combining opposite pairs gives us:
Top and Bottom:
Right and Left:
(Q(x, y + ∆y) − Q(x, y)) ∆x ≈
(P (x + ∆x, y) − P (x, y)) ∆y ≈
∂Q
∂y
∂P
∂x
Adding these last two equations gives
Flux across rectangle boundary ≈
∂P
∂Q
+
∂x
∂y
∆y
∆x
∆x ∆y.
∆x ∆y.
We now divide by ∆x ∆y to estimate the total flux per unit area or flux density for the rectangle:
Flux across rectangle boundary
≈
rectangle area
∂P
∂Q
+
∂x
∂y
.
Finally, we let ∆x and ∆y approach zero to define what we call the flux density of F at the point (x, y).
In mathematics, we call the flux density the divergence of F. The symbol for it is div F, pronounced “divergence
of F” or “div F.”
Definition 3.6.1. [Flux Density or Divergence] The flux density or divergence of a vector field F(x, y) =
P (x, y) i + Q(x, y) j at the point (x, y) in the plane is the scalar
div F = ∇ · F =
∂P
∂Q
+
.
∂x
∂y
(1)
If F(x, y, z) = P (x, y, z)i + Q(x, y, z) j + R(x, y, z), then at the point (x, y, z) in space the flux density or divergence
of F is the scalar
∂P
∂Q
∂R
div F = ∇ · F =
+
+
∂x
∂y
∂z
Intuitively, if water were flowing in to a region through a small hole at the point (x0 , y0 ), the lines of flow would diverge
there (hence the name) and, since water would be flowing out of a small rectangle about (x0 , y0 ), the divergence of
F at (x0 , y0 ) would be positive. If water were draining out of the hold instead of flowing in, the divergence would be
negative.
Exercise 3.6.1. Find the divergence of F(x, y) = (x2 − y) i + (xy − y 2 ) j.
Circulation Density at a Point: The k-component of Curl
The second of the two ideas we need for Green’s Theorem is the idea of circulation density of a vector field F at a
point.
To obtain it, we return to the velocity field
F(x, y) = P (x, y) i + Q(x, y) j
and the rectangle A. The counterclockwise circulation of F around the boundary of A is the sum of
flow rates along the sides. For the bottom edge, the flow rate is approximately
F(x, y) · i ∆x = P (x, y) ∆x.
This is the scalar component of the velocity F(x, y) in the direction of the tangent vector i times the
length of the segment. The rates of flow along the other sides in the counterclockwise direction are
expressed in a similar way. In all, we have
Top:
F(x, y + ∆y) · −i ∆x = −P (x, y + ∆) ∆x
Bottom: F(x, y) · (i) ∆x = P (x, y) ∆x
Right: F(x + ∆x, y) · j ∆y = Q(x + ∆x) ∆y
Left:
F(x, y) · (−j) ∆y = −Q(x, y) ∆y.
We add opposite pairs to get:
Top and Bottom: −(P (x, y + ∆y) − P (x, y)) ∆x ≈ − ∂P
∆y
∆x
∂y Right and Left:
(Q(x + ∆x, y) − Q(x, y)) ∆y ≈ ∂Q
∂x ∆x ∆y.
Adding these last two equations gives
Circulation around rectangle’s boundary ≈
∂Q
∂P
−
∂x
∂y
∆x ∆y.
We now divide by ∆x ∆y to estimate the total flux per unit area or flux density for the rectangle:
Circulation around the rectangle’s boundary
∂P
∂Q
≈
+
.
rectangle area
∂x
∂y
Finally, we let ∆x and ∆y approach zero to define what we call the circulation density of F at the point (x, y).
The positive orientation of the circulation density for the plane is the counterclockwise rotation around the vertical
axis, looking downward on the xy-plane from the tip of the (vertical) unit vector k. The circulation value is actually
the k-component of a more general circulation vector we define later on, called the curl of the vector field F. For
Green’s Theorem, we need only the k-component.
Definition 3.6.2. [k-Component of Circulation Density or Curl] The k-component of the circulation
density or curl of a vector field F(x, y) = P (x, y)i + Q(x, y) j at the point (x, y) is the scalar
curl F · k = ∇ × F · k =
∂Q
∂P
−
.
∂x
∂y
(2)
If water is moving about a region in the xy-plane in a thin layer, then the k-component of the circulation or curl,
at a point (x0 , y0 ) gives a way to measure how fast and in what direction a small paddle wheel will spin if it is put
into the water at point (x0 , y0 ) with it axis perpendicular to the plane, parallel to k.
Exercise 3.6.2. Find the k-component of the curl for the vector field
F(x, y) = (x2 − y) i + (xy − y 2 ) j.
GREEN’S THEOREM
Theorem 3.6.1. [Green’s Theorem in the Plane] Let R be a domain (i.e., an open and connected subset) of
the xy-plane and let γ be a piecewise smooth simple closed curve in R whose interior is also in R. Let P (x, y) and
Q(x, y) be functions defined and continuous and having continuous first partial derivatives in R. Then
I
P (x, y) dx + Q(x, y) dy =
γ
ZZ D
∂Q
∂P
−
∂x
∂y
dA
where D is the closed region (i.e., it consists of a domain together with its boundary curve) bounded by γ.
Two forms for Green’s Theorem
In one form, Green’s Theorem says that under suitable conditions the outward flux of a vector field across a simple
closed curve in the plane equals the double integral of the divergence of the field over the region enclosed by the
curve.
Theorem 3.6.2. [Outward Flux-Divergence or Normal Form of Green’s Theorem] The Outward flux
of a vector field F(x, y) = P (x, y) i + Q(x, y) j across a positively oriented simple close curve γ (with arc-length
parametrization) in the plane equals the double integral of div F over the region D enclosed by the curve γ.
The outward flux of F across γ =
I
γ
F · n ds =
I
γ
−Q(x, y) dx + P (x, y) dy =
ZZ D
∂P
∂Q
+
∂x
∂y
dA.
(3)
Exercise 3.6.3. Calculate the outward Flux of the vector field F(x, y) = x i + y 2 j across the square bounded
by the lines x = ± 1 and y = ± 1 by direct methods and also by using green’s Theorem. Also, draw the positively
oriented piecewise smooth parametrized simple closed curve γ enclosing the region as stated in Green’s Theorem.
In another form, Green’s Theorem says that the counterclockwise circulation of a vector field around a simple closed
curve is the double integral of the k-component of the curl of the field over the region enclosed by the curve.
Theorem 3.6.3. [Counterclockwise Circulation-Curl or Tangential Form of Green’s Theorem] The
counterclockwise circulation of a vector field F(x, y) = P (x, y) i + Q(x, y) j around a positively oriented simple
closed curve γ (with arc-length parametrization) in the plane equals the double integral of the k-component of the
curl of the field (i.e., the double integral of curl F · k) over the region D enclosed by the curve γ.
Counterclockwise Circulation of F around γ =
I
γ
F · T ds =
I
P (x, y) dx + Q(x, y) dy =
γ
ZZ D
∂Q
∂P
−
∂x
∂y
dA
Exercise 3.6.4. Calculate the counterclockwise circulation of the vector field F(x, y) = tan−1 (y/x) i + ln(x2 + y 2 ) j
around the curve γ where γ is the boundary of the region defined by the polar coordinate inequalities 1 6 r 6 2,
0 6 θ 6 π using only Green’s Theorem. Also draw a picture of the region together with its positively oriented
boundary curve γ. What is the length of γ? Use your arc-length formula to verify your answer.
3.7. SURFACE INTEGRALS, STOKE’S THEOREM
SURFACE INTEGRALS OF SCALAR FIELDS
Let G(x, y, z) be a continuous scalar function of three variables (or possibly more) defined on a subset S of R3 .
Suppose S is a surface defined by the equation z = f (x, y) for all (x, y) in a subset D of R2 . We partition D into
n sub-rectangles Ri ; this results in a corresponding partition of the surface S into n surface patches Gi . Choose
sample point (x∗i , yi∗ ) ∈ Ri , and let (x∗i , yi∗ , zi∗ ) = (x∗i , yi∗ , f (x∗i , yi∗ )) be the corresponding point on the surface patch
Gi . Then we define the surface integral of G over the surface S to be the integral given by
ZZ
n
X
G(x, y, z) dS = lim
G(x∗i , yi∗ , zi∗ ) ∆Si (provided this limit exists)
kP k→0
S
i=1
where ∆Si is the area of the ith surface patch Gi . This definition extends in a natural way to non-Rectangular
regions in R2 (by giving G the value 0 outside D).
Please note the difference in this definition of surface integral and the definition of line integral. Pay close attention
to the ∆si (arc-length of curve) and ∆Si (area of surface patch Gi ). The surface integral Generalizes the line integral.
The integral equation in (7) takes on different meanings in different applications. If G(x, y, z) has the constant value
1, then integral gives the area of S. If G(x, y, z) gives the mass density of a thin shell of material modeled by S, the
integral gives the mass of the shell.
Surface integrals behave like other double integrals, the integral of the sum of two functions being the sum of their
integrals and so on. The domain additivity property takes the form
ZZ
ZZ
ZZ
ZZ
G(x, y, z) dS =
G(x, y, z) dS1 +
G(x, y, z) dS2 + · · · +
G(x, y, z) dSn .
S
S1
S2
Sn
The idea is that if S is partitioned by smooth curve into a finite number of non-overlapping patches (i.e., if S is
piecewise smooth), then the integral over S is the sum of the integrals over the patches. Thus, the integral of a
function over the surface of a cube is the sum of the integrals over the faces of the cube.
Theorem 3.7.1. [Evaluation Theorem(s) for surface Integrals] Let S be a piecewise smooth surface given
by z = f (x, y), where (x, y) is in a region Dxy of the xy-plane. If z = f (x, y) has continuous first-order partial
derivatives and G(x, y, z) = G(x, y, f (x, y)) is continuous on Dxy , then
ZZ
ZZ
q
G(x, y, z) dS =
G(x, y, f (x, y)) fx (x, y)2 + fy (x, y)2 + 1 dA.
(4)
S
Dxy
Let S be a piecewise smooth surface given by y = h(x, z), where (x, z) is in a region Dxz of the xz-plane. If
y = h(x, z) has continuous first-order partial derivatives and G(x, y, z) = G(x, h(x, z), z) is continuous on Dxz , then
ZZ
ZZ
p
G(x, y, z) dS =
G(x, h(x, z), z) hx (x, z)2 + hz (x, z)2 + 1 dA.
(5)
S
Dxz
Let S be a piecewise smooth surface given by x = u(y, z), where (y, z) is in a region Dyz of the yz-plane. If
x = u(y, z) has continuous first-order partial derivatives and G(x, y, z) = G(u(y, z), y, z) is continuous on Dyz , then
ZZ
ZZ
q
G(u(y, z), y, z) uy (y, z)2 + uz (y, z)2 + 1 dA.
(6)
G(x, y, z) dS =
S
Dyz
In general, if we view a given surface S as a level surface H(x, y, z) = c and H is a continuous function defined at
the points of the surface S and D is the shadow region of S, then we can write
ZZ
ZZ
k∇H(x, y, z)k
G(x, y, z) dS =
G(x, y, z)
dA.
(7)
| ∇H(x, y, z) · p |
S
D
where p is a unit vector normal to the shadow region D and ∇H · p 6= 0.
In the most popular case where the surface S is given by the equation z = f (x, y), we let H(x, y, z) = z − f (x, y) =
0 = c or H(x, y, z) = f (x, y) − z = 0 = c and the unit vector normal to the shadow region D of S is p = k and
so we get the formula (4) above.
Example 3.7.1. [Integrating over a Surface] Let us use formula(s) from among (4), (5), and (6) and (7) to
evaluate the surface integral
ZZ
y dS,
S
where the surface S is defined by the equation z = f (x, y) = x + y 2 on D = {(x, y) : 0 6 x 6 1, 0 6 y 6 2}.
Solution.
Using formula (4) (which is applicable to our situation z = f (x, y)) we see that must let
G(x, y, z) = y, fx (x, y) = 1, fy (x, y) = 2y so that
q
√ p
fx (x, y)2 + fy (x, y)2 + 1 = 2 2y 2 + 1.
Thus, we have
ZZ
ZZ
q
G(x, y, z) dS =
G(x, y, f (x, y)) fx (x, y)2 + fy (x, y)2 + 1 dA
S
Z ZD √ p
p
√ ZZ
=
y 2 2y 2 + 1 dA = 2
y 2y 2 + 1 dA (All about Calc II&III)
D
D
Z 1 Z 2 p
2y 2 + 1 4y dy dx
= 2−3/2
0
0
√
Z 9
√
13 2
−3/2
= 2
u du =
.
3
1
Using formula (7), we would view the given surface S whose equation z = f (x, y) = x + y 2 as a Level surface
H(x, y, z) = z − f (x, y) = 0 and so H(x, y, z) = z − x − y 2 . Now
∇H(x, y, z) =
∂H ∂H ∂H
,
,
∂x ∂y ∂z
= h−1, −2y, − 1i and k∇H(x, y, z) k = k −1 i −2y j − 1 k k =
√ p
2 2y 2 + 1
and since p = k is a unit vector normal to the shadow region D (of the xy-plane) of the Level
surface H(x, y, z) = 0
√ p
∇H(x,y,z) k
we have that |∇H(x, y, z) · p| = |∇H(x, y, z) · k| = | − 1| = 1 and so k|∇H(x,y,z)·p|
= 2 2y 2 + 1. Thus we can
evaluate the surface integral as follows:
ZZ
S
G(x, y, z) dS =
ZZ
D
G(x, y, z)
k ∇H(x, y, z) k
dA =
|∇H(x, y, z) · p|
ZZ
D
y
√
√ p
13 2
2 2y 2 + 1 dA =
3
which agrees with the result obtained using formula (4). ⊔
⊓
Exercise 3.7.1. Use formula (7) to Evaluate the surface integral
ZZ
xyz dS
S
where S is the surface of the cube cut from the first octant by the planes x = 1, y = 1, and z = 1.
Exercise 3.7.2. Use formula (5) in the cases p = k and p = j above to Evaluate the surface integral
ZZ
xyz dS
S
where S is the portion of the cone z 2 = x2 + y 2 between the planes z = 1 and z = 4. Include a sketch of S and a
sketch of D (the shadow region of S) on separate coordinate axes. Note this problem asks for two evaluation of the
given surface integral!
A TIDBIT ABOUT ORIENTABLE SURFACES!!!
We call a smooth surface S orientable or two-sided if it is possible to define a vector field n = n(x, y, z) of unit
normals vectors on S that varies continuously with position. Any patch or sub-portion of an orientable surface is still
orientable. Spheres and other smooth closed surfaces in space (smooth surfaces that enclose solids) are orientable.
By convention, we choose the unit normal vector field n = n(x, y, z) on a closed surface S to point outward and call
n(x, y, z) the outward unit normal at the point (x, y, z) on the surface S .
Once n has been chosen, we say that we have oriented the surface S, and we call the surface together with its
normal vector field an oriented surface. The vector n at any point on the surface is called the positive direction
at that point.
SURFACE INTEGRALS OF VECTOR FIELDS
Suppose that F(x, y, z) = P (x, y, z) i + Q(x, y, z) j + R(x, y, z) k is a continuous vector field defined over an oriented
(two-sided) surface S and that n is the chosen unit normal field on the surface. We call the surface integral of F · n
over S; the outward flux of F across S. Thus, flux is the integral over the surface S of the scalar component of
F in the direction of the outward unit normal n.
Surface Integral for Flux
Definition 3.7.1. [Outward Flux Across a Surface] The Flux of a three-dimensional vector field F(x, y, z) =
P (x, y, z) i + Q(x, y, z) j + R(x, y, z) k across an oriented surface S in the direction of the chosen outward unit
normal n is given as
ZZ
ZZ
Flux of F across S in the direction of n =
F · dS =
(F · n) dS where dS = n dS
S
S
If we view the vector field F as the velocity field of a three-dimensional fluid flow(such as air), the flux of F across
S is the net rate at which fluid is crossing the surface S in the chosen positive direction.
Note: If S is part of a level surface H(x, y, z) = c, then n may be taken to be one of the two fields
n = n(x, y, z) = +
∇H(x, y, z)
k∇H(x, y, z)k
or n = n(x, y, z) = −
∇H(x, y, z)
k∇H(x, y, z)k
depending on which one gives the preferred direction (you can determine the right n by testing at a convenient point
(x, y, z) on S). The corresponding outward flux across S is
ZZ
Outward Flux of F across S =
F · dS
=
ZSZ
(F · n) dS (Please note the difference between dS and dS)
S
ZZ ± ∇H(x, y, z)
k ∇H(x, y, z) k
F·
dA
k∇H(x, y, z)k | ∇H(x, y, z) · p |
D
ZZ
1
=
(F · ±∇H(x, y, z)) dA
| ∇H(x, y, z) · p |
D
ZZ
1
= ±
(F · ∇H(x, y, z)) dA
| ∇H(x, y, z) · p |
=
(6)
D
Exercise 3.7.3. Find the outward Flux of the vector field F(x, y, z) = yz j + z 2 k across S where S is the portion
cut from the cylinder y 2 + z 2 = 1, z > 0, by the planes x = 0 and x = 1. Include a sketch of S and a sketch of
the shadow region D (This is a projection of the surface S in the xy-plane)
Mass and Moment formulas for very thin shells
Mass:
M =
RR
ρ dS
(ρ = ρ(x, y, z) = density at (x, y, z), mass per unit area)
S
First moments about the coordinate planes:
ZZ
ZZ
Myz =
xρ dS,
Mxz =
yρ dS,
S
Mxy =
S
ZZ
zρ dS
S
Coordinates of Center of mass:
x̄ =
Myz
,
M
ȳ =
Mxz
,
M
z̄ =
Moments of inertia about coordinate axes:
ZZ
ZZ
Ix =
(y 2 + z 2 )ρ dS,
Iy =
(x2 + z 2 )ρ dS,
S
Iz =
S
IL =
ZZ
ZZ
Mxy
M
(x2 + y 2 )ρ dS
S
r2 ρ dS
S
where r = r(x, y, z) is the distance from the point (x, y, z) to the line L
Radius of gyration about a line L :
RL =
r
IL
M
Example 3.7.2. Let’s find the coordinates of the center of mass of a thin hemispherical shell of radius R > 0
whose mass per unit area at each point (x, y, z) on the thin hemispherical shell is the constant κ (i.e., the density
ρ = ρ(x, y, z) = κ at each point (x, y, z) on the thin hemispherical shell is constant).
Solution. We model the shell with hemisphere
H(x, y, z) = x2 + y 2 + z 2 = R2 ,
z >0
The symmetry of the surface about the z-axis tells us that x̄ = ȳ = 0. It remains only to calculate z̄ from the
M
formula z̄ = Mxy . The mass M of the shell is given by
M =
ZZ
ρ dS = κ
S
ZZ
dS = κA(S) =
S
1
(4πR2 )κ = 2πR2 κ.
2
To evaluate the surface integral Mxy we take p to be k so that
k ∇H(x, y, z) k = k 2hx, y, zi k = 2
p
x2 + y 2 + z 2 = 2R
and | ∇H(x, y, z) · p | = | ∇H(x, y, z) · k | = 2 | z | = 2z. Recall that
dS =
Then
Mxy =
ZZ
S
k ∇H(x, y, z) k
2R
R
dA =
dA =
dA.
| ∇H(x, y, z) · p |
2z
z
zρ dS =
ZZ
zκ
R
dA = RκA(D) = Rκ(πR2 ) = πκR3 ;
z
D
and now we have the z-coordinate of the center of mass as
z̄ =
Mxy
πκR3
R
=
= .
M
2πκR2
2
Hence the center of mass of the thin hemispherical shell of radius R > 0 with constant density is (0, 0, R2 ).
⊔
⊓
Exercise 3.7.4. In Example 3.7.2 we avoided direct computation of the x and y coordinates of the center of mass
of the thin shell by use of the phrase “by symmetry of the surface about the z-axis tells us that x̄ = 0 and ȳ = 0.”
Show that we were just in making such a statement.
PARAMETRIC SURFACES !!!!
For a surface S in three-dimensional space we are accustomed to the Explicit forms: z = f (x, y), x = g(y, z) and
y = u(x, z) and the Implicit form: F (x, y, z) = c = 0 (Also known as a level surface for the function F ). There is
also a Parametric form that gives the position of a point (x, y, z) on the surface as a vector function of two variables.
Consider a function r : D → V3 where V3 is the three dimensional real vector space R3 with the standard basis (Oh
NO! there is that Japanese again!) and is defined as:
r = r(u, v) = f (u, v) i + g(u, v) j + h(u, v) k, (u, v) ∈ D
(9)
be a continuous vector function that is defined on a region D in the uv-plane and one-to-one on the interior of
D. We call the range of r the surface S defined or traced out by r. Equation (9) together with the domain
R constitutes a parametrization of the surface S. The variable u and v are the parameters, and D is the
parameter domain. To simplify our discussion, we take D to be the rectangle R defined by inequalities of the
form R = {(u, v) : a 6 u 6 b, c 6 v 6 d}. The requirement that r be one-to-one on the interior of R ensures
that S does not cross itself. Notice that Equation (9) is the vector equivalent of three parametric equations:
x = f (u, v),
y = g(u, v),
z = h(u, v) where (u, v) ∈ R.
PARAMETRIZATIONS OF SOME POPULAR SURFACES
Example 3.7.3. [Parametrizing the Cone] Find a parametrization of the cone
z2
x2
y2
=
+
.
c2
a2
b2
Solution. Cylindrical coordinates provide everything we need. A typical point (x, y, z) on the cone has x = ar cos θ,
y = br sin θ, and z = cr, with − Hc 6 r 6 Hc and 0 6 θ 6 2π. Taking u = r and v = θ in Equation (9) give the
parametrization
r(r, θ) = a(r cos θ) i + b(r sin θ) j + cr k,
−
H
H
6 r 6
,
c
c
0 6 θ 6 2π.
Note: If a = b = c, then we have the familiar RIGHT CIRCULAR CONE. ⊔
⊓
Example 3.7.4. [Parametrizing an Ellipsoid] Find a parametrization of the Ellipsoid
x2
y2
z2
+
+
= 1.
a2
b2
c2
Solution. Spherical coordinates provide what we need. A typical point (x, y, z) on the Ellipsoid has x =
a sin φ cos θ, y = b sin φ sin θ, and z = c cos φ, 0 6 φ 6 π, 0 6 θ 6 2π. Taking u = φ and v = θ in
Equation (9) gives the parametrization
r(φ, θ) = (a sin φ cos θ) i + (b sin φ sin θ) j + c cos φ k,
Note: If a = b = c, then we the familiar SPHERE. ⊔
⊓
0 6 φ 6 π,
0 6 θ 6 2π.
Example 3.7.5. [Parametrizing an Elliptic Cylinder] Find a parametrization of the Cylinder
x2
y2
+
= 1 (It may look like an ellipse; but it is not!).
a2
b2
Solution. Cylindrical coordinates provide what we need. A typical point (x, y, z) on the elliptic cylinder has
x = a cos θ, y = b sin θ, and z = z, 0 6 θ 6 2π, −H 6 z 6 H. Taking u = θ and v = z in Equation (9) gives
the parametrization
r(θ, z) = a cos θ i + b sin θ j + z k, 0 6 θ 6 2π, −H 6 z 6 H, H > 0.
Note: If a = b, then we have the familiar RIGHT CIRCULAR CYLINDER OF HEIGHT 2H.
⊔
⊓
Example 3.7.6. [Parametrizing an Elliptic Paraboloid] Find a parametrization of the Elliptic Paraboloid
z
x2
y2
= 2 + 2 (It may look like an ellipse; but it is not!).
c
a
b
Solution. Cylindrical coordinates provide what we need. A typical
q point (x, y, z) on the Elliptic Paraboloid has
H
2
x = ar cos θ, y = br sin θ, and z = cr , 0 6 θ 6 2π, 0 6 r 6
c . Taking u = r and v = θ in Equation (9)
gives the parametrization
2
r(r, θ) = ar cos θ i + br sin θ j + cr k, 0 6 θ 6 2π, 0 6 r 6
Note: If a = b, then we have the familiar CIRCULAR PARABOLOID.
r
H
, H > 0.
c
⊔
⊓
Example 3.7.7. [Parametrizing a Hyperbolic Paraboloid] Find a parametrization of the Hyperbolic Paraboloid
z
x2
y2
= 2 − 2 (It may look like an ellipse; but it is not!).
c
a
b
Solution. Cylindrical coordinates provide what we need. A typical
q point (x, y, z) on the Elliptic Paraboloid has
H
2
x = ar cosh θ, y = br sinh θ, and z = cr , 0 6 θ 6 2π, 0 6 r 6
c . Taking u = r and v = θ in Equation (9)
gives the parametrization
2
r(r, θ) = ar cosh θ i + br sinh θ j + cr k, 0 6 θ 6 2π, 0 6 r 6
Note: If a = b, then we have the familiar CIRCULAR PARABOLOID.
r
H
, H > 0.
c
⊔
⊓
Here is what I called the trivial Parametrization: Suppose a smooth surface S is given as the level surface
of a function H(x, y, z) = c. The we always have these three natural parametrizations
r(x, y) = x i + y j + f (x, y) k; for (x, y) ∈ Domain (f ) and z = f (x, y).
r(x, z) = x i + g(x, z) j + z k; for (x, z) ∈ Domain (g) and y = g(x, z).
r(y, z) = h(y, z) i + y j + z k; for (y, z) ∈ Domain (h) and x = h(y, z).
Definition 3.7.2. [Smooth Parametrized Surface] A parametrized surface S given by
r(u, v) = f (u, v) i + g(u, v) j + h(u, v) k
is said to be smooth if the first-order partial derivatives ru and rv are continuous and ru × rv is never zero on the
parameter domain. The surface S is said to be piecewise smooth if it is the union of finitely many smooth surfaces.
Area of Parametrized smooth surfaces
Theorem 3.7.2. [Area of a Parametrized smooth surface] The area of the smooth parametrized surface
r(u, v) = f (u, v) i + g(u, v) j + h(u, v) k,
D = {(u, v) | a 6 u 6 b, c 6 v 6 d}
is given by
The Surface Area of S = A(S) =
Z
c
dZ b
a
k ru × rv k du dv
(10)
p
Exercise 3.7.5. Find the area of the closed surface S consisting of the upper-half circular cone z = H
x2 + y 2
R
and the intersecting plane z = H where H and R are positive constants denoting the height and radius of the cone
respectively.
Area of non-parametrized smooth surface
Theorem 3.7.3. [Area of a non-parametrized smooth level surface] If a smooth surface S is given as a
level surface H(x, y, z) = c (non-parametrized), then the Area of S is given by
ZZ
ZZ
1
A(S) =
dS =
k ∇H(x, y, z) k dA
(11)
| ∇H(x, y, z) · p |
S
D
where p is a unit vector normal to the shadow region D of the surface S.
Exercise 3.7.6. Repeat Exercise 3.7.5 using formula (11) (Some of you may have done this before but do not
recognize it! do it again anyway!)
Theorem 3.7.4. [Evaluation Theorem for Surface Integrals over Parametrized surfaces]
If S is a smooth surface defined parametrically as
r(u, v) = f (u, v) i + g(u, v) j + h(u, v) k, a 6 u 6 b, c 6 v 6 d,
and G(x, y, z) is a continuous function defined on S, then the surface Integral of G over S is
ZZ
ZZ
G(x, y, z) dS =
G(r(u, v)) kru × rv k du dv
S
R
=
Z
c
d
Z
a
b
G(f (u, v), g(u, v), h(u, v)) k ru × rv k du dv
where R is the uv-parameter domain.
Exercise 3.7.7. Integrate G(x, y, z) = x2 over the upper-half circular cone z =
p
x2 + y 2 , 0 6 z 6 1.
(12)
Outward Flux for parametrized surfaces
Theorem 3.7.5. [Evaluation Theorem for Outward Flux across a parametrized surface] Suppose that
F(x, y, z) = P (x, y, z) i + Q(x, y, z) j + R(x, y, z) k
is a continuous vector field and that n = n(x, y, z) is the outward unit normal field on smooth parametrized positively
orientable surface S, then the outward Flux of F across S is given by the surface integral
ZZ
ZZ
ZZ
outward flux of F across S =
F · dS =
F · n dS = +
F(r(u, v)) · ru × rv du dv
(13)
S
S
R
Example 3.7.8. Find the outward flux of the vector field F(x, y, z) = hx, y, zi through the
Sphere: r(ϕ, θ) = hR sin ϕ cos θ, R sin ϕ sin θ, R cos ϕi; 0 ≤ ϕ ≤ π, 0 ≤ θ ≤ 2π
of radius R > 0.
Solution. First, F(r(ϕ, θ)) = R hsin ϕ cos θ, sin ϕ sin θ, cos ϕi. Next, rϕ = R hcos ϕ cos θ, cos ϕ sin θ, − sin ϕi
and rθ = R h− sin ϕ sin θ, sin ϕ cos θ, 0i so that rϕ × rθ = R2 hsin2 ϕ cos θ, sin2 ϕ sin θ, sin ϕ cos ϕi. Therefore,
F(r(ϕ, θ)) · rϕ × rθ = R hsin ϕ cos θ, sin ϕ sin θ, cos ϕi · R2 hsin2 ϕ cos θ, sin2 φ sin θ, sin ϕ cos ϕi
= R3 (sin3 ϕ cos2 θ + sin3 ϕ sin2 θ + cos2 ϕ sin ϕ)
= R3 (sin3 ϕ + cos2 ϕ sin ϕ)
= R3 sin ϕ(sin2 ϕ + cos2 ϕ)
= R3 sin ϕ
Now outward flux is
ZZ
D
⊔
⊓
F(r(ϕ, θ)) · rϕ × rθ dϕ dθ = R3
Z
0
2π
Z
0
π
sin ϕ dϕ dθ = 4πR3 = 3 (volume of the sphere)
Example 3.7.9.
Evaluate the surface integral
RR
S
F · dS for the vector field F(x, y, z) = hx, −z, yi where S is the
part of the sphere x2 + y 2 + z 2 = 4 in the first octant. In other words, find the flux of F across
S (this S is not closed!). For closed surfaces, use the positive (outward) orientation.
In the problem we are given the continuous vector field F = x i − z j + y k and the open smooth surface S which
is part of the closed surface (a sphere ) x2 + y 2 + z 2 = 4 in the first octant, with orientation towards the origin.
Solution.
Step 1. Use formulas
Outward Flux of F across S =
ZZ
F · dS =
S
ZZ
S
F · n dS = ±
ZZ
D
1
F · ∇H(x, y, z) dA
| ∇H(x, y, z) · p |
in Definition 3.7.1
Step 2. Choosing the required n: We view the sphere x2 + y 2 + z 2 = 4 as the closed non-parametrized level
surface of H(x, y, z) = c = 0 where H(x, y, z) = x2 + y 2 + z 2 − 4; so that it is enough to choose n by
calculating ± ∇H(x, y, z) = ±2 (x i + y j + z k) (vectors in the direction of n = n(x, y, z)). At the point
(0, 0, 2) on the surface we see that ∇H(0, 0, 2) = 4 k (which points away from the origin). Since the problem
requires the opposite orientation we can choose n as a unit vector pointing in the direction of −∇H(x, y, z).
Thus, we would choose n in the direction of
− ∇H(x, y, z) = − (2x i + 2y j + 2z k).
Based on this choice of n our formula in step 1 above becomes
ZZ
ZZ
ZZ
1
F · dS =
F · n dS = −
(F · ∇H(x, y, z)) dA
| ∇H(x, y, z) · p |
S
S
D
Step 3. Since portion of the sphere x2 + y 2 + z 2 = 4 that we are interested in lies in the first octant, then we
know that z > 0 and the projection of S in the xy-plane is the quarter disk
p
D = {(x, y) | 0 6 x 6 2, 0 6 y 6 4 − x2 }.
Thus a unit vector normal to the shadow
p region D is k (i.e. p = k.) Hence, | ∇H(x, y, z) · p | = | ∇H(x, y, z) ·
k | = 2 | z | = 2 z (Recall that z = 4 − (x2 + y 2 ) > 0).
Step 4. We calculate
−
This gives us
−
1
(F · ∇H(x, y, z)) dA.
| ∇H(x, y, z) · p |
1
x2
x2
1
dA
(F·∇H(x, y, z)) dA = −
hx, −z, yi·h2x, 2y, 2zi dA = −
dA = − p
| ∇H(x, y, z) · p |
2z
z
4 − (x2 + y 2 )
Step 5. We can now integrate as follows
ZZ
ZZ
1
F · dS = −
(F · ∇H(x, y, z)) dA.
| ∇H(x, y, z) · p |
S
D
ZZ
x2
p
= −
dA (where D = {(x, y)| x > 0, y > 0 and 0 6 x2 + y 2 6 4})
4 − (x2 + y 2 )
D
Z π/2 Z 2 2
r cos2 θ
√
= −
r dr dθ (In polar coordinates)
4 − r2
0
0
!
Z
Z
π/2
= −
2
cos2 θ dθ
0
0
r3 (4 − r2 )−1/2 dr
Step 6. To get a result, we must now calculate the definite integrals
Z π/2
Z π/2
Z π/2
1
1
π
π
cos2 θ dθ =
dθ +
cos 2θ 2 dθ =
+0 =
2 0
4 0
4
4
0
and
Z
2
r2
r dr (We must be careful, because the integral is an improper integral !!)
4 − r2
0
Try letting τ = 4 − r2 so that dτ = −2r dr and r2 = 4 − τ . So the integral becomes
Z 0
Z
1
4−τ
1 44−τ
√
√
−
dτ =
dτ
2 4
τ
2 0
τ
Z 4
1
−1/2
1/2
lim
(4τ
− τ ) dτ
=
2 λ→0+ λ
√
1
2 3/2
16
=
lim 16 −
−8 λ+ λ
2 λ→0+
3
3
16
=
3
√
Step 7. Finally, we arrive at an answer (you check my calculations for errors !) Thus the Flux of F across S
in the direction of n (towards the origin) is given as
ZZ
ZZ
1
F · dS = −
(F · ∇H(x, y, z)) dA
| ∇H(x, y, z) · p |
S
D
ZZ
p
x2
p
dA (where D = {(x, y) | 0 6 x 6 2, 0 6 y 6 4 − x2 })
= −
4 − (x2 + y 2 )
D
Z π/2 Z 2 2
r cos2 θ
√
= −
r dr dθ (In polar coordinates)
4 − r2
0
0
!
Z
Z
π/2
= −
2
cos2 θ dθ
0
0
r3 (4 − r2 )−1/2 dr
4π
π 16
= − ·
= −
.
4 3
3
⊔
⊓
Example 3.7.10. Consider the same problem in Example 3.7.9. We ask what if we now parametrize S instead?
Solution. The surface S is part of a sphere and in Example 3.7.4 know how to parametrize a sphere. Here is a
parametrization of S
r(u, v) = 2 sin u cos v i + 2 sin u sin v j + 2 cos u k,
0 6 u 6 π/2, 0 6 v 6 π/2.
An outward unit normal n points in the direction of the vector ru × rv . The problem requires a unit normal in the
opposite direction (i.e., in the direction of rv × ru = −(ru × rv )). Thus the flux of F across S is given by
ZZ
ZZ
F · dS =
F · n dS
S
S
Z
= −
Z
π/2
0
π/2
F(r) · ru × rv du; dv
0
Step 1. F(r) = 2 sin u cos v i − 2 cos u j + 2 sin u sin v k
Step 2. ru = 2 cos u cos v i + 2 cos u sin v j − 2 sin u k and rv = −2 sin u sin v i + 2 sin u cos v j + 0 k
Step 3. ru × rv = 4 sin2 u cos v i + 4 sin2 u sin v j + 4 sin u cos u k
Step 4. F(r) · ru × rv = 8 sin3 u cos2 v
Step 5.
ZZ
S
F · n dS = −
Z
0
= −8
= −8 ·
⊔
⊓
π/2
Z
π/2
0
Z
π/2
F(r) · ru × rv du; dv
3
sin u cos u du
0
2 π
4π
·
= −
3 4
3
!
·
Z
0
π/2
2
cos v dv
!
Example 3.7.11. Find the flux of the vector field F(x, y, z) = hxy, yz, zxi across the surface S which is part of
the paraboloid z = 4 − x2 − y 2 that lies above the square D = {(x, y)| 0 6 x 6 1, 0 6 y 6 1}, and has upward
orientation.
Solution. We use formula (6) (I will deal with the orientation of S later): First, view S as the level surface
H(x, y, z) = x2 + y 2 + z − 4 = 0 so that ∇H(x, y, z) = h2x, 2y, 1i and also that
F · ∇H(x, y, z) = hxy, yz, zxi · h2x, 2y, 1i
= 2x2 y + 2y 2 z + zx
= 2x2 y + 2y 2 (4 − x2 − y 2 ) + x(4 − x2 − y 2 )
= 2x2 y + 8y 2 − 2x2 y 2 − 2y 4 + 4x − x3 − xy 2 .
Since the surface S is given as a function z = g(x, y) which lies above the unit square D (the shadow region of S)
we immediately have that the unit vector p is h0, 0, 1i = k. Thus | ∇H(x, y, z) · k | = | − 1 | = 1. Now the double
integral
ZZ
ZZ
1
F · n dS = ±
F · ∇H(x, y, z) dA
| ∇H(x, y, z) · p |
S
ZDZ
= ±
(2x2 y + 8y 2 − 2x2 y 2 − 2y 4 + 4x − x3 − xy 2 ) dA
D
Z
1
Z
1
(2x2 y + 8y 2 − 2x2 y 2 − 2y 4 + 4x − x3 − xy 2 ) dx dy
0
0
Z 1
2 2
7
1 2
713
2
2
4
dy = ±
= ±
y + 8y − y − 2y +
− y
.
3
3
4
2
180
0
= ±
Now there is the issue of n: We want the upward (very bad wording here!) orientation (i.e., we want S oriented
so that its gradient field ∇H(x, y, z) is outward). To choose the proper n, I choose a convenient point on S, let’s
say P (0, 0, 4). Now calculate ∇H(0, 0, 4) = h0, 0, 1i = k; which points outward away from S. Thus we want n to
∇H(x,y,z)
2y, 1i
√h2x,
point in the same direction as ∇H(x, y, z). This tells us that the required n = + k ∇H(x,y,z)
and
k =
2
2
4x + 4y + 1
so finally we have that the flux of F across S is 713/180. ⊔
⊓
Exercise 3.7.8. Calculate the outward Flux of the vector field p
F(x, y, z) = −x i − y j + z 2 k across a smooth
parametrized surface S where S is the portion of the cone z =
x2 + y 2 between the planes z = 1 and z = 2.
The unit normal is away from the z-axis. Also, setup the integral of the non parametrized version of this problem
(But Do Not Evaluate!).
STOKE’S THEOREM
Stoke’s Theorem says that, under conditions normally met in practice, the circulation of a vector field around the
boundary of an oriented surface in space in the direction counterclockwise with respect to the surface’s unit normal
vector field n equals the integral of the normal component of the circulation density of the field over the surface.
Theorem 3.7.6. [Stoke’s Theorem] Let S be an orientable surface (i.e., S is a two-sided surface) with a
continuously varying unit normal vector field n = n(x, y, z). Let the boundary of S (which we denote by ∂S) be a
piecewise smooth, simple closed curve, oriented consistently with n (i.e., γ = ∂S is a positively oriented piecewise
smooth simple closed curve). Suppose F(x, y, z) = P (x, y, z); i + Q(x, y, z) j + R(x, y, z) k is a vector field with
P (x, y, z), Q(x, y, z), and R(x, y, z) having continuous first-order partial derivatives on S and its boundary curve
∂S. If T denotes the unit tangent vector to γ = ∂S, then the Circulation of F around the boundary curve
γ = ∂S is given by the surface integral
circulation of F around γ =
ZZ
curl F · dS =
S
i
∂
Recall that curl F = ∇ × F = ∂x
P
j
∂
∂y
Q
ZZ
curl F · n dS =
S
k ∂ ∂R
∂z =
∂y −
R
∂Q
∂z
i+
∂P
∂z
−
∂R
∂x
I
γ
F · T ds =
j+
∂Q
∂x
−
I
∂P
∂y
γ
F · d~γ
k.
Example 3.7.12. Calculate the work done by the “force field” given by
F(x, y, z) = hxx + z 2 , y y + x2 , z z + y 2 i
when a particle moves under its influence around the edge of the part of the sphere x2 + y 2 + z 2 = 32 that lies in
the first octant, in a counterclockwise direction as viewed from above.
Solution.
R Observe that we are being
R asked to calculate a work integral in space. that is we must calculate a line
integral ∂S F · d∂S or if you prefer γ F · d~γ where γ = ∂S. Doing this calculation directly might not be a wise
idea (but you can certainly try!). So we will use Stoke’s Theorem (equivalently Green’s Theorem in Space). So we
calculate curl F as
i
∂
curl F(x, y, z) = ∇ × F(x, y, z) = ∂x
x
x + z2
j
∂
∂y
y y + x2
∂
∂z
= h2y, 2z, 2xi or 2hy, z, xi.
z
2
z +y k
Observe that curl F 6= 0 so that F is a non-conservative force field. Next, we must decide on whether we want to
parametrize S or not. I choose to parametrize S as follows:
Parametrized S := r(u, v) = h3 sin u cos v, 3 sin u sin v, 3 cos ui; 0 6 u 6 π/2, 0 6 v 6 π/2;
which naturally gives us the outer unit normal and the positively oriented boundary curve ∂S or γNow ru =
h3 cos u cos v, 3 cos u sin v, −3 sin ui and rv = h−3 sin u sin v, 3 sin u cos v, 0i and so
ru × rv = 9 hsin2 u cos v, sin2 u sin v, sin u cos ui
and curl F(r(u, v)) = 6 hsin u sin v, cos u, sin u cos vi. Now we have that
curl F(r(u, v)) · ru × rv = 54 (sin3 u sin v cos v + sin2 u cos u sin v + sin2 u cos u cos v).
Therefore,
Work done by F =
I
γ
=
ZZ
F · d~γ =
I
γ
F · T ds =
ZZ
curl F · dS
S
curl F(r(u, v)) · ru × rv du dv
D
Z
π/2
Z
π/2
(sin3 u sin v cos v + sin2 u cos u(sin v + cos v)) du dv
0
0
! Z
!
! Z
Z π/2
Z π/2
π/2
3
2
= 54
sin u du
sin v cos v dv + 54
sin u cos u du
= 54
0
0
0
0
π/2
sin v + cos v dv
!
u=π/2
1
1
1
− cos u + cos3 u
+ 54 · (1 + 1)
2
3
3
u=0
1
1
1
= 54 ·
1−
+ 54 · (1 + 1)
2
3
3
= 54 ·
= 54 work-energy units
⊔
⊓
H
Exercise 3.7.9. Use Stoke’s theorem to evaluate γ F · T ds, where F = 2z i + (8x − 3y) j + (3x + y) k where γ
is the triangular positively oriented curve with vertices at (0, 0, 2), (0, 1, 0), and (1, 0, 0). Include a sketch of S and
its boundary curve in your work.
DIVERGENCE THEOREM AND A UNIFIED THEORY
The divergence form of Green’s Theorem in the plane states that the net outward flux of a vector field across a
simple closed curve can be calculated by integrating the divergence of the field over the region enclosed by the curve.
The corresponding theorem in three dimensions, called the Divergence Theorem, states that the net outward flux of
a vector field across a closed surface in space can be calculated by integrating the divergence of the field over the
region enclosed by the surface.
Divergence Theorem
The Divergence Theorem says that: Under suitable conditions, the outward flux of a vector field across a closed
surface (oriented outward) equals the triple integral of the divergence of the field over the solid region enclosed by
the surface.
Theorem 3.7.7. [Gauss’s Divergence Theorem] Suppose F(x, y, z) = P (x, y, z); i + Q(x, y, z) j + R(x, y, z) k
is a vector field with P , Q, and R having continuous first-order partial derivatives on a solid E with boundary surface
S (E is the solid region in space enclosed by S). If n denotes the outward unit normal to the surface S = ∂E , then
Outward flux of F across closed surface S =
ZZ
F · dS =
S
ZZ
F · n dS =
S
ZZZ
div F dV
E
Example 3.7.13. [Verifying the divergence theorem] Evaluate both sides of the equation in the Divergence
Theorem for the field F = hx, y, zi over the sphere x2 + y 2 + z 2 = R2 .
RR
Solution. Calculating Outward flux using the surface integral
F · n dS:
S
2 hx, y, zi
1
=
hx, y, zi
The outward unit normal field is n = p
2
2
2
R
4 (x + y + z )
and so
F · n dS =
x2 + y 2 + z 2
R2
dS =
dS = R dS
R
R
since x2 + y 2 + z 2 = R2 on the surface. Therefore,
ZZ
F · n dS =
S
Next, we calculate
R dS = R
S
RRR
E
ZZ
dS = R(4πR2 ) = 4πR3 = 3 (Volume of E ).
S
div F dV . Since F = hx, y, zi, then the divergence of F is
div F =
Therefore,
ZZ
ZZZ
E
∂
∂
∂
(x) +
(y) +
(z) = 1 + 1 + 1 = 3.
∂x
∂y
∂z
div F dV = 3
ZZZ
dV = 3 (volume of E ) = 3
E
4 3
πR
3
⊔
⊓
Problem on This theorem will be on Final Exam! So study it well
Exercise 3.7.10. Use the Divergence Theorem to Evaluate the surface integral
ZZ
S
(2x + 2y + z 2 )dS
= 4πR3
where S is the sphere x2 + y 2 + z 2 = a2 , a > 0.
MISCELLANEOUS EXERCISES
Problem 1. Let S denote the portion of the parabolic cylinder z = 1 − x2 ; 0 6 y 6 2 which lies in the first
octant. Let γ be piecewise smooth curve denoting the boundary of S, which is positively oriented when viewed
from above. Consider now the continuous vector field F(x, y, z) = h1, 0, y 2 i defined every where. Calculate the
circulation of F around γ two ways: (1) Directly as a line integral and (2) Directly as flux of F through S.
Problem 2. Let E denote the solid enclosed by the circular paraboloid z = x2 + y 2 and the circular disk x2 + y 2 = 4
at z = 4. Let F(x, y, z) = hx, y, 1i be a continuous vector field and the surface S is the boundary of E . Verify the
Divergence Theorem.
2
2
Problem 3. Let D denote the
H plane region bounded x + y − y = 0, y = 0 and y = −x. Let γ denote
the boundary of D. Calculate γ P (x, y) dx + Q(x, y) dy where P (x, y) = −y and Q(x, y) = x by verifying the
counterclockwise circulation form of Green’s theorem.
Problem√4. Calculate the area of the portion of the sphere x2 + y 2 + z 2 = 22 that lies between the planes z = 0
and z = 2.
Problem 5. Consider the continuous vector field F(x, y, z) = h2z, ez , 2x + yez i.
(a) Does the vector field have a potential? (Explain by finding it or say why it does not).
R
(b) Evaluate γ F · d~γ where γ is the space curve given by y = 1 − x2 , z = 0 with 0 6 x 6 1 and oriented so
that x is increasing.
Problem 6. Let E be the solid contained between the surfaces z = 4 − x2 − y 2 and z = 0. Let S denote the
boundary of E positively oriented (i.e., oriented with the outward
unit normal field). Consider the continuous vector
RR
field F(x, y, z) = hx, y, 1i. Evaluate the surface integral
F · dS as a double and Triple integral.
S
Problem 7. Let E be the solid inside the cylinder x2 + y 2 = 1 bounded by the RR
planes z = 0 and z = 2. Let
F(x, y, z) = hzy, zx, y 2 + z 2 i. Setup; but do not Evaluate the surface integral
F · dS both as a double and
a Triple integral. Clearly state the limits of integration for both integrals.
S
Problem 8. Let γ be the curve of intersection of the plane z = 3x − 7 and the right circular cylinder x2 + y 2 = 1
and has the clockwise orientationH as viewed from above. Let F be the continuous vector field given by F(x, y, z) =
h4z − 1, 2x, 5y + 1i. Calculate γ F · d~γ (a) directly as a line integral (impossible for your level of understanding
thus far) and (b) calculate using (Stoke’s Theorem) (Of course, I could be wrong about this! At any rate, I will just
assume that the above wording is what was given)
Solution. After many hours of thought about this problem, I finally came to the conclusion that the difficulty
with the problem lies totally in its wording. The inventor of the problem did not intend for us to do both the line
integral directly (without the use of Stoke’s Theorem) and the surface integral directly (even though the problem
stated it that way). This was precise the point I tried to make in class; it is worth it to take a minute or so to analyze
some of the questions, you will find that they contain “logical flaws” in the sense that they asked you to do the same
thing twice! even though the intent is for you to do two different things altogether (this is usually the tester’s fault
not yours). Anyway, real intent was that you do just of the two integrals (one of them is easy and the other is rather
difficult (but not impossible!)
as you would have discovered, so take your pick of only one!). The easy integral is
RR
the surface integral (∇ × F · n) dS (from Stoke’s Theorem). The surface S is the plane z = f (x, y) = 3x − 7 or
S
H(x, y, z) = −3x + z + 7 = 0 (Here we view S as a level surface of H). So let’s calculate this surface integral by
the numbers:
curl F(x, y, z) = ∇ × F(x, y, z) = i
j
k
∂
∂x
∂
∂y
∂
∂z
4z − 1
2x
5y + 1
= h5, 4, 2i or 5i + 4j + 2k.
Next, since we are traversing γ clockwise as viewed from above, it means that the surface S is to our right as we
∇H(x,y,z)
√1
travel around γ, in other words, we negatively orient the surface and so n = k ∇H(x,y,z)
h−3, 0, 1i.
k =
10
However our orientation is negative is we use −n as our unit normal. Thus, by our formula for surface integrals we
have
I
ZZ
ZZ
1
13 √
F · T ds =
(curl F(x, y, z) · −n) dS = −
−√
10 dA = 13π
10
γ
D | ∇H(x, y, z) · k |
D
where D is the shadow region of S (in our case, this is the unit disk centered at the origin) and p = k is a unit
normal to the shadow region.
R
Here’s the correct wording: Evaluate γ F · d~γ , for the vector field F(x, y, z) = h4z − 1, 2x, 5y + 1i, where
the curve γ is the intersection of the circular cylinder x2 + y 2 = 1 and the plane z = 3x − 7, oriented so that it is
traversed clockwise when viewed from high up on the positive z-axis. ⊔
⊓
NEATLY WRITE UP THE EXERCISES AND TURN IT IN FOR A GRADE NO LATER THAN
DECEMBER 12th, 2008!!
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
Antony L Foster
Name (Printed)_______________________________________________
Antony Foster
Name (Signed)_______________________________________________
MATH 39200 C FINAL EXAMINATION
December 22, 2008
1
10
2
10
3
10
4
10
5
10
6
10
7
10
8, 9 OR 10
10
9, 10 OR 11
10
10, 11 OR 12
10
TOTAL
10
POSSIBLE
100 (120)
Instructions: Show all work. Calculators and other electronic devices must be out of sight, turned off, and not used.
Answer 5 questions from part I and 2 questions from part II.
PART I: Answer 5 complete questions from this part. (14 points each)
1 4
1
1. (a) Find the inverse of the matrix 𝐴𝐴 = �2 7 3�.
1 7
−1
(b) Use the matrix 𝐴𝐴
5
𝑥𝑥 + 4𝑦𝑦 + 𝑧𝑧 = 1
2𝑥𝑥
that you found in (a) to solve the system � + 7𝑦𝑦 + 3𝑧𝑧 = −2
𝑥𝑥 + 7𝑦𝑦 + 5𝑧𝑧 = −1
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
No credit for any other method!
Show your work on this page and on the page to the left.
MY ANSWER:
1 4 1 1 0 0
(a) (𝑨𝑨|𝑰𝑰𝟑𝟑 ) = �2 7 3 0 1 0� reduce to reduced row echelon form using just a few elementary row
1 7 5 0 0 1
operations to get. I did not state my operations here!
⎛1 0
⎜0 1
0 0
⎝
0
0
1
−𝟐𝟐
𝟏𝟏
−𝟏𝟏
𝟏𝟏𝟏𝟏
𝟕𝟕
𝟒𝟒
− 𝟕𝟕
𝟑𝟑
𝟕𝟕
𝟓𝟓
− 𝟕𝟕
𝟏𝟏 ⎞
from which we have the desire inverse of 𝐴𝐴 as:
𝟕𝟕 ⎟
𝟏𝟏
𝟕𝟕 ⎠
−𝟐𝟐
⎛
𝑨𝑨−𝟏𝟏 = ⎜ 𝟏𝟏
⎝
−𝟏𝟏
𝟏𝟏𝟏𝟏
𝟕𝟕
𝟒𝟒
− 𝟕𝟕
𝟑𝟑
𝟕𝟕
𝟓𝟓
− 𝟕𝟕
𝟏𝟏 ⎞
.
𝟕𝟕 ⎟
𝟏𝟏
𝟕𝟕 ⎠
(b) Since the coefficient matrix 𝐴𝐴 is nonsingular then the solution to the matrix equation 𝐴𝐴𝒙𝒙 = 𝐵𝐵 is
therefore
−2
13
𝑥𝑥
⎛
4
𝒙𝒙 = �𝑦𝑦� = 𝐴𝐴−1 𝐵𝐵 = ⎜ 1 − 7
𝑧𝑧
3
−1
7
⎝
7
5
−7
1
−5
�
�
=
�
−2
2 �.
7⎟
−1
−2
1
7⎠
1⎞
Therefore, the unique solution to the given linear non-homogeneous system 𝐴𝐴𝒙𝒙 = 𝐵𝐵 is:
𝒙𝒙 = −𝟓𝟓,
𝒚𝒚 = 𝟐𝟐,
𝒛𝒛 = −𝟐𝟐
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
2
7 −1
(a) Find all eigenvalues and eigenvectors of the matrix �
�
3
3
𝑦𝑦1′ (𝑡𝑡) = 7𝑦𝑦1 (𝑡𝑡) − 𝑦𝑦2 (𝑡𝑡)
(b) Use your answer to (a) to solve �
for 𝑦𝑦1 (𝑡𝑡) and 𝑦𝑦2 (𝑡𝑡) subject to initial conditions
′ (𝑡𝑡)
𝑦𝑦2
= 3𝑦𝑦1 (𝑡𝑡) + 3𝑦𝑦2 (𝑡𝑡)
𝑦𝑦1 (0) = 2 and 𝑦𝑦2 (0) = 1.
MY ANSWER
(a) We seek real values 𝜆𝜆 and non zero vectors 𝒙𝒙 such that 𝐴𝐴𝒙𝒙 = 𝜆𝜆𝒙𝒙 holds.
𝜆𝜆 − 7
1
𝜆𝜆 − 7
1
𝜆𝜆𝜆𝜆 − 𝐴𝐴 = �
� and so 𝑝𝑝(𝜆𝜆) = �
� = 𝜆𝜆2 − 10𝜆𝜆 + 24 = (𝜆𝜆 − 6)(𝜆𝜆 − 4) = 0.
−3 𝜆𝜆 − 3
−3 𝜆𝜆 − 3
Therefore the eigenvalues of the matrix 𝐴𝐴 are 𝝀𝝀 = 𝟔𝟔 𝑎𝑎𝑎𝑎𝑎𝑎 𝝀𝝀 = 𝟒𝟒.
For the eigenvalue 𝜆𝜆 = 6, we solve the homogeneous linear system (6𝐼𝐼 − 𝐴𝐴)𝒙𝒙 = 0 for 𝒙𝒙 which tells us
𝟏𝟏
that the vector � � is the eigenvector corresponding to the eigenvalue 𝜆𝜆 = 6.
𝟏𝟏
For the eigenvalue 𝜆𝜆 = 4, we solve the homogeneous linear system (4𝐼𝐼 − 𝐴𝐴)𝒙𝒙 = 0 for 𝒙𝒙 which tells us
𝟏𝟏
that � � is the eigenvector corresponding to the eigenvalue 𝜆𝜆 = 4.
𝟑𝟑
(b) The given system of first-order linear differential equations has general solution (in matrix form)
𝒑𝒑𝟏𝟏𝟏𝟏
𝒑𝒑𝟏𝟏𝟏𝟏
𝒚𝒚 (𝒕𝒕)
𝟏𝟏
𝟏𝟏
𝒀𝒀(𝒕𝒕) = � 𝟏𝟏 � = 𝒄𝒄𝟏𝟏 �𝒑𝒑 � 𝒆𝒆𝝀𝝀𝟏𝟏 𝒕𝒕 + 𝒄𝒄𝟐𝟐 �𝒑𝒑 � 𝒆𝒆𝝀𝝀𝟐𝟐 𝒕𝒕 = 𝒄𝒄𝟏𝟏 � � 𝒆𝒆𝟔𝟔𝟔𝟔 + 𝒄𝒄𝟐𝟐 � � 𝒆𝒆𝟒𝟒𝟒𝟒
𝒚𝒚𝟐𝟐 (𝒕𝒕)
𝟑𝟑
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
𝟏𝟏
where the constants 𝑐𝑐1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐2 are to be determined. Now imposing the initial conditions
𝑦𝑦1 (0) = 2 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦2 (0) = 1 we get that
𝑐𝑐1 + 𝑐𝑐2 = 2 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐1 + 3𝑐𝑐2 = 1 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎ℎ𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠)
from which it follows that
5
1
𝑐𝑐1 = 2 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐2 = − 2
and so the solution to the initial value system is
5 1
1 1
𝑦𝑦 (𝑡𝑡)
𝑌𝑌(𝑡𝑡) = � 1 � = � � 𝑒𝑒 6𝑡𝑡 − � � 𝑒𝑒 4𝑡𝑡
𝑦𝑦2 (𝑡𝑡)
2 1
2 3
𝒚𝒚𝟏𝟏 (𝒕𝒕) =
𝟓𝟓 𝟔𝟔𝟔𝟔 𝟏𝟏 𝟒𝟒𝟒𝟒
𝟓𝟓
𝟑𝟑
𝒆𝒆 − 𝒆𝒆 𝒂𝒂𝒂𝒂𝒂𝒂 𝒚𝒚𝟐𝟐 (𝒕𝒕) = 𝒆𝒆𝟔𝟔𝟔𝟔 − 𝒆𝒆𝟒𝟒𝟒𝟒
𝟐𝟐
𝟐𝟐
𝟐𝟐
𝟐𝟐
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
1 −4 9 −7
3. (a) Find the rank of the matrix 𝐴𝐴 = �−1
2 −4 1 �
5 −6 10 7
𝑥𝑥 − 4𝑦𝑦 + 9𝑧𝑧 − 7𝑤𝑤 = 0
−𝑥𝑥
+ 2𝑦𝑦 − 4𝑧𝑧 + 𝑤𝑤 = 0 . Indicate clearly all row operations that you use.
(b) Solve the linear system �
5𝑥𝑥 − 6𝑦𝑦 + 10𝑧𝑧 + 7𝑤𝑤 = 0
3
(c) Find the determinant of the matrix � 0
1
1000
4
1
2
2000
0
2
3
3000
6
1
�
4
4001
MY ANSWER
(a) Bring the matrix to reduced-row echelon form and then count the number of rows with leading one’s.
1
�−1
5
−4 9 −7
1
2 −4 1 � 𝑟𝑟1 + 𝑟𝑟2 → � 𝟎𝟎
−6 10 7
5
1
7𝑟𝑟2 + 𝑟𝑟3 → � 0
𝟎𝟎
−4
−2
𝟎𝟎
−4
−𝟐𝟐
−6
9 −7
1
𝟓𝟓 −𝟔𝟔� − 5𝑟𝑟1 + 𝑟𝑟3 → � 0
10 7
𝟎𝟎
9 −7
𝟏𝟏
5 −6� − 2𝑟𝑟2 + 𝑟𝑟1 → � 0
𝟎𝟎 𝟎𝟎
0
𝟎𝟎
−2
0
−4
−2
𝟏𝟏𝟏𝟏
𝟏𝟏
−𝟏𝟏 𝟓𝟓
1
5 −6� − 𝑟𝑟2 → � 0
2
0 0
0
9 −7
5 −6 �
−𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒
0
𝟏𝟏
0
−1
5
−𝟓𝟓/𝟐𝟐 𝟑𝟑 �
0
0
From the last matrix (which is in reduced row echelon form) we can see that the Rank of the original matrix 𝐴𝐴 is
2, we have two pivots. Furthermore, the row of zeroes tells us that the homogeneous linear system has
infinitely many solutions. So let’s write them down.
(b) The above information is useful in solving the given homogeneous linear system since the matrix in part
(a) is the coefficient matrix of the system. We would use the same elementary row operations to achieve
𝒙𝒙 − 𝒛𝒛 + 𝟓𝟓𝟓𝟓 = 𝟎𝟎
𝟓𝟓
the equivalent system �𝒚𝒚 − 𝟐𝟐 𝒛𝒛 + 𝟑𝟑𝟑𝟑 = 𝟎𝟎 from which we let 𝑧𝑧 = 𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎 𝑤𝑤 = 𝑡𝑡 and then determine that the
𝒛𝒛 = 𝒛𝒛
𝒘𝒘 = 𝒘𝒘
system has infinitely many solutions which we write in matrix form as
𝟏𝟏
𝒙𝒙
−𝟓𝟓
𝟓𝟓
𝒚𝒚
−𝟑𝟑
� 𝒛𝒛 � = � 𝟐𝟐 � 𝒔𝒔 + � 𝟎𝟎 � 𝒕𝒕 where 𝒔𝒔, 𝒕𝒕 ∈ ℝ.
𝟏𝟏
𝒘𝒘
𝟏𝟏
𝟎𝟎
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
By the way the solutions of the system 𝐴𝐴𝒙𝒙 = 𝟎𝟎 which we express as linear combination of two
𝟏𝟏
−𝟓𝟓
𝟓𝟓
−𝟑𝟑
independent vectors 𝒗𝒗1 = � 𝟐𝟐 � 𝑎𝑎𝑛𝑛𝑛𝑛 𝒗𝒗2 = � 𝟎𝟎 � spans a fundamental 2-dimensional subspace of ℝ4
𝟏𝟏
𝟏𝟏
𝟎𝟎
called the Nullspace of the matrix 𝐴𝐴. An important theorem in linear algebra says that the rank of 𝑨𝑨 plus
the nullity of 𝑨𝑨 equals the dimension of 𝑨𝑨. The rank of 𝑨𝑨 is the dimension of the row-space of A, this is
the number of leading one’s in the row echelon form of 𝐴𝐴. In our case, the rank of 𝐴𝐴 = 2. The null-space
of 𝑨𝑨 is the solution set to the homogeneous system 𝐴𝐴𝒙𝒙 = 𝟎𝟎. The dimension of the nullspace is called the
nullity of 𝑨𝑨. In our case, the nullity of 𝐴𝐴 = 2.
(c) Use row reduction method to bring the matrix to upper-triangular from (not necessarily row echelon
form ) as follows (recalling that only 2 of the 3 elementary row operations have an effect on the
determinant):
𝟑𝟑
𝟒𝟒
𝟏𝟏
𝟎𝟎
� 𝟏𝟏
𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
𝟏𝟏
− � 𝟎𝟎
𝟎𝟎
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏 𝟐𝟐
𝟎𝟎 𝟏𝟏
� 𝟎𝟎
𝟎𝟎
𝟎𝟎 𝟎𝟎
𝟏𝟏
𝟐𝟐
𝟎𝟎
𝟔𝟔
𝟐𝟐
𝟏𝟏
𝟏𝟏
� = − � 𝟎𝟎
𝟑𝟑
𝟑𝟑
𝟒𝟒
𝟒𝟒
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒
𝟐𝟐
𝟑𝟑
𝟒𝟒
𝟏𝟏
𝟐𝟐
𝟏𝟏
�= −
−𝟐𝟐
−𝟗𝟗
−𝟔𝟔
𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒
𝟑𝟑 𝟒𝟒
𝟐𝟐 𝟏𝟏
� = (𝟏𝟏)(𝟏𝟏)(𝟓𝟓)(𝟏𝟏) = 𝟓𝟓
𝟓𝟓 𝟒𝟒
𝟎𝟎 𝟏𝟏
3
det � 0
1
1000
4
1
2
2000
0
2
3
3000
𝟏𝟏
�𝟎𝟎
𝟎𝟎
𝟎𝟎
𝟐𝟐
𝟏𝟏
−𝟐𝟐
𝟎𝟎
𝟑𝟑
𝟒𝟒
𝟐𝟐
𝟏𝟏
�=
𝟎𝟎
𝟔𝟔
𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒
𝟑𝟑
𝟐𝟐
−𝟗𝟗
𝟎𝟎
𝟏𝟏
𝟒𝟒
𝟏𝟏
𝟎𝟎
� = − � 𝟎𝟎
−𝟔𝟔
𝟎𝟎
𝟏𝟏
Thus we get that
𝟏𝟏
6
1
𝟎𝟎
� = det � 𝟎𝟎
4
𝟎𝟎
4001
𝟐𝟐
𝟏𝟏
𝟎𝟎
𝟎𝟎
𝟑𝟑
𝟐𝟐
𝟓𝟓
𝟎𝟎
𝟐𝟐 𝟑𝟑
𝟒𝟒
𝟏𝟏 𝟐𝟐
𝟏𝟏
�=
𝟎𝟎 −𝟓𝟓 −𝟒𝟒
𝟎𝟎 𝟎𝟎
𝟏𝟏
𝟒𝟒
𝟏𝟏
�=5
𝟒𝟒
𝟏𝟏
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
4. (a) Find the length of the parametrized curve with position vector
𝜸𝜸(𝑡𝑡) = 〈cos 𝑡𝑡, cos t, −√2 sin t〉; 0 ≤ t ≤ 2π.
(b) Let 𝜸𝜸 be the intersection curve of the surfaces 𝑧𝑧 = 𝑥𝑥 2 + 𝑦𝑦 2 and 𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 = 2, oriented
counterclockwise as seen from above. Find. ∫𝜸𝜸 𝑦𝑦 𝑑𝑑𝑑𝑑 − 𝑥𝑥 𝑑𝑑𝑑𝑑 + 𝑥𝑥 2 𝑦𝑦 3 𝑧𝑧 5 𝑑𝑑𝑑𝑑
MY ANSWER
(a) Let 𝐿𝐿 denote the length of the curve 𝛾𝛾 then we calculate 𝐿𝐿 as follows:
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
𝑳𝑳 = � ‖𝜸𝜸′ (𝒕𝒕) ‖ 𝒅𝒅𝒅𝒅 = � √𝟐𝟐 𝒅𝒅𝒅𝒅 = 𝟐𝟐𝟐𝟐√𝟐𝟐 𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖.
𝟎𝟎
And the arc-length parameter 𝑠𝑠 =
𝑡𝑡
∫0 ‖𝜸𝜸′ (𝜏𝜏)‖
𝟎𝟎
𝑡𝑡
1
𝑑𝑑𝜏𝜏 = ∫0 √2 𝑑𝑑𝜏𝜏 = √2 𝑡𝑡 or that 𝑡𝑡 = 𝑠𝑠. This means
√2
could parameterize the curve as
1
1
1
𝛾𝛾(𝑠𝑠) = 〈cos � 𝑠𝑠� , sin � 𝑠𝑠� , −√2 sin � 𝑠𝑠�〉 ,
√2
√2
√2
0 ≤ 𝑠𝑠 ≤ 2√2 𝜋𝜋
(b) The curve 𝜸𝜸 of intersection of the two given surfaces is 𝑥𝑥 2 + 𝑦𝑦 2 = 1; 𝑧𝑧 = 1 whose usual
parametrization is 𝜸𝜸(𝑡𝑡) = 〈cos 𝑡𝑡, sin 𝑡𝑡, 1〉; 0 ≤ 𝑡𝑡 ≤ 2𝜋𝜋 (This is a simple closed space curve in the plane
𝑧𝑧 = 1).
Define 𝑭𝑭(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 〈𝑦𝑦, −𝑥𝑥, 𝑥𝑥 2 𝑦𝑦 3 𝑧𝑧 5 〉 as the vector field (though it is not necessary!). We are being
asked to calculate the line integral ∮𝜸𝜸 𝑭𝑭 ⋅ 𝑑𝑑𝜸𝜸 so as usual
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
� 𝑭𝑭 ⋅ 𝒅𝒅𝒅𝒅 = � 𝑭𝑭�𝜸𝜸(𝒕𝒕)� ⋅ 𝜸𝜸′ (𝒕𝒕) 𝒅𝒅𝒅𝒅 = � −(𝐬𝐬𝐬𝐬𝐬𝐬𝟐𝟐 𝒕𝒕 + 𝐜𝐜𝐜𝐜𝐜𝐜𝟐𝟐 𝒕𝒕) 𝒅𝒅𝒅𝒅 = −𝟐𝟐𝟐𝟐
𝜸𝜸
𝟎𝟎
𝟎𝟎
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
5. Let 𝑆𝑆 be the surface 𝑧𝑧 = �𝑥𝑥 2 + 𝑦𝑦 2 ; 𝑧𝑧 ≤ 9, and let 𝑃𝑃 be the point (3,4,5) on 𝑆𝑆.
(a) Find an equation of the tangent plane to the surface 𝑆𝑆 at point 𝑃𝑃.
(b) Find the surface area of 𝑆𝑆.
MY ANSWER
(a) Let 𝐻𝐻(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = �𝑥𝑥 2 + 𝑦𝑦 2 − 𝑧𝑧 = 0 where (𝑥𝑥, 𝑦𝑦) ∈ 𝐷𝐷 = {(𝑥𝑥, 𝑦𝑦): 𝑥𝑥 2 + 𝑦𝑦 2 ≤ 81} (A circular disk of
radius 9).
𝑥𝑥
𝒚𝒚
Now ∇𝐻𝐻(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 〈𝐻𝐻𝑥𝑥 , 𝐻𝐻𝑦𝑦 , 𝐻𝐻𝑧𝑧 〉 = 〈 2 2 , 2 2 , −1〉. The equation of the tangent plane to the given
�𝑥𝑥 +𝑦𝑦
surface 𝑆𝑆 at the point (3,4,5) is
𝟑𝟑
�𝑥𝑥 +𝑦𝑦
𝟒𝟒
𝛁𝛁𝑯𝑯(𝟑𝟑, 𝟒𝟒, 𝟓𝟓) ⋅ 〈𝒙𝒙 − 𝟑𝟑, 𝒚𝒚 − 𝟒𝟒, 𝒛𝒛 − 𝟓𝟓〉 = 𝟓𝟓 (𝒙𝒙 − 𝟑𝟑) + 𝟓𝟓 (𝒚𝒚 − 𝟒𝟒) − (𝒛𝒛 − 𝟓𝟓) = 𝟎𝟎
𝐨𝐨𝐨𝐨
𝟑𝟑𝟑𝟑 + 𝟒𝟒𝟒𝟒 − 𝟓𝟓𝟓𝟓 = 𝟎𝟎
(b) Based on the information in part (a), there is perhaps no need to parametrize the surface.
Calculate the surface Area as follows (From Math 203)
� ‖𝛁𝛁𝐻𝐻(𝑥𝑥, 𝑦𝑦, 𝑧𝑧)‖ 𝒅𝒅𝒅𝒅 = √𝟐𝟐 � 𝒅𝒅𝒅𝒅 = √𝟐𝟐 𝑨𝑨(𝑫𝑫) = 𝟖𝟖𝟖𝟖√𝟐𝟐 𝝅𝝅 𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒔𝒔𝟐𝟐
𝑫𝑫
𝑫𝑫
Just in case you parametrized the circular cone 𝑆𝑆:
𝒓𝒓(𝑢𝑢, 𝑣𝑣) = 〈𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑣𝑣, 𝑢𝑢 sin 𝑣𝑣, 𝑢𝑢〉;
0 ≤ 𝑢𝑢 ≤ 9, 0 ≤ 𝑣𝑣 ≤ 2𝜋𝜋.
Then 𝒓𝒓𝒖𝒖 (𝑢𝑢, 𝑣𝑣) = 〈cos 𝑣𝑣, sin 𝑣𝑣, 1〉 𝑎𝑎𝑎𝑎𝑎𝑎 𝒓𝒓𝒗𝒗 (𝑢𝑢, 𝑣𝑣) = 〈−𝑢𝑢 sin 𝑣𝑣, 𝑢𝑢 cos 𝑣𝑣, 0〉 𝑎𝑎𝑎𝑎𝑎𝑎 𝒓𝒓𝒖𝒖 × 𝒓𝒓𝒗𝒗 = 〈−𝑢𝑢 cos 𝑣𝑣, −𝑢𝑢 sin 𝑣𝑣, 𝑢𝑢〉
‖𝑟𝑟𝑢𝑢 × 𝑟𝑟𝑣𝑣 ‖ = √2 𝑢𝑢. Now surface area is
� 𝒅𝒅𝒅𝒅 = � ‖𝒓𝒓𝒖𝒖 × 𝒓𝒓𝒗𝒗 ‖ 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 = �
𝑺𝑺
𝑫𝑫
𝟐𝟐𝟐𝟐
𝟎𝟎
𝟗𝟗
� ‖𝒓𝒓𝒖𝒖 × 𝒓𝒓𝒗𝒗 ‖𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 = √𝟐𝟐 �
𝟎𝟎
𝟐𝟐𝟐𝟐
𝟎𝟎
𝟗𝟗
� 𝒖𝒖 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 = 𝟖𝟖𝟖𝟖√𝟐𝟐 𝝅𝝅 𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒔𝒔𝟐𝟐 .
𝟎𝟎
There is a nice formula for the surface area of a right circular cone of radius 𝑅𝑅 and height 𝐻𝐻 is given as:
𝜋𝜋𝜋𝜋√𝑅𝑅 2 + 𝐻𝐻 2 .
This should agree with the surface area of the right circular cone given above whose radius and height is the
same number which is 9 = 𝑅𝑅 = 𝐻𝐻.
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
6. (a) Let 𝑭𝑭 be the vector field defined by 𝑭𝑭(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 〈2𝑥𝑥𝑥𝑥, 3𝑦𝑦, 𝑥𝑥 2 〉. Explain why the value of the line
integral ∫𝛾𝛾 𝑭𝑭 ⋅ 𝑑𝑑𝜸𝜸 is the same for all curves 𝜸𝜸 from 𝑃𝑃(−1,1,0) to 𝑄𝑄(1,2,2) and find the value of this
integral.
b) Let 𝑆𝑆 be the part of the surface 𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 = 4 with 𝑥𝑥 ≤ 0 and 𝑧𝑧 ≥ 0. Evaluate the surface
integral ∬𝑆𝑆 𝑧𝑧 3 𝑑𝑑𝑑𝑑
MY ANSWER
(a) If the line integral ∫𝜸𝜸 𝑭𝑭 ⋅ 𝑑𝑑𝜸𝜸 has the same value for all paths or curves from 𝑃𝑃(−1,1,0) 𝑡𝑡𝑡𝑡 𝑄𝑄(1,2,2), then it
must be that it is independent of path and so it would mean that the given vector field is the gradient field
of a potential function 𝝋𝝋(𝒙𝒙, 𝒚𝒚, 𝒛𝒛) = 𝑪𝑪, 𝒊𝒊. 𝒆𝒆. , 𝛁𝛁𝝋𝝋(𝒙𝒙, 𝒚𝒚, 𝒛𝒛) = 𝑭𝑭 which in turn means that 𝛁𝛁 × 𝛁𝛁𝜑𝜑(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) =
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑭𝑭 = 𝟎𝟎. But this needs to be checked in order to receive credit as shown below.
𝒊𝒊
𝒋𝒋
𝒌𝒌
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 F = ∇ × 𝑭𝑭 = �
Now we find the potential for 𝑭𝑭 as follows:
𝜕𝜕
𝜕𝜕𝜕𝜕
2𝑥𝑥𝑥𝑥
3𝑦𝑦
𝜑𝜑(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = ∫ (2𝑥𝑥𝑥𝑥)𝑑𝑑𝑑𝑑 = 𝑥𝑥 2 𝑧𝑧 + ℎ(𝑦𝑦, 𝑧𝑧) and
3
immediately that ℎ(𝑦𝑦, 𝑧𝑧) = 2 𝑦𝑦 2 + 𝑔𝑔(𝑧𝑧) and so
𝜕𝜕𝜕𝜕 (𝑥𝑥,𝑦𝑦,𝑧𝑧)
𝜕𝜕𝜕𝜕
𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕
𝜕𝜕𝜕𝜕
2
𝑥𝑥
� = 〈0,0,0〉 = 𝟎𝟎
𝜕𝜕𝜕𝜕 (𝑥𝑥,𝑦𝑦,𝑧𝑧)
𝜕𝜕𝜕𝜕
=
𝜕𝜕ℎ(𝑦𝑦 ,𝑧𝑧)
𝜕𝜕𝜕𝜕
= 3𝑦𝑦. This tells us
3
𝜑𝜑(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑥𝑥 2 𝑧𝑧 + 𝑦𝑦 2 + 𝑔𝑔(𝑧𝑧)
2
= 𝑥𝑥 2 + 𝑔𝑔′ (𝑧𝑧) = 𝑥𝑥 2 which implies that 𝑔𝑔(𝑧𝑧) 𝑖𝑖𝑖𝑖 𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. Finally, we have
𝟑𝟑
𝑐𝑐 = 𝝋𝝋(𝒙𝒙, 𝒚𝒚, 𝒛𝒛) = 𝒙𝒙𝟐𝟐 𝒛𝒛 + 𝟐𝟐 𝒚𝒚𝟐𝟐 as the potential function. Now we use the potential function to calculate the line
integral ∫𝜸𝜸 𝑭𝑭 ⋅ 𝑑𝑑𝜸𝜸 as follows (That was the point of part (a), to spare you the laboring task of calculating
the line integral directly!)
� 𝑭𝑭 ⋅ 𝒅𝒅𝒅𝒅 = 𝑸𝑸(𝟏𝟏, 𝟐𝟐, 𝟐𝟐) − 𝑸𝑸(−𝟏𝟏, 𝟏𝟏, 𝟎𝟎) = 𝟖𝟖 −
𝜸𝜸
𝟑𝟑 𝟏𝟏𝟏𝟏
=
𝟐𝟐
𝟐𝟐
(b) Perhaps it is best to parameterize the given portion of the sphere 𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 = 4; 𝑥𝑥 ≤ 0, 𝑧𝑧 ≥ 0 as:
So that
𝜋𝜋
𝒓𝒓(𝑢𝑢, 𝑣𝑣) = 〈2 sin 𝑢𝑢 cos 𝑣𝑣, 2 sin 𝑢𝑢 sin 𝑣𝑣, 2 cos 𝑢𝑢〉; 0 ≤ 𝑢𝑢 ≤ 2 ,
𝜋𝜋
2
≤ 𝑣𝑣 ≤
3𝜋𝜋
2
.
𝒓𝒓𝑢𝑢 = 〈2 cos 𝑢𝑢 cos 𝑣𝑣, 2 cos 𝑢𝑢 sin 𝑣𝑣, −2 sin 𝑢𝑢〉 𝑎𝑎𝑎𝑎𝑎𝑎 𝒓𝒓𝑣𝑣 = 〈−2 sin 𝑢𝑢 sin 𝑣𝑣, 2 sin 𝑢𝑢 cos 𝑣𝑣, 0〉
and 𝒓𝒓𝑢𝑢 × 𝒓𝒓𝑣𝑣 = 〈4 cos 𝑣𝑣 sin2 𝑢𝑢, 4 sin 𝑣𝑣 sin2 𝑢𝑢, 4 sin 𝑢𝑢 cos 𝑢𝑢〉 𝑎𝑎𝑎𝑎𝑎𝑎 ‖𝑟𝑟𝑢𝑢 × 𝑟𝑟𝑣𝑣 ‖ = 4 sin 𝑢𝑢
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
Now let 𝐺𝐺(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑧𝑧 3 so that 𝐺𝐺�𝒓𝒓(𝑢𝑢, 𝑣𝑣)� = 𝐺𝐺(2 sin 𝑢𝑢 cos 𝑣𝑣, 2 sin 𝑢𝑢 sin 𝑣𝑣, 2 cos 𝑢𝑢) = 8 cos 3 𝑢𝑢 and then
𝐺𝐺�𝒓𝒓(𝑢𝑢, 𝑣𝑣)� ‖𝑟𝑟𝑢𝑢 × 𝑟𝑟𝑣𝑣 ‖ = −32 cos3 𝑢𝑢 (− sin 𝑢𝑢) . Hence we can evaluate the surface integral over a
scalar field in the usual way:
𝟑𝟑
� 𝑮𝑮(𝒙𝒙, 𝒚𝒚, 𝒛𝒛)𝒅𝒅𝒅𝒅 = � 𝒛𝒛 𝒅𝒅𝒅𝒅 = � 𝑮𝑮�𝒓𝒓(𝒖𝒖, 𝒗𝒗)�‖𝒓𝒓𝒖𝒖 × 𝒓𝒓𝒗𝒗 ‖𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 = 𝟑𝟑𝟑𝟑 �
𝑺𝑺
𝑺𝑺
𝑹𝑹
𝟑𝟑𝟑𝟑
𝟐𝟐
𝝅𝝅
𝟐𝟐
𝟏𝟏
� 𝒘𝒘𝟑𝟑 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 = 𝟖𝟖𝟖𝟖
𝟎𝟎
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
7. Let 𝑅𝑅 be the region in the 𝑥𝑥𝑥𝑥-plane contained between the curves 𝑦𝑦 = 𝑥𝑥 + 2 and 𝑦𝑦 = 𝑥𝑥 2 − 2𝑥𝑥 + 4. Let 𝜸𝜸 be
the boundary curve of 𝑅𝑅, oriented counterclockwise. Find ∫𝛾𝛾 𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑 − (𝑥𝑥 + 𝑦𝑦) 𝑑𝑑𝑑𝑑
(a) directly, as a line integral AND
(b) as a double integral, by using Green’s Theorem
MY ANSWER
(a) Observe that the boundary curve is a piecewise smooth simple closed curve made up of two smooth
parametrized curves 𝜸𝜸𝟏𝟏 (𝑡𝑡) = 〈𝑡𝑡 + 1, 𝑡𝑡 2 + 3〉; 0 ≤ 𝑡𝑡 ≤ 1 𝑎𝑎𝑎𝑎𝑎𝑎 𝜸𝜸𝟐𝟐 (𝑡𝑡) = 〈−𝑡𝑡 + 2, −𝑡𝑡 + 4〉; 0 ≤ 𝑡𝑡 ≤ 1.
1
� 𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑 − (𝑥𝑥 + 𝑦𝑦) 𝑑𝑑𝑑𝑑 = � (−𝑡𝑡3 − 𝑡𝑡2 − 5𝑡𝑡 + 3)𝑑𝑑𝑑𝑑 = −
𝜸𝜸𝟏𝟏
0
1
12
1
� 𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑 − (𝑥𝑥 + 𝑦𝑦)𝑑𝑑𝑑𝑑 = � (−𝑡𝑡2 + 4𝑡𝑡 − 2)dt = −
𝜸𝜸𝟐𝟐
0
� 𝒙𝒙𝒙𝒙 𝒅𝒅𝒅𝒅 − (𝒙𝒙 + 𝒚𝒚)𝒅𝒅𝒅𝒅 = −
𝜸𝜸
𝟓𝟓
𝟏𝟏 𝟏𝟏
− =−
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏 𝟑𝟑
1
3
(b) Evaluation using Green’s Theorem (Circulation or Curl form) straight from the notes:
𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖 𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 𝜸𝜸 = �𝑷𝑷(𝒙𝒙, 𝒚𝒚)𝒅𝒅𝒅𝒅 + 𝑸𝑸(𝒙𝒙, 𝒚𝒚)𝒅𝒅𝒅𝒅 = � �
𝟐𝟐
=� �
𝟏𝟏
𝒙𝒙+𝟐𝟐
𝒙𝒙𝟐𝟐 −𝟐𝟐𝟐𝟐+𝟒𝟒
�
𝜸𝜸
𝝏𝝏
𝟓𝟓
𝝏𝝏
(−𝒙𝒙 − 𝒚𝒚) −
(𝒙𝒙𝒙𝒙)� 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 = −
𝝏𝝏𝝏𝝏
𝟏𝟏𝟏𝟏
𝝏𝝏𝝏𝝏
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
𝑫𝑫
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
−
� 𝒅𝒅𝒅𝒅
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
8. Let 𝑆𝑆 be the triangle with vertices 𝑃𝑃(0,0,0), 𝑄𝑄(1,2,3), 𝑅𝑅(2,1,2). Let 𝜸𝜸 be the boundary curve of 𝑆𝑆,
oriented clockwise as seen from above. Calculate ∫𝜸𝜸 𝑦𝑦 𝑑𝑑𝑑𝑑 − 𝑥𝑥 𝑑𝑑𝑑𝑑 + 𝑧𝑧 𝑑𝑑𝑑𝑑
(a) Directly as a line integral AND
(b) as a double integral, by using Stoke’s Theorem.
MY ANSWER
(a) We parametrize the three line segments as follows:
𝜸𝜸𝟏𝟏 (𝑡𝑡) = (1 − 𝑡𝑡)〈0,0,0〉 + 𝑡𝑡〈1,2,3〉 = 〈𝑡𝑡, 2𝑡𝑡, 3𝑡𝑡〉; 0 ≤ 𝑡𝑡 ≤ 1.
⎧
⎪
𝜸𝜸𝟐𝟐 (𝑡𝑡) = (1 − 𝑡𝑡)〈1,2,3〉 + 𝑡𝑡〈2,1,2〉 = 〈1 + 𝑡𝑡, 2 − 𝑡𝑡, 3 − 𝑡𝑡〉;
⎨
⎪
⎩
0 ≤ 𝑡𝑡 ≤ 1.
𝜸𝜸𝟑𝟑 (𝑡𝑡) = (1 − 𝑡𝑡)〈2,1,2〉 + 𝑡𝑡〈0,0,0〉 = 〈2 − 2𝑡𝑡, 1 − 𝑡𝑡, 2 − 2𝑡𝑡〉; 0 ≤ 𝑡𝑡 ≤ 1.
𝟏𝟏
𝟏𝟏
𝟏𝟏
𝟏𝟏
� 𝒚𝒚 𝒅𝒅𝒅𝒅 − 𝒙𝒙 𝒅𝒅𝒅𝒅 + 𝒛𝒛 𝒅𝒅𝒅𝒅 = � 𝟗𝟗𝟗𝟗 𝒅𝒅𝒅𝒅 + � 𝒕𝒕 𝒅𝒅𝒅𝒅 + � 𝟒𝟒𝟒𝟒 − 𝟒𝟒 𝒅𝒅𝒅𝒅 = � 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟒𝟒 𝒅𝒅𝒅𝒅 = 𝟕𝟕𝒕𝒕𝟐𝟐 − 𝟒𝟒𝟒𝟒|𝟏𝟏𝟎𝟎 = 𝟑𝟑
𝜸𝜸
𝟎𝟎
𝟎𝟎
𝟎𝟎
1
𝟎𝟎
(b) ∫𝜸𝜸 𝑦𝑦 𝑑𝑑𝑑𝑑 − 𝑥𝑥 𝑑𝑑𝑑𝑑 + 𝑧𝑧 𝑑𝑑𝑑𝑑 = ∬𝑆𝑆 𝛁𝛁 × 𝑭𝑭 ⋅ 𝒏𝒏 𝑑𝑑𝑑𝑑 = ∬𝐷𝐷 |∇𝐻𝐻(𝑥𝑥,𝑦𝑦,𝑧𝑧)⋅𝑘𝑘| �𝛁𝛁 × 𝑭𝑭 ⋅ −𝛁𝛁𝐻𝐻(𝑥𝑥, 𝑦𝑦, 𝑧𝑧)�𝑑𝑑𝑑𝑑
where 𝐻𝐻(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = −𝑥𝑥 − 4𝑦𝑦 + 3𝑧𝑧 = 0 and 𝑭𝑭(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 〈𝑦𝑦, −𝑥𝑥, 𝑧𝑧〉. Now −∇𝐻𝐻(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 〈1,4, −3〉 and 𝒑𝒑 = 𝒌𝒌
and 𝐷𝐷 is the triangle in the 𝑥𝑥𝑥𝑥-plane with vertices at (0,0), (1,2)𝑎𝑎𝑎𝑎𝑎𝑎 (2,1) so that |∇𝐻𝐻(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) ⋅ 𝒌𝒌| = 3 and
𝛁𝛁 × 𝑭𝑭 = 〈0,0, −2〉 and 𝛁𝛁 × 𝑭𝑭 ⋅ −𝛁𝛁𝐻𝐻(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 6. So our formula above becomes
𝟏𝟏
(−𝟔𝟔) 𝒅𝒅𝒅𝒅 = 𝟐𝟐 � 𝒅𝒅𝒅𝒅 = 𝟐𝟐(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 𝒐𝒐𝒐𝒐 𝑫𝑫) = ‖〈𝟏𝟏, 𝟐𝟐, 𝟎𝟎〉 × 〈𝟐𝟐, 𝟏𝟏, 𝟎𝟎〉‖ = ‖〈𝟎𝟎, 𝟎𝟎, −𝟑𝟑〉‖ = 𝟑𝟑
𝑫𝑫 𝟑𝟑
𝑫𝑫
�𝛁𝛁 × 𝑭𝑭 ⋅ 𝒅𝒅𝒅𝒅 = − �
𝑺𝑺
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
9. Let 𝑇𝑇 be the solid contained between the surfaces 𝑧𝑧 = 4 − 𝑥𝑥 2 − 𝑦𝑦 2 and 𝑧𝑧 = 𝑥𝑥 2 + 𝑦𝑦 2 − 4. Let 𝑭𝑭 be the
vector field defined by 𝑭𝑭(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 〈𝑥𝑥, 𝑦𝑦, 1〉. Use the outward pointing unit normal vector to calculate
∬𝑆𝑆 𝑭𝑭 ⋅ 𝒏𝒏 𝑑𝑑𝑑𝑑
(a) directly as a surface integral AND
(b) as a triple integral, by using the Divergence Theorem.
MY SOLUTION
(a) Let 𝑆𝑆1 denote the surface 𝑧𝑧 = 4 − (𝑥𝑥 2 + 𝑦𝑦 2 ) or 𝐻𝐻1 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 = 4. Then
∇𝐻𝐻1 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 〈2𝑥𝑥, 2𝑦𝑦, 1〉 and |∇𝐻𝐻1 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) ⋅ 𝒌𝒌| = 1 and ∇𝐹𝐹(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) ⋅ ∇𝐻𝐻1 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 2𝑥𝑥 2 + 2𝑦𝑦 2 + 1
where 𝐷𝐷 = {(𝑥𝑥, 𝑦𝑦): 𝑥𝑥 2 + 𝑦𝑦 2 ≤ 4}
So since the surface 𝑆𝑆1 is positively oriented, then we use
1
�∇𝐹𝐹(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) ⋅ ∇𝐻𝐻1 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧)�𝑑𝑑𝑑𝑑
𝐷𝐷 |∇𝐻𝐻1 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) ⋅ 𝒌𝒌|
� 𝑭𝑭 ⋅ 𝒏𝒏 𝑑𝑑𝑆𝑆1 = �
𝑆𝑆1
=�
2𝜋𝜋
0
2
� (2𝑟𝑟 2 + 1)𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 20𝜋𝜋
0
Let 𝑆𝑆2 denote the surface 𝑧𝑧 = 𝑥𝑥 2 + 𝑦𝑦 2 − 4 or 𝐻𝐻2 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = −𝑥𝑥 2 − 𝑦𝑦 2 + 𝑧𝑧 = −4. Then
−∇𝐻𝐻2 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 〈2𝑥𝑥, 2𝑦𝑦, −1〉 and |∇𝐻𝐻2 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) ⋅ −𝒌𝒌| = 1 and
∇𝐹𝐹(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) ⋅ −∇𝐻𝐻2 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 2𝑥𝑥 2 + 2𝑦𝑦 2 + 1 where 𝐷𝐷 = {(𝑥𝑥, 𝑦𝑦): 𝑥𝑥 2 + 𝑦𝑦 2 ≤ 4}
So since the surface 𝑆𝑆2 is positively oriented, then we use
� 𝑭𝑭 ⋅ 𝒏𝒏 𝑑𝑑𝑆𝑆2 =
𝑆𝑆2
1
�∇𝐹𝐹(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) ⋅ ∇𝐻𝐻2 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧)�𝑑𝑑𝑑𝑑
𝐷𝐷 |∇𝐻𝐻2 (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) ⋅ 𝒌𝒌|
−�
= −�
2𝜋𝜋
0
2
� (1 − 2𝑟𝑟 2 )𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 12𝜋𝜋
0
Finally we get ∬𝑺𝑺 𝑭𝑭 ⋅ 𝒏𝒏 𝒅𝒅𝒅𝒅 = ∬𝑺𝑺 𝑭𝑭 ⋅ 𝒏𝒏 𝒅𝒅𝑺𝑺𝟏𝟏 + ∬𝑺𝑺 𝑭𝑭 ⋅ 𝒏𝒏 𝒅𝒅𝑺𝑺𝟐𝟐 = 𝟐𝟐𝟐𝟐𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟑𝟑𝟑𝟑𝟑𝟑
𝟏𝟏
(b) By the Divergence Theorem we have:
𝟐𝟐
� 𝑭𝑭 ⋅ 𝒏𝒏 𝒅𝒅 = � 𝒅𝒅𝒅𝒅𝒅𝒅 𝑭𝑭 𝒅𝒅𝒅𝒅 = 𝟐𝟐 � 𝒅𝒅𝒅𝒅 = 𝟐𝟐 (𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽 𝒐𝒐𝒐𝒐 𝑻𝑻) = 𝟐𝟐 �
𝑺𝑺
𝑻𝑻
𝑻𝑻
𝟐𝟐𝟐𝟐
𝟎𝟎
𝟐𝟐
� �
𝟎𝟎
𝟒𝟒−𝒓𝒓𝟐𝟐
𝒓𝒓𝟐𝟐 −𝟒𝟒
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
𝒅𝒅𝒅𝒅 𝒓𝒓 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 = 𝟑𝟑𝟑𝟑𝟑𝟑
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
END OF THE EXAMINATION. Make sure that you answered 5 complete questions from Part I and 2
complete questions from Part II.
Post Solution Commentary: First, I just got through solve this final Exam above using the Class Notes (not your official textbook). There is
nothing in this final that you should have difficulty with; unless you completely ignored the class notes (which was designed for the working student
who has no time for long drawn out discussion of the ideas) or you just don’t know how to double integrate and triple integrate, or even successfully
carry out a u-substitution etc. In those cases, you should have learned these things in Math 20300 like your classmate Mr. Wei Pan (a former student
from my last semester Math 20300 course who took Math 39200 with you) whom I must congratulate for the scoring the highest among your class on
the final Exam! Now Mr. Wei Pan is an ordinary student like all of you, he did not come to see me for help all semester long. Besides, I felt that he
should not need too much help from me anyway. I believe what made the difference was that he had the unique experience of the integration
project, the unconventional methods that I used last semester which force everyone have some firsthand knowledge of some of the important ideas
of calculus. Furthermore, he was taught how to study for Exams! You too, can have similar success on important test, if you make a conscious effort
to learn your material well (not always trying to avoid learning your subject matter).
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
SOLUTIONS TO THIS December 22, 2008 FINAL EXAM WAS DONE TO BENEFIT MY STUDENTS TAKING MATH 39200 C
(Spring 2008). NOT TO BE SOLD FOR MONETARY GAIN BUT YOU MAY DISTRIBUTE IT FREELY. © 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)