ELEMENTARY LINEAR ALGEBRA AND VECTOR CALCULUS
Transcript of ELEMENTARY LINEAR ALGEBRA AND VECTOR CALCULUS
ELEMENTARY LINEARALGEBRA AND VECTORCALCULUS
MATHEMATICS FOR UNDERGRADUATES
Antony L. FosterDepartment of Mathematics (NAC 6/273)The City College of New YorkNew 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 studyโsuch as mathematics, physics,biology, chemistry, economics, all phases of engineering, operations research, the social sciences, and so forthโis thatof solving a system of linear equations. The equation
b = a1x1 + a2x2 + ยท ยท ยท + anxn, (1)
which expresses b in terms of the variables x1, x2, . . . , xn and the constants a1, a2, . . . , an, is called a linearequation. 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 whenx1 = s1, x2 = s2, . . . , xn = sn are substituted in (1). Thus x1 = 2, x2 = 3, and x3 = โ4 is a solution to thelinear equation
6x1 โ 3x2 + 4x3 = โ13,
because6(2) โ 3(3) + 4(โ4) = โ13.
More generally, a system of m linear equations in n unknowns, or a linear system, is a set of m linearequations each in n unknowns called x1, x2, . . . , xn.
A linear system can be conveniently written as
a11x1 + a12x2 + ยท ยท ยท + a1nxn = b1
a21x1 + a22x2 + ยท ยท ยท + a2nxn = b2...
......
...am1x1 + am2x2 + ยท ยท ยท + amnxn = bm.
(2)
Thus the ith equation in the system is
ai1x1 + ai2x2 + ยท ยท ยท + ainxn = bi.
In (2) the aij are known constants. Given the values b1, b2, . . . , bn, we want to find values of x1, x2, . . . , xn thatwill 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 eachequation 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. Ifb1 = b2 = ยท ยท ยท = bn = 0, then (2) is called a homogeneous system. The solution x1 = x2 = ยท ยท ยท = xn = 0to a homogeneous system is called the trivial solution. A solution to a homogeneous system in which not all ofx1, x2, . . . , xn are zero is called a nontrivial solution.
Consider another system of r linear equations in n unknowns:
c11x1 + c12x2 + ยท ยท ยท + c1nxn = d1
c21x1 + c22x2 + ยท ยท ยท + c2nxn = d2...
......
...cr1x1 + cr2x2 + ยท ยท ยท + crnxn = dr.
(3)
We say that (2) and (3) are equivalent if they both have exactly the same solutions.
Example 1.1.1. The linear systemx1 โ 3x2 = โ72x1 + x2 = 7
(4)
has only the solution x1 = 2 and x2 = 3. The linear system
8x1 โ 3x2 = 73x1 โ 2x2 = 0
10x1 โ 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, weeliminate some variables by adding a multiple of one equation to another equation. Elimination merely amountsto 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, linearsystems having as many equations as unknowns. In this course we shall broaden our outlook by dealing with systemsin 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 systemx1 โ 3x2 = โ32x1 + 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 systemx1 โ 3x2 = โ72x1 + 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 sameconclusion 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 = 62x1 โ 3x2 + 2x3 = 143x1 + x2 โ x3 = โ2.
(8)
To eliminate x1, we subtract twice the first equation from the second and three times the first equation from thethird, obtaining
โ7x2 โ 4x3 = 2โ5x2 โ 10x3 = โ20.
(9)
This is a system of two equations in the unknowns x2 and x3. We divide the second equation of (9) by -5, obtaining
โ7x2 โ 4x3 = 2x2 + 2x3 = 4,
which we write, by interchanging equations as
x2 + 2x3 = 4โ7x2 โ 4x3 = 2.
(10)
We now eliminate x2 in (10) by adding 7 times the first equation to the second one, to obtain
10x3 = 30,
orx3 = 3. (11)
Substituting this value of x3 into the first equation of (10), we find that x2 = โ2. 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 procedurehas actually produced the linear system
x1 + 2x2 + 3x3 = 6x2 + 2x3 = 4
x3 = 3.(12)
obtained by using the first equations of (8) and (10) as well as (11). The importance of the procedure is that althoughthe 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 โ 3x3 = โ4
2x1 + x2 โ 3x3 = 4.(13)
Eliminating x1, we subtract twice the first equation from the second equation, to obtain
โ3x2 + 3x3 = 12. (14)
We must now solve (14). A solution isx2 = x3 โ 4,
where x3 can be any real number. Then from the first equation of (13),
x1 = โ4 โ 2x2 + 3x3
= โ4 โ 2(x3 โ 4) + 3x3
= x3 + 4.
Thus a solution to the linear system (13) is
x1 = x3 + 4
x2 = x3 โ 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 obtainanother solution to (13). Thus, if x3 = 1, then
x1 = 5, x2 = โ3, and x3 = 1
is a solution, while if x3 = โ2, then
x1 = 2, x2 = โ6, and x3 = โ2
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.
a1x1 + a2x2 = c1
b1x1 + b2x2 = c2
(15)
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 ledgeometrically to the same three possibilities mentioned above in (Figure 1.1).
x1
x2
x1
x2
x1
x2
(a) A unique solution. (b) No solution. (c) Infinitely many solutions.
l2
l1
l1
l2
l1
l2
x2 x2
If we examine the method of elimination more closely, we find that it involves three manipulations that can beperformed 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, replaceai1x1 + ai2x2 + ยท ยท ยท + ainxn = bi
by(ai1 + caj1)x2 + (ai2 + caj2)x2 + ยท ยท ยท + (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 constantc, obtaining the linear system
a11x1 + a12x2 + ยท ยท ยท + a1nxn = b1
a21x1 + a22x2 + ยท ยท ยท + a2nxn = b2...
......
...cai1x1 + cai2x2 + ยท ยท ยท + cainxn = cbi
......
......
am1x1 + am2x2 + ยท ยท ยท + amnxn = bm.
(16)
If x1 = s1, x2 = s2, ยท ยท ยท xn = sn is a solution to (2), then it is a solution to all the equations in (16) exceptpossibly for the ith equation. For the ith equation we have
c(ai1s1 + ai2s2 + ยท ยท ยท + ainsn) = cbi
orcai1s1 + cai2s2 + ยท ยท ยท + cainsn = 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 offon these exercises until later when we discuss the method of Gaussian Elimination.)
Exercise 1.1.1.x1 + 2x2 = 83x1 โ 4x2 = 4
.
Exercise 1.1.2.2x1 โ 3x2 + 4x3 = โ12
x1 โ 2x2 + x3 = โ53x1 + x2 + 2x3 = 1
.
Exercise 1.1.3.3x1 + 2x2 + x3 = 24x1 + 2x2 + 2x3 = 8
x1 โ x2 + x3 = 4.
Exercise 1.1.4.x1 + x2 = 5
3x1 + 3x2 = 10.
Exercise 1.1.5.2x1 + 4x2 + 6x3 = โ122x1 โ 3x2 โ 4x3 = 153x1 + 4x2 + 5x3 = โ8.
Exercise 1.1.6.x1 + x2 โ 2x3 = 5
2x1 + 3x2 + 4x3 = 2.
Exercise 1.1.7.x1 + 4x2 + 6x3 = 123x1 + 8x2 โ 2x3 = 4.
Exercise 1.1.8.3x1 + 4x2 โ x3 = 8
6x1 + 8x2 โ 2x3 = 3.
Exercise 1.1.9.x1 + x2 + 3x3 = 12
2x1 + 2x2 + 6x3 = 6.
Exercise 1.1.10.x1 + x2 = 12x1 โ x2 = 5
3x1 + 4x2 = 2.
Exercise 1.1.11.2x1 + 3x2 = 13x1 โ 2x2 = 3
5x1 + 2x2 = 27.
Exercise 1.1.12.x1 โ 5x2 = 63x1 + 2x2 = 15x1 + 2x2 = 1
.
Exercise 1.1.13.x1 + 3x2 = โ42x1 + 5x2 = โ8x1 + 3x2 = โ5.
Exercise 1.1.14.2x1 + 3x2 โ x3 = 6
2x1 โ x2 + 2x3 = โ83x1 โ x2 + x3 = โ7.
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 anotherequation 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 greatlysimplify 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 thenumbers in front of the unknowns x1, x2, . . . , xn are being changed as we perform the steps in the method ofelimination. Thus we might think of looking for a way of writing a linear system without having to carry along theunknowns. Matrices enables us to do thisโthat is, to write linear systems in a compact form that makes it easierto automate the elimination method on an electronic computer in order to obtain a fast and efficient procedure forfinding solutions. Their use is not, however, merely that of a convenient notation. We now develop operations onmatrices and will work with matrices according to the rules they obey; this will enable us to solve systems of linearequations and to do other computational problems in a fast and efficient manner. Of course, as any good definitionshould do, the notion of a matrix provides not only a new way of looking at old problems but also gives rise to agreat 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 a12 ยท ยท ยท a1j ยท ยท ยท a1n
a21 a22 ยท ยท ยท a2j ยท ยท ยท a2n
......
... ยท ยท ยท...
ai1 ai2 ยท ยท ยท aij ยท ยท ยท ain
......
... ยท ยท ยท...
am1 am2 ยท ยท ยท amj ยท ยท ยท amn
.
Unless stated otherwise, we assume that all our matrices are composed entirely of real numbers. The ith row of Ais
( ai1 ai2 ยท ยท ยท ain ) (1 6 i 6 m)
while the jth column of A is
a1j
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 ร n) matrix. If m = n, we say that A isa square matrix of order n and that the elements a11, a22, . . . , ann are on the main diagonal of A. We referto 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 ]mรn
We shall also write Amรn to indicate that A has m rows and n columns. If A is n ร n, we merely write An.
Example 1.2.1. The following are matrices:
A =
1 2 34 5 67 8 9
, B = ( 1 โ2 9 )
C =
2โ134
, and D =
(0 3โ1 โ2
)
.
In matrix A, the entry a32 = 8; in matrix C, the entry c41 = 4. Here A is a 3 ร 3 square matrix, B is a 1 ร 3,matrix C is a 4ร 1 matrix, and matrix D is a 2ร 2 square matrix. In A, the entries a11 = 1, a22 = 5 and a33 = 9are on the main diagonal.
Whenever a new object is introduced in mathematics, one must determine when two such objects are equal. Forexample, in the set of all rational numbers, the numbers 2
3 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, wenow have the following definition.
EQUALITY OF MATRICES
Definition 1.2.2. Two m ร n matrices A = [aij ] and B = [bij ] are equal if they agree entry by entry, that is, ifaij = bij for i = 1, 2, . . . , m and j = 1, 2, . . . , n.
Example 1.2.2. The matrices
1 2 โ12 โ3 40 โ4 5
and B =
1 2 w2 x 4y โ4 z
are equal if and only if w = โ1, x = โ3, 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 tocompute with the matrices and not deal with the equations from which they arise. These operations are also usefulin the applications of matrices.
ADDITION OF MATRICES
Definition 1.2.3. If A = [aij ] and B = [bij ] are both m ร n matrices, then their matrix sum A + B is anmร n matrix C = [cij ] defined by cij = aij + bij , i = 1, 2, . . . , m; j = 1, 2, . . . , n. Thus, to obtain the sum of Aand B, we merely add corresponding (i, j) entries.
Example 1.2.3. Let
A =
(1 โ2 32 โ1 4
)
and B =
(0 2 11 3 โ4
)
.
Then
A + B =
(1 โ2 32 โ1 4
)
+
(0 2 11 3 โ4
)
=
(1 + 0 โ2 + 2 3 + 12 + 1 โ1 + 3 4 โ 4
)
=
(1 0 43 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 numberof rows and the same number of columns, that is, only when A and B are of the same size. We now establish theconvention that when A + B is formed, both A and B are of the same size. The basic properties of matrix additionare 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 ร n matrix and ฮฑ is a real number, then the scalar multiple of A by ฮฑ,is ฮฑA, is the m ร n matrix C = [cij ], where cij = ฮฑ aij , i = 1, 2, . . . , m, j = 1, 2, . . . , n; that is, the matrix C isobtained by multiplying each entry of A by ฮฑ.
Example 1.2.4. We have
2
(4 โ2 โ37 โ3 2
)
=
(8 โ4 โ614 โ6 4
)
.
SUBTRACTION: If A and B are m ร n matrices, we write A + (โ1)B as A โ B and call this the differencebetween A and B.
We shall sometimes use the summation notation, and we now review this useful and compact notation.
Bynโ
k=1
rkak
we mean r1a1 + r2a2 + ยท ยท ยท + rnan. The letter k is called the index of summation; it is a dummy variable thatcan be replaced by another letter. Hence we can write
nโ
k=1
rkak =nโ
j=1
rjaj =nโ
i=1
riai.
Thus4โ
i=1
riai = r1a1 + r2a2 + r3a3 + r4a4.
The summation notation satisfies the following properties:
1.
nโ
i=1
(ri + si)ai =
nโ
i=1
(riai + siai) =
nโ
i=1
riai +
nโ
i=1
siai
2.
nโ
i=1
ฮฑ (riai) = ฮฑ
nโ
i=1
riai
3.mโ
j=1
nโ
i=1
aij =nโ
i=1
mโ
j=1
aij
Property 3 can be interpreted as follows. If we add up the entries in each row of a matrix and then add the resultingnumbers, we obtain the same result as when we add up the entries in each column of the matrix and then add theresulting numbers.
MULTIPLICATION OF MATRICES
Definition 1.2.5. If A = [aij ] is an mร n matrix and B = [bij ] is an nร p matrix, then the matrix product ofA and B, denoted C = AB = [cij ], is an m ร p matrix defined y
cij =
nโ
k=1
aikbkj = ai1b1j + ai2b2j + ยท ยท ยท + ainbnj
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 ofrows of B. We also observe that the (i, j) entry of C = AB is obtained by using the ith row of A and the jthcolumn of B. Thus
a11 a12 ยท ยท ยท a1j ยท ยท ยท a1n
a21 a22 ยท ยท ยท a2j ยท ยท ยท a2n
......
... ยท ยท ยท...
ai1 ai2 ยท ยท ยท aij ยท ยท ยท ain
......
... ยท ยท ยท...
am1 am2 ยท ยท ยท amj ยท ยท ยท amn
b11 b12 ยท ยท ยท b1j ยท ยท ยท b1n
b21 b22 ยท ยท ยท b2j ยท ยท ยท b2n
......
... ยท ยท ยท...
bi1 bi2 ยท ยท ยท bij ยท ยท ยท bin
......
... ยท ยท ยท...
bm1 bm2 ยท ยท ยท bmj ยท ยท ยท bmn
=
c11 c12 ยท ยท ยท c1j ยท ยท ยท c1n
c21 c22 ยท ยท ยท c2j ยท ยท ยท c2n
......
... ยท ยท ยท...
ci1 ci2 ยท ยท ยท cij ยท ยท ยท cin
......
... ยท ยท ยท...
cm1 cm2 ยท ยท ยท cmj ยท ยท ยท cmn
Example 1.2.5. Let
A =
(1 2 โ13 1 4
)
and B =
โ2 54 โ32 1
.
Then
AB =
((1)(โ2) + (2)(4) + (โ1)(2) (1)(5) + (2)(โ3) + (โ1)(1)(3)(โ2) + (1)(4) + (4)(2) (3)(5) + (1)(โ3) + (4)(1)
)
=
(4 โ26 16
)
The basic properties of matrix multiplication are considered in the following section. However, we note here thatmultiplication of matrices requires much more care than their addition, since the algebraic properties of matrixmultiplication differ from those satisfied by the real numbers. Part of the problem is due to the fact that AB isdefined only when the number of columns of A is the same as the number of rows of B. Thus, if A is an mรn matrixand B is an n ร p matrix, then AB is an m ร p matrix. What about BA? Three different situations may occur:
1. BA may not be defined. This will take place if p 6= m.
2. If BA is defined, BA will be an n ร n matrix and AB will be an m ร m matrix, and if m 6= n, AB and BAare of different sizes.
3. 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 ร 3 matrix and let B be a 3 ร 4 matrix. Then AB is 2 ร 4 and BA is not defined.
Example 1.2.7. Let A be 2 ร 3 and let B be 3 ร 2. Then AB is 2 ร 2 and BA is 3 ร 3.
Example 1.2.8. Let
A =
(1 2โ1 3
)
and B =
(2 10 1
)
.
Then
AB =
(2 3โ2 2
)
while BA =
(1 7โ1 3
)
.
Thus AB 6= BA.
One might ask why matrix equality and matrix addition are defined in such a natural way while matrix multiplicationappears to be much more complicated. Only a thorough understanding of the composition of functions and therelationship that exists between matrices and what are called linear transformations would show that the definitionof 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 twomatrices. It is not difficult to show that the jth column of the matrix product AB is equal to the matrix productABj , where Bj is the jth column of B.
Example 1.2.9. Let
A =
1 23 2โ1 5
and B =
(โ2 3 43 2 1
)
.
Then the second column of AB is
AB2 =
1 23 4โ1 5
(32
)
=
7177
We now return to the linear system (2) in Section 1.1 and define the following matrices:
A =
a11 a12 ยท ยท ยท a1n
a21 a22 ยท ยท ยท a2n...
......
...am1 am2 ยท ยท ยท amn
, X =
x1
x2...
xn
, B =
b1
b2...
bm
We can then write the linear system (2) as AX = B. The matrix A is called the coefficient matrix of the systemand the matrix
(A | B) =
a11 a12 ยท ยท ยท a1n | b1
a21 a22 ยท ยท ยท a2n | b2
......
...... |
...am1 am2 ยท ยท ยท amn | bm
is called the augmented matrix of the system. The coefficient and augmented matrices of a linear system will playkey roles in our methods of solving linear systems.
Example 1.2.10. Consider the following linear system:
2x1 + 3x2 โ 4x3 + x4 = 5โ2x1 + x2 + x3 + x4 = 7
3x1 + 2x2 + x3 โ 4x4 = 3.
We can write this in matrix form as
2 3 โ4 1โ2 0 1 0
3 2 0 โ4
x1
x2
x3
x4
=
573
.
The coefficient matrix of this system is
2 3 โ4 1โ2 0 1 03 2 0 โ4
and the augmented matrix is
2 3 โ4 1 | 5โ2 0 1 0 | 73 2 0 โ4 | 3
.
Example 1.2.11. The matrix (2 โ1 3 | 43 0 2 | 5
)
.
is the augmented matrix of the linear system
2x1 โ x2 + 3x3 = 43x1 + 0x2 + 2x3 = 5.
TRANSPOSE OF MATRICES
Definition 1.2.6. If A = [aij ] is an m ร n matrix, then the transpose of A, denoted by AT = [aTij ] is an n ร 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 โ1
โ3 2 7
)
, then AT =
1 โ32 2
โ1 7
.
TRACE OF A MATRIX
Definition 1.2.7. If A = [aij ] is an nร n matrix, then the trace of A, denoted by Tr(A), is defined as the sum ofall elements on the main diagonal of A (these are elements of A of the form aii). Thus
Tr(A) =
nโ
i=1
aii.
Example 1.2.13. If
C =
1 2 34 5 67 8 9
, then Tr(C) =
3โ
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:
A =
(1 2 32 1 4
)
, B =
1 02 13 2
, C =
3 โ1 34 1 52 1 3
, D =
(3 โ22 5
)
, and E =
2 โ4 50 1 43 2 1
.
Exercise 1.2.1. If possible, compute:(a) C + E.
(b) AB and BA.
(c) 2C โ 3E.
(d) CB + D.
(e) AB + D2, where D2 = DD.
(f) (3)(2A) and 6A.
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) BT AT .
(e) (C + E)T and CT + ET .
(f) A(2B) and 2(AB).
Exercise 1.2.4.(a) Let A be an mรn matrix with a row consisting entirely of zeros. Show that if B is an nร p matrix, then the
matrix product AB has a row of zeros.
(b) Let A be an m ร n matrix with a column consisting entirely of zeros and let B be p ร m matrix. Prove thatthe matrix product BA has a column of zeros.
Exercise 1.2.5. Let A =
(1 23 2
)
and B =
(2 โ1โ3 4
)
. Show that AB 6= BA.
Exercise 1.2.6. Consider the following linear system
2x1 + 3x2 โ 3x3 + x4 + x5 = 73x1 + 2x3 + 3x5 = โ22x1 + 3x2 โ4x4 = 3
(a) Find the coefficient matrix.
(b) Write the linear system in matrix form.
(c) Find the augmented matrix.
Exercise 1.2.7. Write the linear system whose augmented matrix is
โ2 โ1 0 4 | 5โ3 2 7 8 | 31 0 0 2 | 43 0 1 3 | 6
.
Exercise 1.2.8. If
(a + b c + dc โ d a โ b
)
=
(4 610 2
)
, find a, b, c, and d.
Exercise 1.2.9. Write the following linear system in matrix form.
2x1 + 3x2 = 03x2 + x3 = 02x1 โ x3 = 0.
Exercise 1.2.10. Write the linear system whose augmented matrix is
(a)
2 1 3 4 | 03 โ1 2 0 | 3
โ2 1 โ4 3 | 2
(b)
2 1 3 4 | 03 โ1 2 0 | 3
โ2 1 โ4 3 | 20 0 0 0 | 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 ร n matrix and ฮฑ is a real number, then prove that
(a) Tr(ฮฑA) = ฮฑ 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.
(a)
(1 02 3
)
(b)
2 2 32 4 43 โ2 โ5
. (c)
1 0 00 1 00 0 1
.
Exercise 1.2.14. Show that there are no 2 ร 2 matrices A and B such that AB โ BA =
(1 00 1
)
.
Exercise 1.2.15. Show that the jth column of the matrix product AB is equal to the matrix product ABj , whereBj 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 = ฮฑX1 + ฮฒX2, where ฮฑ + ฮฒ = 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 aresimilar to the familiar properties holding for the real numbers. However, there will be striking differences between theset R of real numbers and the set Mmรn(R) of all mรn 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, whoseproof will be left as exercises.
Theorem 1.3.1. If A and B are in Mmรn(R), then A + B = B + A.
Proof. LetA = [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 arereal numbers, we have aij + bij = bij + aij , which implies that cij = dij for all i, j. โโ
Example 1.3.1. We have
(1 2 3โ2 1 4
)
+
(3 2 โ13 โ1 2
)
=
(4 4 21 0 6
)
=
(3 2 โ13 โ1 2
)
+
(1 2 3โ2 1 4
)
.
Theorem 1.3.2. If A, B, and C are in Mmรn(R), then A + (B + C) = (A + B) + C.
Proof. โโ
Example 1.3.2. We have
(1 23 4
)
+
((2 13 โ2
)
+
(3 1โ2 1
))
=
(6 44 3
)
=
((1 23 4
)
+
(2 13 โ2
))
+
(3 1โ2 1
)
.
Theorem 1.3.3. There exists a unique 0mรn in Mmรn(R) such that A + 0mรn = 0mรn + A = A for any A โMmรn(R).
Proof. Let U = [uij ]. Then A + U = A if and only if aij + uij = aij , which holds if and only if uij = 0. ThusU is the m ร n matrix all of whose entries are zero; U is denoted by 0. โโ
We call 0mรn the m ร n zero matrix. When m = n, we write 0n. When m and n are understood, we shall write0mรn merely as 0.
Theorem 1.3.4. For any A in Mmรn(R), there exists B in Mmรn(R) such that A + B = 0mรn.
Proof. โโ
We can now show that B is unique and that it is (โ1)A, which we have already agreed to write as โA, and call itthe negative of A.
Example 1.3.3. If
A =
(1 3 โ2โ2 4 3
)
, then โ A =
(โ1 โ3 22 โ4 โ3
)
Theorem 1.3.5. If A is in Mmรn(R) and B is in Mnรp(R), and C is in Mpรq(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 completelyanalogous.
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
fij =
pโ
k=1
dikckj =
pโ
k=1
(nโ
r=1
airbrk
)
ckj
and
gij =
nโ
r=1
airerj =
nโ
r=1
air
(pโ
k=1
brkckj
)
.
Then
fij =
pโ
k=1
(ai1b1k + ai2b2k + ยท ยท ยท + ainbnk)ckj
= ai1
pโ
k=1
b1kckj + ai2
pโ
k=1
b2kckj + ยท ยท ยท + ain
pโ
k=1
bnkckj
=nโ
r=1
air
(pโ
k=1
brkckj
)
= gij .
โโ
Example 1.3.4. Let
A =
(5 2 32 โ3 4
)
, B =
2 โ1 1 00 2 2 23 0 โ1 3
, and C =
1 0 22 โ3 02 1 0
.
Then
A(BC) =
(5 2 32 โ3 4
)
0 3 78 โ4 69 3 3
=
(43 16 5612 30 8
)
and
(AB)C =
(19 โ1 6 1316 โ8 โ8 6
)
1 0 22 โ3 00 0 32 1 0
=
(43 16 5612 30 8
)
.
Recall Example 8 in Section 1.2, which shows that AB need not always equal BA. This is the first significantdifference between multiplication of matrices and multiplication of real numbers.
Theorem 1.3.6. Let A and B be in Mmรn(R) and C be in Mnรp(R), then(a) (A + B) C = AC + BC.
(b) If C is in Mmรn(R) and A and B are both in Mnรp(R), then
C (A + B) = CA + CB.
Proof. An easy exercise! โโ
Example 1.3.5. Let
A =
(2 2 33 โ1 2
)
, B =
(0 0 12 3 โ1
)
, and C =
1 02 23 โ1
.
Then
(A + B)C =
(2 2 45 2 1
)
1 02 23 โ1
=
(18 012 3
)
and
AC + BC =
(15 17 โ4
)
+
(3 โ15 7
)
=
(18 012 3
)
.
Theorem 1.3.7. If ฮฑ and ฮฒ are real numbers, A is m ร n matrix, and B is an n ร p matrix, then(a) ฮฑ(ฮฒ A) = (ฮฑฮฒ) A = ฮฒ(ฮฑ A).
(b) A(ฮฑB) = ฮฑ (AB).
Proof. An Easy Exercise. โโ
Example 1.3.6. Let
A =
(4 2 32 โ3 4
)
and B =
3 โ2 12 0 โ10 1 2
.
Then
2(3A) = 2
(12 6 96 โ9 12
)
=
(24 12 1812 โ18 24
)
= 6A.
We also have
A(2B) =
(4 2 32 โ3 4
)
=
6 โ4 24 0 โ20 2 4
=
(32 โ10 160 0 26
)
= 2(AB).
Theorem 1.3.8. If ฮฑ and ฮฒ are real numbers and A is in Mmรn(R), then
(ฮฑ + ฮฒ) A = ฮฑA + ฮฒA.
Proof. An easy exercise. โโ
Theorem 1.3.9. If A and B are both in Mmรn(R) and ฮณ is any real number, then
ฮณ(A + B) = ฮณA + ฮณB.
Proof. An easy exercise. โโ
So far we have seen that multiplication and addition of matrices have much in common with multiplication andaddition of real numbers. We now look at some properties of the transpose.
Theorem 1.3.10. If A is in Mmรn(R), then (AT )T = A
Proof. An easy exercise. โโ
Theorem 1.3.11. If A and B are both in Mmรn(R) and ฮณ is any real number, then(a) (ฮณA)T = ฮณAT .
(b) (A + B)T = AT + BT .
Proof. An easy exercise. โโ
Example 1.3.7. Let
A =
(1 2 3โ2 0 1
)
and B =
(3 โ1 23 2 โ1
)
.
Then
AT =
1 โ22 03 1
and BT =
3 3โ2 22 โ1
.
Also
A + B =
(4 1 51 2 0
)
and (A + B)T =
4 11 25 0
.
Now
AT + BT =
4 11 25 0
= (A + B)T .
Theorem 1.3.12. If A is in Mmรn(R) and B is in Mnรp(R), then
(AB)T = BT AT .
Proof. Let A = [aij ] and B = [bij ]; let AB = C = [cij ]. We must prove that cTij is the (i, j) entry in BT AT .
Recall from matrix multiplication that
cij =nโ
k=1
aikbkj , where i = 1, 2, . . . , m and j = 1, 2, . . . , p.
Thus
cTij = cji =
nโ
k=1
ajkbki =nโ
k=1
aTkjb
Tik =
nโ
k=1
bTikaT
kj = the (i, j) entry in BT AT .
โโ
Example 1.3.8. Let
A =
(1 3 22 โ1 3
)
and B =
0 12 23 โ1
.
Then
AB =
(12 57 โ3
)
and (AB)T =
(12 75 โ3
)
.
On the other hand,
AT =
1 23 โ12 3
and BT =
(0 2 31 2 โ1
)
,
and then
BT AT =
(12 75 โ3
)
= (AB)T .
We also note two other peculiarities of matrix multiplication. If ฮฑ and ฮฒ are real numbers, then ฮฑฮฒ = 0 can holdonly if ฮฑ = 0 or ฮฒ = 0. However, this is not true for matrices. This means it is possible for the matrix product ABbeing zero without neither A nor B being zero matrices (very strange indeed!).
Example 1.3.9. If A =
(1 22 4
)
and B =
(4 โ6โ2 3
)
, then neither A nor B is the zero matrix but the matrix
product AB =
(0 00 0
)
is the zero matrix.
The Cancelation Law for Real numbers: If ฮฑ, ฮฒ, and ฮณ are real numbers for which ฮฑฮฒ = ฮฑฮณ and ฮฑ 6= 0, itfollows that ฮฒ = ฮณ. That is, we can cancel out the nonzero factor ฮฑ. However, the cancelation law does not hold formatrices, as the following example shows.
Example 1.3.10. If
A =
(1 22 4
)
B =
(2 13 2
)
, and C =
(โ2 75 โ1
)
,
then
AB = AC =
(8 516 10
)
,
but B 6= C.
ALGEBRAIC NOTION OF A RING
A Ring is a set R together with two binary operations โ : RรR โ R called addition, and โ : R รR โ R calledmultiplication. 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 โ B = B โ A for all A and B in R (commutative law of addition).
2) (A โ B) โ C = A โ (B โ C) for all A, B, and C in R (associative law of addition).
3) There exist an element 0R โ R such that A โ 0R = 0R โ A for all A in R (existence of an additive identity).
4) For each A in R there corresponds an element โA in R such that A โ (โA) = (โA) โ A = 0R (existenceof additive inverses).
5) (A โ B) โ C = A โ (B โ C) for all A, B, and C in R (associative law of multiplication).
6) A โ (B โ C) = (A โ B) โ (A โ C) (left distributive law)
7) (B โ C) โ A = (B โ A) โ (C โ A) (right distributive law).
8) There exists an element IR in R (IR 6= 0R) such that A โ IR = IR โ A = A for all A in R (existence of amultiplicative 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. IfR is a ring with unity and AโB = B โA for all A, B โ R, then we say that R is a Commutative Ring with unity.Otherwise, if there exists A, B โ R for which A โ B 6= B โ A, then R is a Non-Commutative Ring with Unity.
Theorem 1.3.13. The set Mmรn(R) of all m ร n matrices equipped with addition and multiplication is a non-commutative ring (without unity whenever m 6= n).
Proof. See the theorems and examples of this section. โโIn this section we have developed a number of properties about matrices and their transposes. If a future problemeither asks a question about these ideas or involves these concepts, refer to these properties to help answer thequestion. 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 32 โ1
)
, B =
(โ1 3 21 โ3 4
)
, and C =
1 03 โ11 2
.
Exercise 1.3.4. Prove Theorem 1.3.6.
Exercise 1.3.5. Verify Theorem 1.3.6(b) for the following matrices:
A =
(2 โ3 23 โ1 โ2
)
, B =
(0 1 21 3 โ2
)
, and C =
(1 โ3
โ3 4
)
.
Exercise 1.3.6. Prove Theorem 1.3.7.
Exercise 1.3.7. Verify Theorem 1.3.7(b) for the following matrices:
A =
(1 32 โ1
)
, B =
(โ1 3 21 โ3 4
)
, and ฮฑ = โ3.
Exercise 1.3.8. Find a pair of unequal nonzero matrices A and B in M2ร2(R), other than those given in Exam-ple 1.3.9, such that AB = O2ร2.
Exercise 1.3.9. Find two different solutions in M2ร2(R) of the matrix equation A2 =
(0 00 0
)
= O2 (Recall
A2 = AA).
Exercise 1.3.10. Prove Theorem 1.3.8.
Exercise 1.3.11. Verify Theorem 1.3.8 for ฮฑ = 4, ฮฒ = โ2 and A =
(2 โ34 2
)
.
Exercise 1.3.12. Find two different solutions in M2ร2(R) of the matrix equation A2 =
(1 00 1
)
= I2.
Exercise 1.3.13. Prove Theorem 1.3.9.
Exercise 1.3.14. Verify Theorem 1.3.9 for ฮณ = โ3 and
A =
4 21 โ33 2
and B =
0 24 3
โ2 1
.
Exercise 1.3.15. Find A, B โ M2ร2(R) such that A 6= B and AB =
(1 00 1
)
= I2.
Exercise 1.3.16. Prove Theorem 1.3.10.
Exercise 1.3.17. Find matrices A, B, and C all in M3ร3(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 22 1 โ3
)
, B =
(4 2 โ1
โ2 1 5
)
, and ฮณ = โ4.
Exercise 1.3.20. Verify Theorem 1.3.12 for A =
(1 3 22 1 โ3
)
and B =
3 โ12 41 2
.
Exercise 1.3.21. Let A be in Mmรn(R) and ฮณ is a real number. Show that if ฮณA = O, then ฮณ = 0 or A = O.
Exercise 1.3.22. Determine all A in M2ร2(R) such that AB = BA for any B in M2ร2(R).
1.4. SPECIAL TYPES OF MATRICES
We have already introduced one special type of matrix, the zero m ร n matrix, denoted by Omรn whose entriesconsists entirely of zeros. We now consider several other types of matrices whose structure is rather specialized andfor which it will be convenient to have special names
DIAGONAL MATRICES
Definition 1.4.1. An n ร n matrix A = [aij ] is called a diagonal matrix if aij = 0 for i 6= j. Thus, for adiagonal 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 nรn identitymatrix.
Example 1.4.1. Let
A =
1 0 00 2 00 0 3
, B =
2 0 00 2 00 0 2
, and I3 =
1 0 00 1 00 0 1
.
Then A, B, and I3 are diagonal matrices; B and I3 are scalar matrices; and I3 is the 3 ร 3 identity matrix.
It is easy to show that if A is any m ร n matrix, then
AIn = A and ImA = A.
Also, if A is a scalar matrix, then A = ฮฑ In for some scalar ฮฑ.
Definition 1.4.4. Suppose that A is a square matrix, that is, A is in Mnรn(R) = Mn(R) the set of all matriceswith the same number of rows and column with real entries. If p is any positive integer, then we define
Ap = A ยท A ยท ยท ยท ยท ยท A๏ธธ ๏ธท๏ธท ๏ธธ
p factors of A
.
If A is in Mn(R), then we also defineA0 = In.
Definition 1.4.5. For nonnegative integers p and q, the familiar law of exponents for the real numbers can alsobe proved for matrix multiplication (see Exercise 1.4.25) of a square matrix A:
ApAq = Ap+q
and(Ap)q = Apq.
It should be noted that the rule(AB)p = ApBp
does not hold for square matrices unless AB = BA (see Exercise 1.4.26).
UPPER AND LOWER TRIANGULAR MATRICES
Definition 1.4.6. An nรn matrix A = [aij ] is called upper triangular if aij = 0 for i > j. It is called lowertriangular if aij = 0 for i < j.
Example 1.4.2. The matrix
A =
1 3 30 3 50 0 2
is upper triangular and
B =
1 0 02 3 03 5 2
is lower triangular.
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 = โA.
Example 1.4.3. A =
1 2 32 4 53 5 6
is a symmetric matrix.
Example 1.4.4. B =
0 2 3โ2 0 โ4โ3 4 0
is a skew symmetric matrix.
We can make a few observations about symmetric and skew symmetric matrices; the proofs of most of these statementswill be left as exercises. It follows from the definitions above that if A is symmetric or skew symmetric, then A is asquare matrix. If A is a symmetric matrix, then the entries of A are symmetric with respect to the main diagonalof A. Also, A is symmetric if and only if aij = aji and A is skew symmetric if and only if aij = โaji. Moreover, ifA is skew symmetric, then the entries on the main diagonal of A are all zero. An important property of symmetricand skew symmetric matrices is the following.
Theorem 1.4.1. If A is an n ร 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 = ST + KT = S โ K. Thus we have the expressions
A = S + K
AT = S โ K.
Adding these two expressions, we obtain A + AT = 2S, so
S =1
2(A + AT ).
Subtracting instead of adding leads to
K =1
2(A โ AT ).
It is easy to verify that A = S + K, that S is symmetric, and that K is skew symmetric. Thus we have shown thatsuch a representation is possible and that the expressions for S and K are unique. โโ
Example 1.4.5. Let A =
1 3 โ24 6 25 1 3
. Then
S =1
2(A + AT ) =
1 72
32
72 6 3
2
32
32 3
,
K =
0 โ 12 โ 7
2
12 0 1
2
72 โ 1
2 0
,
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 nonzeroreal number.
Definition 1.4.9. An nร n matrix A is called nonsingular, or invertible, if there exists an nร n matrix B suchthat AB = BA = In. Otherwise, A is called singular, or noninvertible; B is called an inverse of A.
Example 1.4.6. Let A =
(2 32 2
)
and let B =
(โ1 3
21 โ1
)
. Since AB = BA = I2, we conclude that B is an
inverse of A.
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 = InC = C, which prove that the inverse of a matrix, if it exits, is unique. โโ
We now write the inverse of a nonsingular matrix A, as Aโ1. Thus
AAโ1 = Aโ1A = In.
Example 1.4.7. Let
A =
(1 23 4
)
.
To find Aโ1 if it exists, we let
Aโ1 =
(a bc d
)
.
Then we must have
AAโ1 =
(1 23 4
) (a bc d
)
= I2 =
(1 00 1
)
,
so that (a + 2c b + 2d3a + 4c 3b + 4d
)
=
(1 00 1
)
.
Equating corresponding entries of these two matrices, we obtain the linear systems
a + 2c = 1 b + 2d = 0and
3a + 4c = 0 3b + 4d = 1.
The solutions are (you should verify this) a = โ1, c = 32 , b = 1, and d = โ 1
2 . Moreover, since the matrix
(a bc d
)
=
(โ2 1
32 โ 1
2
)
also satisfies the property that(โ2 1
32 โ 1
2
) (1 23 4
)
=
(1 00 1
)
,
we conclude that A is nonsingular and that
Aโ1 =
(โ2 1
32 โ 1
2
)
.
Example 1.4.8. Let
A =
(1 22 4
)
.
To find Aโ1 if it exists, we let
Aโ1 =
(a bc d
)
.
Then we must have
AAโ1 =
(1 22 4
) (a bc d
)
= I2 =
(1 00 1
)
,
so that (a + 2c b + 2d2a + 4c 2b + 4d
)
=
(1 00 1
)
.
Equating corresponding entries of these two matrices, we obtain the linear systems
a + 2c = 1 b + 2d = 0and
2a + 4c = 0 2b + 4d = 1.
These linear systems have no solutions, so A has no inverse.
We next establish several properties of inverses of matrices.
Theorem 1.4.3. If A and B are both nonsingular matrices in Mn(R), then the matrix product AB is nonsingularand (AB)โ1 = Bโ1Aโ1.
Proof. We have (AB)(Bโ1Aโ1) = A(BBโ1)Aโ1 = (AIn)Aโ1 = AAโ1 = In. Similarly, (Bโ1Aโ1)(AB) = In.Therefore, AB is nonsingular. Since the inverse of a matrix is unique, we conclude that (AB)โ1 = Bโ1Aโ1. โโ
Corollary 1.4.1. If A1, A2, . . . , Ar are nonsingular matrices in Mn(R), then the matrix product A1A2 ยท ยท ยท An isnonsingular and (A1A2 ยท ยท ยทAr)
โ1 = Aโ1r Aโ1
rโ1 ยท ยท ยทAโ11 .
Proof. An easy exercise. โโ
Theorem 1.4.4. If A is a nonsingular matrix in Mn(R), then Aโ1 is nonsingular and (Aโ1)โ1 = A.
Proof. An easy exercise. โโ
Theorem 1.4.5. If A is a nonsingular matrix in Mn(R), then AT is nonsingular and (AT )โ1 = (Aโ1)T .
Proof. We have AAโ1 = In. Taking transposes of both sides, we obtain (Aโ1)T AT = ITn = In. Taking
transposes of both sides of the equation Aโ1A = In, we find, similarly, that (AT )(Aโ1)T = In. These equationsimply that (Aโ1)T = (AT )โ1. โโ
Example 1.4.9. If
A =
(1 23 4
)
,
then from Example 1.4.7
Aโ1 =
(โ2 1
32 โ 1
2
)
and (Aโ1)T =
(โ2 3
21 โ 1
2
)
.
Also (you should verify this!)
AT =
(1 32 4
)
and (AT )โ1 =
(โ2 3
21 โ 1
2
)
.
It follows from Theorem 1.4.5 that if A is a symmetric nonsingular matrix, then Aโ1 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 impliesthat B = On (see Exercise 1.4.11).
LINEAR SYSTEMS AND INVERSES
If A is an n ร n matrix, then the linear system A X = B is a system of n equations in n unknowns. Suppose thatA is nonsingular. Then Aโ1 exists and we can multiply A X = B by Aโ1 on both sides, obtaining
Aโ1(A X) = Aโ1B,
orIn X = Aโ1B.
Moreover, X = Aโ1B is clearly a solution to the given linear system. Thus, if A is nonsingular, we have a uniquesolution.
Applications. This observation is useful in industrial problems. Many physical models are described by linearsystems. This means that if n values are used as inputs (which can be arranged as the n ร 1 matrix X), then mvalues are obtained as outputs (which can be arranged as the m ร 1 matrix B) by the rule A X = B. The matrixA 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 internalstructure of the process does not interest us. The problem frequently encountered in systems analysis is that ofdetermining the input to be used to obtain a desired output. That is, we want to solve the linear system A X = Bfor X as we vary B. If A is a nonsingular square matrix, an efficient way of handling this is as follows. ComputeAโ1 once; then whenever we change B, we find the corresponding solution X by forming Aโ1B.
Example 1.4.10. Consider an industrial process whose matrix is the matrix A of Example 1.4.7. If B is the outputmatrix (
86
)
,
then the input matrix X is the solution to the linear system A X = B. Using the result from Example 1.4.7, wehave
X = Aโ1B =
(โ2 1
32 โ 1
2
) (86
)
=
(โ10
9
)
.
On the other hand, if B is the output matrix(
1020
)
,
then
X = Aโ1
(1020
)
=
(05
)
.
SUGGESTED EXERCISES FOR SECTION 1.4
Exercise 1.4.1.(a) Show that if A is in Mmรn(R), then ImA = A and AIn = A.
(b) Show that if A is any scalar matrix in Mn(R), then A = ฮณ In for some real number ฮณ.
Exercise 1.4.2. Prove that the sum, product, and scalar multiple of diagonal, scalar, and upper (lower) triangularmatrices 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 = โ 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 ร n matrix, then(a) AAT and AT A are symmetric.
(b) A + AT is symmetric.
(c) A โ 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.
Exercise 1.4.7. Write the matrix A =
3 โ2 15 2 3โ1 6 2
as a sum of a symmetric and a skew symmetric ma-
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 35 2
)
.
Exercise 1.4.10.
Exercise 1.4.11. Show that if A is nonsingular and AB = On for an n ร n matrix B, then B = On.
Exercise 1.4.12. Let A =
(a bc d
)
. Show that A is nonsingular if and only if ad โ bc 6= 0
Exercise 1.4.13. Consider the linear system A X = B, where A is the matrix defined in Exercise 1.4.9.
(a) Find a solution if B =
(34
)
.
(b) Find a solution if B =
(56
)
.
Exercise 1.4.14. Prove that if A is symmetric and nonsingular, then Aโ1 is symmetric.
Exercise 1.4.15. Consider the homogeneous system A X = O, where A is n ร n. If A is nonsingular, show thatthe only solution is the trivial one, X = O.
Exercise 1.4.16. Prove that if one row (column) of the nรn matrix A consists entirely of zeros, then A is singular.(Hint: Assume that A is nonsingular, that is, there exists an nร n matrix B such that AB = BA = In. Establisha 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 =
(2 34 6
)
is singular.
Exercise 1.4.21. Let
A =
3 2 โ10 โ4 30 0 0
and B =
6 โ3 20 2 40 0 3
.
Verify that A + B and AB are upper triangular.
Exercise 1.4.22. Find two 2 ร 2 singular matrices whose sum is nonsingular.
Exercise 1.4.23. Find two 2 ร 2 nonsingular matrices whose sum is singular.
Exercise 1.4.24. If A is a nonsingular matrix whose inverse is
(2 14 1
)
, find A.
Exercise 1.4.25. Let p and q be nonnegative integers and let A be a square matrix. Show that
ApAq = Ap+q and (Ap)q = Apq.
Exercise 1.4.26. If AB = BA, and p is a nonnegative integer, show that (AB)p = ApBp.
Exercise 1.4.27. If p is a nonnegative integer and ฮฑ is a scalar, show that
(ฮฑ A)p = ฮฑpAp.
Exercise 1.4.28. Consider an industrial process with associated linear system A X = B, where A is nรn. Supposethat
Aโ1 =
(1 21 3
)
.
Find the input matrix for each of the following output matrices: (a)
(46
)
, (b)
(815
)
.
Exercise 1.4.29. If
D =
4 0 00 โ2 00 0 3
.
Find Dโ1.
Exercise 1.4.30. If
Aโ1 =
(3 21 3
)
and Bโ1 =
(2 53 โ2
)
,
find (AB)โ1.
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 itby introducing the language of matrices. This will result in two methods for solving a system of m linear equationsin 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 originallinear system). The important point here is that the latter linear system can be solved very easily.
For example, if
1 2 0 | 30 1 1 | 20 0 1 | โ1
represents the augmented matrix of a linear system, then the solution is easily found from the corresponding equations
x1 + 2x2 = 3x2 + x3 = 2
x3 = โ1.
The task of this section is to manipulate the augmented matrix representing a given linear system into a form fromwhich the solution can easily be found.
Definition 1.5.1. An m ร n matrix A is said to be in reduced row echelon form (r.r.e.f) if it satisfies thefollowing 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 mustbe 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 theleading 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 maybe 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) andcolumn echelon form (c.e.f).
Example 1.5.1. The following are matrices in row echelon form:
A =
1 5 0 2 โ2 40 1 0 3 4 80 0 0 1 7 โ20 0 0 0 0 00 0 0 0 0 0
, B =
1 0 0 00 1 0 00 0 1 00 0 0 1
,
and
C =
0 0 1 3 5 7 90 0 0 0 1 โ2 30 0 0 0 0 0 10 0 0 0 0 0 0
Example 1.5.2. The following are matrices in reduced row echelon form:
B =
1 0 0 00 1 0 00 0 1 00 0 0 1
, D =
1 0 0 0 โ2 40 1 0 0 4 80 0 0 1 7 โ20 0 0 0 0 00 0 0 0 0 0
and
E =
1 2 0 0 10 0 1 2 30 0 0 0 0
.
The following matrices are not in reduced row echelon form. (Why not?)
E =
1 2 0 40 0 0 00 0 1 โ3
, G =
1 0 3 40 2 โ2 50 0 1 2
,
H =
1 0 3 40 1 โ2 50 1 2 20 0 0 0
, J =
1 2 3 40 1 โ2 50 0 1 20 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 ฮฑ 6= 0.
(c) Add ฮฑ 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 areequivalent, respectively, to interchanging tow equations, multiplying an equation by a nonzero constant, and addinga multiple of one equation to another equation.
Definition 1.5.3. An mร n matrix A is said to be row (column) equivalent to an mร n matrix B if B can beobtained by applying a finite sequence of elementary row (column) operations to A.
Example 1.5.3. The matrix
A =
1 2 4 32 1 3 21 โ1 2 3
is row equivalent to
D =
2 4 8 61 โ1 2 34 โ1 7 8
,
because if we add twice row 3 of A to row 2 of A, we obtain
B =
1 2 4 34 โ1 7 81 โ1 2 3
.
Interchanging rows 2 and 3 of B, we obtain
C =
1 2 4 31 โ1 2 34 โ1 7 8
,
Multiplying row 1 of C by 2, we obtain D.
We can easily show (see Exercise 1.5.1) that:
(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 โA is row equivalent to Bโ and โB is row equivalent to Aโ can be replaced by โAand B are row equivalent.โ A similar statement holds for column equivalence.
Theorem 1.5.1. Every nonzero m ร 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 elementaryrow operations we can transform A into a matrix in row echelon form. A completely analogous proof using elementarycolumn 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 saythat this first nonzero entry in column j occurs in row i; that is, entry aij 6= 0. Now interchange (ifnecessary), 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 โchj times row 1; all elementsin column j, rows 2, 3, . . . , m are zero. Denote the resulting matrix by D. Note that we have used onlyelementary row operations.
Step 4: Next, consider the (m โ 1) ร n submatrix A1 of D obtained by mentally deleting the firstrow of D.
We now repeat the four step procedure above with submatrix A1 instead of matrix A. Continuing this way, weobtain a matrix H in row echelon form which is row equivalent to A. โโ
Example 1.5.4. Let
A =
0 2 3 โ4 10 0 2 3 42 2 โ5 2 42 0 โ6 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 interchangethe first and third rows of A to produce a matrix B given as
B =
2 2 โ5 2 40 0 2 3 40 2 3 โ4 12 0 โ6 9 7
.
Step 2: Multiply the first row of B by 1b11
= 12 , to produce a matrix C given as
C =
1 1 โ 52 1 2
0 0 2 3 40 2 3 โ4 12 0 โ6 9 7
Step 3: Add โ2 times the first row of C to the fourth row of C to produce a matrix D in which the only nonzeroentry in the first column is d11 = 1.
D =
1 1 โ 52 1 2
0 0 2 3 40 2 3 โ4 10 โ2 โ1 7 3
Step 4: Identify A1 as the submatrix A1 obtained by mentally deleting the first row of D; do not physically erasethe first row of D.
Repeat the four steps above using the submatrix A1 which we write below with the first row of D separated fromits submatrix A1.
1 1 โ 52 1 2
A1 =
0 0 2 3 40 2 3 โ4 10 โ2 โ1 7 3
Interchange the first and Second rows of A1 to obain
1 1 โ 52 1 2
B1 =
0 2 3 โ4 10 0 2 3 40 โ2 โ1 7 3
Multiply the first row of B1 by 1/2 to obain
1 1 โ 52 1 2
C1 =
0 1 32 โ2 1
20 0 2 3 40 โ2 โ1 7 3
Add two times the first row of C1 to the third row to obain
1 1 โ 52 1 2
D1 =
0 1 32 โ2 1
20 0 2 3 40 0 2 3 4
Identify A2 as the submatrix A2 obtained by mentally deleting the first row of D1; do not physically erase the firstrow of D1.
Repeat the four steps above using the submatrix A2 which we write below with the first row of D1 separated fromits submatrix A2. No row of A2 needs to be interchanged; So then we have B2 = A2.
1 1 โ 52 1 2
0 1 32 โ2 1
2
A2 =
(0 0 2 3 40 0 2 3 4
)
= B2 Multiply the first row of B2 by 1/2 to obtain
1 1 โ 52 1 2
0 1 32 โ2 1
2
C2 =
(0 0 1 3
2 20 0 2 3 4
)
Finally, add -2 times the first row of C2 to its second row to obtain
1 1 โ 52 1 2
0 1 32 โ2 1
2
D2 =
(0 0 1 3
2 20 0 0 0 0
)
.
The matrix
H =
1 1 โ 52 1 2
0 1 32 โ2 1
2
0 0 1 32 2
0 0 0 0 0
is in row echelon form and is row equivalent to A. โโWhen doing hand computations, it is sometimes possible to avoid messy fractions by suitably modifying the stepsin the procedure.
Theorem 1.5.2. Every nonzero m ร 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.Letโs suppose this leading one is in column j which we denote by cj . Then c1 < c2 < ยท ยท ยท < cj < . . . < cr wherer (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 Hpreceding row i to make all entries in column cj and rows iโ 1, i โ 2, . . . , 1 of H equal to zero; except the leadingones. The result is a matrix K in reduced row echelon form which has been obtained from H by elementary rowoperations 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 inreduced column echelon form. โโ
It can be shown, with some difficulty, that:
For a given nonzero mร n matrix A,there is only one matrix B in reduced row (column) echelon formthat 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 thematrix A of Example 1.5.4. Starting with the matrix H obtained there, we add โ1 times the second row to the firstrow, obtaining
1 0 โ4 3 32
0 1 32 โ2 1
20 0 1 3
2 20 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. Thisyields
1 0 0 9 192
0 1 0 โ 174 โ 5
20 0 1 3
2 20 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 theaugmented matrices (A | B) and (C | D) are row equivalent, then the linear systems are equivalent; that is, theyhave exactly the same solutions.
Proof. This follows from the definition of row equivalence and from the fact that the three elementary rowoperations on the augmented matrix are the three manipulations on linear systems, discussed in Section 1.1, whichyield equivalent linear systems. We also note that if one system has no solution, then the other system has nosolution. โโ
Corollary 1.5.1. If A and B are row equivalent m ร n matrices, then the homogeneous systems A X = O andB X = O are equivalent.
Proof. An easy exercise. โโ
We now pause to observe that we have developed the essential features of two very straightforward methods forsolving linear systems. The idea consists of starting with the linear system A X = B, then obtaining a partitionedmatrix (C | D) in either row echelon form or reduced row echelon form that is row equivalent to the augmentedmatrix (A | B). Now (C | D) represents the linear system C X = D, which is quite simple to solve because of thestructure of (C | D), and the set of solutions to this system gives precisely the set of solutions to A X = B. Themethod where (C | D) is in row echelon form is called Gaussian elimination; the method where (C | D) is inreduced row echelon form is called Gauss-Jordan reduction. These methods are used often and computer codesof their implementation are widely available.
We thus consider the linear system C X = D, where C is m ร n, and (C | D) is in row echelon form. Then forexample, (C | D) is of the following form:
1 c12 c13 . . . c1n | d1
0 0 1 c24 . . . c2n | d2
...... |
...0 0 . . . 0 1 c(kโ1)n | dkโ1
0 . . . 0 1 | dk
0 . . . 0 | dk+1
...... |
...0 . . . 0 | dm
.
This augmented matrix represents the linear system
x1 + c12x2 + c13x3 + . . . + c1nxn = d1
x3 + c24x4 + . . . + c2nxn = d2...
xnโ1 + c(kโ1)nxn = dkโ1
xn = dk
0x1 + . . . + 0xn = dk+1
......
0x1 + . . . + 0xn = dm.
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 = ยท ยท ยท = dm = 0, we then obtain xn = dk, xnโ1 = dkโ1 โ c(kโ1)nxn =dkโ1 โ c(kโ1)ndk, and continue using backward substitution to find the remaining unknowns corresponding to theleading entry in each row. Of course, in the solution, some of the unknowns may be expressed in terms of othersthat can take on any values whatsoever. This merely indicates that C X = D has infinitely may solutions. On theother 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
(C... D) =
1 2 3 4 5 | 60 1 2 3 โ1 | 70 0 1 2 3 | 70 0 0 1 2 | 9
.
Thenx4 = 9 โ 2x5
x3 = 7 โ x3 โ 2x4 โ 3x5 = 7 โ 2(9 โ 2x5) โ 3x5 = โ11 + x5
x2 = 7 โ 2x3 โ 3x4 + x5 = 2 + 5x5
x1 = 6 โ 2x2 โ 3x3 โ 4x4 โ 5x5 = โ1 โ 10x5
x5 = any real number.
Thus all solutions X are of the form
x1 = โ1 โ 10r
x2 = 2 + 5r
x3 = โ11 + r
x4 = 9 โ 2r
x5 = r, any real number.
or X =
โ12
โ1190
+
โ1051โ21
r, where r is any real number.
Since r can be assigned any real number, the given linear system as infinitely many solutions.
Example 1.5.7. If
(C | D) =
1 2 3 4 | 50 1 2 3 | 60 0 0 0 | 1
is the augmented matrix in row echelon form of the linear system A X = B,then linear system C X = D has nosolution, 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
(C | D) =
1 2 3 4 | 50 1 2 3 | 60 0 1 2 | 70 0 0 1 | 8
,
thenx1 = 0
x2 = 0
x3 = โ9
x4 = 8. or X =
00
โ98
.
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, ofcourse, it takes more effort to put a matrix in reduced row echelon form than in row echelon form.
Example 1.5.9. If
(C | D) =
1 0 0 0 | 50 1 0 0 | 60 0 1 0 | 70 0 0 1 | 8
,
then
X =
5678
.
Example 1.5.10. If
(C | D) =
1 1 2 0 โ 52 | 3
20 0 0 1 1
2 | 12
0 0 0 0 0 | 0
,
then
x4 =1
2โ 1
2x5
x1 =2
3โ x2 โ 2x3 +
5
2x5,
where x2, x3, an x5 can take on any real numbers. Thus a solution is of the form
x1 =2
3โ r โ 2s +
5
2t
x2 = r
x3 = s
x4 =1
2โ 1
2t
x5 = t,
or in matrix form as
X =
2300120
+
โ11000
r +
โ20100
s +
5200
โ 121
t 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 + 2x2 + 3x3 = 62x1 โ 3x2 + 2x3 = 143x1 + x2 โ x3 = โ2.
Solution. We form the augmented matrix of the system as
1 2 3 | 62 โ3 2 | 143 1 โ1 | โ2
.
Next, we start the Gaussian elimination method by applying the elementary row operations to the above augmentedmatrix.
1 2 3 | 62 โ3 2 | 143 1 โ1 | โ2
Add -2 times the first row to the second row to obtain
1 2 3 | 60 โ7 โ4 | 23 1 โ1 | โ2
Add -3 times the first row to the third row to obtain
1 2 3 | 60 โ7 โ4 | 20 โ5 โ10 | โ20
Multiply the third row by โ1/5 and interchange the second and third rows to obtain
1 2 3 | 60 1 2 | 40 โ7 โ4 | 2
Add 7 times the second row to the third row to obtain
1 2 3 | 60 1 2 | 40 0 10 | 30
Multiply the third row by 1/10 to obtain
1 2 3 | 60 1 2 | 40 0 1 | 3
.
The above matrix is in row echelon form. This means that x3 = 3 and from the second row
x2 + 2x3 = 4,
sox2 = 4 โ 2(3) = โ2.
From the first rowx1 + 2x2 + 3x3 = 6,
which implies thatx1 = 6 โ 2x2 โ 3x3 = 6 โ 2(โ2) โ 3(3) = 1.
Thus x1 = 1, x2 = โ2, and x3 = 3 is the solution to the system. This gives the solution by Gaussian elimination.โโIf, instead, we wish to use Gauss-Jordan reduction, we would transform the last matrix into reduced row echelonform by the following steps:
Add -2 times the second row to the first row:
1 0 โ1 | โ20 1 2 | 40 0 1 | 3
Add -2 times the third row to the second row to obtain
1 0 โ1 | โ20 1 0 | โ20 0 1 | 3
Add the third row to the first row to obtain
1 0 0 | 10 1 0 | โ20 0 1 | 3
.
The solution is x1 = 1, x2 = โ2, and x3 = 3, as before.
NOW WE CONSIDER A HOMOGENEOUS SYSTEM A X = O CONSISTING OF m EQUATIONSIN n UNKNOWNS.
(A | O) =
a11 a12 ยท ยท ยท a1n | 0a21 a22 ยท ยท ยท a2n | 0...
......
... |...
am1 am2 ยท ยท ยท amn | 0
Example 1.5.12. Consider the homogeneous system whose augmented matrix is
1 0 0 0 2 | 00 0 1 0 3 | 00 0 0 1 4 | 00 0 0 0 0 | 0
.
Since the augmented matrix is in reduced row echelon form, the solution is easily seen as
x1 = โ2rx2 = s
x3 = โ3rx4 = โ4rx5 = r,
or X =
โ20
โ3โ41
r +
01000
s, where r and s are any real numbers.
In Example 1.5.12 we solved a homogeneous system of m (= 4) linear equations in n(= 5) unknowns, where m < nand the augmented matrix A was in reduced row echelon form. We can ignore any row of the augmented matrix thatconsists entirely of zeros. Thus let rows 1, 2, . . . , r of A be the nonzero rows, and let the 1 in row i occur in columnci. We are then solving a homogeneous system of r equations in n unknowns, r < n, and in this special case (A isin reduced row echelon form) we can solve for xc1
, xc2, . . . , xcr
in terms of the remaining n โ r unknowns. Since thelatter can take on any real values, there are infinitely many solutions to the system A X = O; in particular, thereis a nontrivial solution. We now show that this situation holds whenever we have m < n; A does not have to be inreduced row echelon form.
Theorem 1.5.4. Any homogeneous system A X = O consisting of m linear equations in n unknowns always has anontrivial 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 systemsA 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 ofthe remaining n โ r unknowns, the latter being free to take on any values we please. Thus B X = O, and henceA X = O has a nontrivial solution. โโ
We shall soon use this result in the following equivalent form: If A is m ร n and A X = O has only the trivialsolution, then m > n.
Example 1.5.13. Consider the homogeneous system
x1 + x2 + x3 + x4 = 0x1 x4 = 0x1 + 2x2 + x3 = 0.
The augmented matrix
A =
1 1 1 1 | 01 0 0 1 | 01 2 1 0 | 0
is row equivalent to
1 0 0 1 | 00 1 0 โ1 | 00 0 1 1 | 0
.
Hence the solution isx1 = โrx2 = r
x3 = โrx4 = r
or X =
โ11
โ11
r, any real number.
A useful property of matrices in reduced row echelon form (see Exercise 1.5.3) is that if A is an n ร n matrix inreduced 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
A =
0 0 โ1 2 30 2 3 4 50 1 3 โ1 20 3 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 ร n matrix in reduced row echelon form. Prove that if A 6= In, then A has a rowconsisting entirely of zeros.
Exercise 1.5.4. Let
A =
1 โ2 0 22 โ3 โ1 51 3 2 51 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 + x2 + 2x3 = โ1x1 โ 2x2 + x3 = โ53x1 + x2 + x3 = 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 + x2 + 2x3 + 3x4 = 13x1 โ 2x2 + x3 + x4 = 83x1 + x2 + x3 โ x4 = 1.
(b)x1 + x2 + x3 = 1x1 + x2 โ x3 = 32x1 + x2 + x3 = 2.
(c)
2x1 + x2 + x3 โ 2x4 = 13x1 โ 2x2 + x3 โ 6x4 = โ2x1 + x2 โ x3 โ x4 = โ16x1 + x3 โ 9x4 = โ25x1 โ x2 + 2x3 โ 8x4 = 3.
Exercise 1.5.7. In Exercises 7, 8, and 9, solve the linear system, if it is consistent, with given augmented matrix.
(a)
1 1 1 | 01 1 0 | 30 1 1 | 1
(b)
1 2 3 | 01 1 1 | 01 1 2 | 0
(c)
1 2 3 | 01 1 1 | 05 7 9 | 0
(d)
(1 2 3 | 01 2 1 | 0
)
Exercise 1.5.8. (a)
1 2 3 1 | 81 3 0 1 | 71 0 2 1 | 3
(b)
1 1 3 โ3 | 00 2 1 โ3 | 31 0 2 โ1 | โ1
Exercise 1.5.9. (a)
1 2 1 | 72 0 1 | 41 0 2 | 51 2 3 | 112 1 4 | 11
(b)
1 2 1 | 02 3 0 | 00 1 2 | 02 1 4 | 0
Exercise 1.5.10. Let A =
(a bc d
)
and X =
(x1
x2
)
. Show that the linear system A X = O has only the trivial
solution if and only if ad โ bc 6= 0.
Exercise 1.5.11. Show that A =
(a bc d
)
is row equivalent to I2 if and only if ad โ bc 6= 0.
Exercise 1.5.12. In the following linear system, determine all values of ฮป for which the resulting linear system has:(1) No solution.
(b) A unique solution.
(c) Infinitely may solutions.x1 + x2 โ x3 = 2x1 + 2x2 + x3 = 3x1 + x2 + (ฮป2 โ 5)x3 = ฮป.
Exercise 1.5.13. Repeat Exercise 1.5.12 for the linear system
x1 + x2 + x3 = 22x1 + 3x2 + 2x3 = 52x1 + 3x2 + (ฮป2 โ 1)x3 = ฮป + 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 ร n nonzero matrix is column equivalent to a matrix in column echelon form.
Exercise 1.5.15. Prove that every m ร n nonzero matrix is column equivalent to a matrix in reduced columnechelon form.
Exercise 1.5.16. Repeat Exercise 1.5.12 for the linear system
x1 + x2 = 3x1 + (ฮป2 โ 8)x2 = ฮป.
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 + x2 + x3 = 2x1 + 2x2 + x3 = 3x1 + x2 + (ฮป2 โ 5)x3 = ฮป.
Exercise 1.5.19. Repeat Exercise 1.5.17 for the matrix
1 2 3 4 52 1 3 โ1 23 1 2 4 1
.
Exercise 1.5.20. Show that if the homogeneous system
(a โ r)x1 + dx2 = 0cx1 + (b โ r)x2 = 0
has a nontrivial solution, then r satisfies the equation (a โ r)(b โ r) โ 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 toA X = B and Y1 is a solution to A X = O. [Hint: Let X = X1 + (X โ 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 Aโ1
In this section we develop a method for finding the inverse, Aโ1 of a square matrix A if it exists. This method issuch that we do not have to first find out whether Aโ1 exists. Give the square matrix A, we start to find Aโ1, ifin the course of the computation we hit a certain condition, then we know that Aโ1 does not exist. Otherwise, weproceed to the end and obtain Aโ1. 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 nรn matrix E is called an elementary matrix if it was obtained by performing elementaryrow or elementary column operations on In.
Example 1.6.1. The following are elementary matrices:
E1 =
0 0 10 1 01 0 0
, E2 =
1 0 00 โ2 00 0 1
,
E3 =
1 2 00 1 00 0 1
, and E4 =
1 0 30 1 00 0 1
.
Matrix E1 was obtain by interchanging the first and third rows of I3; E2 was obtained by multiplying the secondrow 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 ร 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 elementaryrow (column) operation as was performed on A. Then B = EA (B = AE).
Proof. An easy exercise. โโ
Theorem 1.6.1 says that an elementary row operation on A can be achieved by pre-multiplying A (multiplying Aon the left) by the corresponding elementary matrix E; an elementary column operation on A can be obtained bypost-multiplying A (multiplying A on the right) by the corresponding elementary matrix.
Example 1.6.2. Let A =
1 3 2 1โ1 2 3 4
3 0 1 2
and let B result from A by adding -2 times the third row of A to its
first row. Thus B =
โ5 3 0 โ3โ1 2 3 43 0 1 2
. Now let E be the elementary matrix that is obtained from I3 by adding
-2 times the third row of I3 to the its first row. Thus E =
1 0 โ20 1 00 0 1
. It is easy to verify that B = EA.
Theorem 1.6.2. If A and B are nonzero m ร n matrices, then A is row (column) equivalent to B if and only ifB = EkEkโ1 ยท ยท ยท , E2E1A (B = AE1E2 ยท ยท ยท Ekโ1Ek), where E1, E2, ยท ยท ยท , Ekโ1, 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 asequence of elementary row operations. This implies that there exist elementary matrices E1, E2, . . . , Ek such thatB = EkEkโ1 ยท ยท ยท E2E1A. Conversely, if B = EkEkโ1 ยท ยท ยท , E2E1A, where the Eโฒ
is are elementary matrices, then Bresults from A by a sequence of elementary row operations, which implies that A is row equivalent to B. โโ
Theorem 1.6.3. An elementary matrix E is nonsingular and its inverse, Eโ1 is an elementary matrix of the sametype.
Proof. An easy exercise. โโ
Thus an elementary row operation can be โundoneโ by another elementary row operation of the same type.
We now obtain an algorithm for finding the inverse, Aโ1 (if it exists) of a given square matrix A; first, we prove thefollowing lemma.
Lemma 1.6.1. Let A be an n ร n matrix and let the homogeneous system A X = O have only the trivial solutionX = 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 systemA 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 isthe number of nonzero rows of B, then the homogeneous system B X = O is equivalent to the homogeneous systemwhose coefficient matrix consists of the nonzero rows of B and is therefore rรn. Since this last homogeneous systemonly has the trivial solution, we conclude from Theorem 1.5.4 that r > n. Since B is n ร n, r โค n. Hence r = n,which means that B has no zero rows. Thus B = In. โโ
Theorem 1.6.4. An n ร n matrix A is nonsingular if and only if A is a product of elementary matrices (a finitenumber).
Proof. If A is a product of finitely many elementary matrices E1, E2, . . . , Ek, then A = E1E2 ยท ยท ยท , Ekโ1Ek.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 Aโ1(A X) = Aโ1(O) = O, soIn X = O or X = O. Thus A X = O has only the trivial solution. Lemma 1.6.1 then implies that A is rowequivalent to In. This means that there exists elementary matrices E1, E2, . . . , Ek such that
In = EkEkโ1 ยท ยท ยท E2E1A.
It then follows that A = (EkEkโ1 ยท ยท ยท E2E1)โ1 = Eโ1
1 Eโ12 ยท Eโ1
kโ1Eโ1k . Since the inverse of an elementary matrix
is an elementary matrix, we have established the result. โโ
Corollary 1.6.1. An n ร 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 = EkEkโ1 ยท ยท ยท E2E1A, where the Ei are all elementary matrices.Therefore, it follows that A = Eโ1
1 Eโ12 ยท ยท ยท Eโ1
kโ1Eโ1k . 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 = E (a product of finitely manyelementary matrices). Now A = A In = EIn which implies that A is row equivalent to In. โโ
Theorem 1.6.5. The homogeneous system of n linear equations in n unknowns A X = O has a nontrivial solutionif 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, whereA is an n ร n matrix, has only the trivial solution X = O, then A is nonsingular. Conversely, consider A X = O,where A is nรn, and let A be nonsingular. Then Aโ1 exists and we form Aโ1(A X) = Aโ1O = O. Thus X = O,which means that the homogeneous system has only the trivial solution. โโ
Example 1.6.3. Let A =
(1 22 4
)
be the matrix defined in Example 1.4.8, which is singular; i.e., Aโ1 does not
exist. Consider the homogeneous system AX = 0; that is,
(1 22 4
) (x1
x2
)
=
(00
)
. The reduced row echelon form
of the augmented matrix is
(1 2 | 00 0 | 0
)
, and so a solution is
x1 = โ2t
x2 = t or X =
โ2
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
A = Eโ11 Eโ1
2 ยท ยท ยท Eโ1kโ1E
โ1k ,
from which it follows that
Aโ1 = (Eโ11 Eโ1
2 ยท ยท ยท Eโ1kโ1E
โ1k )โ1 = EkEkโ1 ยท ยท ยท E2E1.
This now provides an algorithm for finding Aโ1. Thus we perform elementary row operations on A until we get In;the product of the elementary matrices EkEkโ1 ยท ยท ยท E2E1 then gives Aโ1.
An Algorithm for find Aโ1: A convenient way of organizing the computing process is to write down the augmentedmatrix (A | In). Then
(EkEkโ1 ยท ยท ยท E2E1)(A | In) = (EkEkโ1 ยท ยท ยท E2E1A | EkEkโ1 ยท ยท ยท E2E1) = (In | Aโ1).
In other words, for a given n ร n matrix A the algorithm for finding Aโ1 as follows:
Step 1. Form the augmented n ร 2n matrix (A | In) where In is the n ร 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 | Aโ1)and Aโ1 is the n ร n submatrix which remains after deleting In from (In | Aโ1).
Otherwise, there is either a row consisting entirely of zeros or else there is a leading one in a columnbeyond the nth column (Remember the matrix (A | In) has 2n columns!). In this case A is consideredto be singular.
Example 1.6.4. For the 3 ร 3 matrix A =
1 1 10 2 35 5 1
. Find Aโ1 if exists.
Solution.
Step 1. Form the 3 ร 6 augmented matrix (A | I3)
(A | I3) =
1 1 1 | 1 0 00 2 3 | 0 1 05 5 1 | 0 0 1
.
Step 2. Bring the augmented matrix above to reduced row echelon form. Here are the elementary row operationsthat were used:
1. โ5R1 + R4.
2. 12R2.
3. โR2 + R1.
4. โ 14R3.
5. โ 32R3 + R2.
6. 12R3 + R1.
The reduced row echelon form of (A | I3) is
(I3 | Aโ1) =
1 0 0 | 138 โ 1
2 โ 18
0 1 0 | โ 158
12
38
0 0 1 | 54 0 โ 1
4
.
Step 3. Hence
Aโ1 =
138 โ 1
2 โ 18
โ 158
12
38
54 0 โ 1
4
.
It is easy to verify that AAโ1 = Aโ1A = I3.
The question that arise at this point is how can the algorithm tell us if A is singular? that is, when the abovealgorithm for finding Aโ1 fails. In step 3 of the algorithm we said if the reduced row echelon form of (A | In) hasat least one row consisting entirely of zeros or else there is a column beyond the nth column (counting from left toright) containing a leading one. This really means that A is row equivalent to some matrix B containing a row ofzeros. Thus the answer is that A is singular if and only if A is row equivalent to a matrix B having at least one rowthat consists entirely of zeros. We now prove this result.
Theorem 1.6.6. An n ร n matrix A is singular if and only if A is row equivalent to a matrix B that has a row ofzeros.
Proof. First, let A be row equivalent to a matrix B that has a row consisting entirely of zeros. From Exercise 1.4.16in section 4 it follows that B is singular. Now B = EkEkโ1 ยท ยท ยท E2E1A, where E1, E2, . . . , Ekโ1, Ek are elementarymatrices. 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 amatrix B 6= In, which is in reduced row echelon form. From Exercise 1.5.3 of section 5 it follows that B must havea row of zeros. โโ
This means that in order to find Aโ1, we do not have to determine, in advance, whether or not it exists. We merelystart to calculate Aโ1; if at any point in the computation we find a matrix B that is row equivalent to A and has arow of zeros, then Aโ1 does not exist.
Example 1.6.5. Consider the matrix A =
1 2 โ31 โ2 15 โ2 โ3
. Determine if A is singular or not.
Solution. Apply the algorithm for calculating Aโ1. After applying the three elementary row operations โR1 +
R2, โ5R1 + R3, โ3R2 + R3 to (A | I3) we find that A is row equivalent to the matrix B =
1 2 โ30 โ4 40 0 0
.
Since B has a row of zeros, we stop and conclude that A is a singular matrix. โโ
In section 4 we defined an n ร n matrix B to be the inverse of the n ร n matrix A if AB = In and BA = In. Wenow show that one of these equations follows from the other.
Theorem 1.6.7. If A and B are n ร n matrices such that AB = In, then BA = In. Thus B = Aโ1.
Proof. We first show that if AB = In, then A is nonsingular. Suppose that A is singular. Then A is row equivalentto a matrix C with a row of zeros. Now C = EkEkโ1 ยท ยท ยท E2E1A, where all the Ei are elementary matrices. ThenCB = EkEkโ1 ยท ยท ยท E2E1AB, so AB is row equivalent to CB. Since CB has a row of zeros, we conclude fromTheorem 1.6.6 that AB is singular. Then AB = In is impossible, because In is nonsingular. This contradictionshows that A is nonsingular, and so Aโ1 exists. Multiplying both sides of the equation AB = In by Aโ1 on the left,we then obtain Aโ1(AB) = (Aโ1A)B = InB = B = Aโ1In = Aโ1. Thus B = Aโ1 and BA = Aโ1A = In
holds. โโThis theorem tells us that when using multiplication to verify that matrix B is the inverse of A it is enough to verifyonly 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 ร 3 matrix. Find the elementary matrix E, which as a premultiplier of A, that is asEA, 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 3ร 4 matrix. Find the elementary matrix F , which as a postmultiplier of A, that is asAF , 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 variouselementary row operations.)
Exercise 1.6.5. Find the inverse, if it exists, of
(a)
1 1 11 2 30 1 1
(b)
1 1 1 11 2 โ1 21 โ1 2 11 3 3 2
(c)
1 1 1 11 3 1 21 2 โ1 15 9 1 6
(d)
1 2 11 3 21 0 1
(e)
1 2 21 3 21 1 3
Exercise 1.6.6. Let A =
(a bc d
)
be a 2 ร 2 matrix and denote by det(A) the real number ad โ bc. Then the
statement A is nonsingular if and only if det(A) 6= 0 holds. Show that
Aโ1 =1
det(A)
(d โb
โc a
)
.
provided that the statement holds.
Exercise 1.6.7. Let A =
2 3 โ11 0 30 2 โ3
โ2 1 3
. Find the elementary matrix that as a postmultiplier of A performs the
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.
Exercise 1.6.8. Prove that A =
1 2 30 1 21 0 3
is nonsingular and write it as a product of elementary matrices.
(Hint: First, write the inverse as a product of elementary matrices then use Theorem 1.6.3.)
Exercise 1.6.9. Which of the following homogeneous systems have a nontrivial solution?
a)x1 + 2x2 + 3x3 = 0
2x2 + 2x3 = 0x1 + 2x2 + 3x3 = 0.
(b)2x1 + x2 โ x3 = 0x1 โ 2x2 โ 3x3 = 0
โ3x1 โ x2 + 2x3 = 0.
(c)3x1 + x2 + 3x3 = 0โ2x1 + 2x2 โ 4x3 = 02x1 โ 3x2 + 5x3 = 0.
Exercise 1.6.10. Find out which of the following matrices are singular. For the nonsingular ones find the inverse.
(a)
(1 32 6
)
(b)
(1 3
โ2 6
)
(c)
1 2 31 1 20 1 2
(d)
1 2 31 1 20 1 1
.
Exercise 1.6.11. Find the inverse of A =
(1 32 4
)
.
Exercise 1.6.12. Find the inverse of A =
1 2 30 2 31 2 4
.
Exercise 1.6.13. Show that A =
(1 23 4
)
is nonsingular and write it as a product of elementary matrices (See
the hint in Exercise 1.6.8.)
Exercise 1.6.14. Find the inverse (if possible) of the following matrices:
(a)
1 2 โ3 1โ1 3 โ3 โ22 0 1 53 1 โ2 5
(b)
3 1 22 1 21 2 2
(c)
1 2 31 1 21 1 0
(d)
2 1 30 1 21 0 3
.
Exercise 1.6.15. Find the inverse of A =
1 1 2 10 โ2 0 01 2 1 โ20 3 2 1
Exercise 1.6.16. If A is a nonsingular matrix whose inverse is
(4 21 1
)
, find A.
Exercise 1.6.17. Prove that two mรn matrices A and B are row equivalent if and only if there exists a nonsingularmatrix P such that B = PA (Hint: Use Theorem 1.6.2 and Theorem 1.6.4.)
Exercise 1.6.18. Let A and B be row equivalent n ร n matrices. Prove that A is nonsingular if and only if B isnonsingular.
Exercise 1.6.19. Let A and B be n ร n matrices. Show that if AB is nonsingular, then A and B must benonsingular. (Hint: Use Theorem 1.6.5.)
Exercise 1.6.20. Let A be an m ร n matrix. Show that A is row equivalent to Omรn if and only if A = Omรn.
Exercise 1.6.21. Let A and B be mร n matrices. Show that A is row equivalent to B if and only if AT is columnequivalent to BT .
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 ฮป for which the inverse of
A =
1 1 01 0 01 2 ฮป
exists. What is Aโ1?
Exercise 1.6.24.(a) Is (A + B)โ1 = Aโ1 + Bโ1?
(b) Is (ฮณ A)โ1 = 1ฮณ Aโ1?
Exercise 1.6.25. For what value(s) of ฮป does the homogeneous system
(ฮป โ 1)x1 + x2 = 02x1 + (ฮป โ 1)x2 = 0
have a nontrivial solution?
1.7. EQUIVALENT MATRICES
โ WE WILL NOT DISCUSS IT IN CLASS โ
Chapter 2.
REAL VECTOR SPACES
VECTORS IN THE PLANEโA REVIEW
In many applications we deal with measurable quantities, such as pressure, mass, and speed, which can be completelydescribed by giving their magnitude. They are called scalars and will be denoted by lowercase Latin letters. Thereare many other measurable quantities, such as velocity, force, and acceleration, which require for their descriptionnot 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 elementaryphysics 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-axisis usually taken in a horizontal position. The other line, the y-axis, is then taken in a vertical position. We nowchoose 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 andpositive directions on the x- and y-axes. Frequently, but not always, these points are chosen so that they are bothequidistant from O, that is, so that the same unit of length is used for both axes. The x and y axes together arecalled coordinate axes and they form a rectangular coordinate system or a Cartesian coordinate system.
x1
x2
each point P in the plane we associate an ordered pair (x, y) of real numbers, its coordinates. Conversely, we canassociate a point in the plane with each ordered pair of real numbers. The point P with coordinates (x, y) is denotedP (x, y), or simply as (x, y). The set of all points in the plane is denoted by R
2, it is called 2-space.
Consider the 2 ร 1 matrix
ฮฑ =
[xy
]
,
where x and y are real numbers (note the use of square brackets instead of round ones used in the previous chapter).With ฮฑ 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 byโโOP ; 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 itshead. 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 segmentโโOP with tail O(0, 0) and head P (x, y) we
can associate the 2 ร 1 matrix[
xy
]
.
A vector in the plane is a 2ร1 matrix ฮฑ =
[xy
]
, where x and y are real numbers, which are called the components
of ฮฑ. 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 segmentwe can associate a vector. Frequently, the notions of directed line segment and vector are used interchangeably, anda directed line segment is called a vector. Since a vector is a matrix, the vectors
ฮฑ1 =
[x1
y1
]
and ฮฑ2 =
[x2
y2
]
are said to be equal if x1 = x2 and y1 = y2. That is, two vectors are equal if their respective components are
equal. With each vector ฮฑ =
[xy
]
we can associate the unique point P (x, y); conversely, with each point P (x, y) we
associate the unique vector
[xy
]
. Thus we also write the vector ฮฑ as (x, y). Of course, this association is carried out
by means of the directed line segmentโโOP , 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 dependingupon the context, we sometimes take R
2 as the set of all ordered pairs (x, y) and sometimes as the set of all 2 ร 1
matrices
[xy
]
(or directed line segments).
Let
ฮฑ1 =
[x1
y1
]
and ฮฒ =
[x2
y2
]
be two vectors in the plane. The sum of the vectors ฮฑ and ฮฒ is the vector
ฮฑ + ฮฒ =
[x1 + y1
x2 + y2
]
.
We can interpret vector addition geometrically, as follows. In the figure below, the directed line segment ฮณ is parallelto ฮฒ, it has the same length as ฮฒ, and its tail is the head (x1, y1) of ฮฑ so its head is (x1 + y1, y1 + y2). Thus thevector with tail at 0 and head at (x1 + y1, x2 + y2) is ฮฑ + ฮฒ. We can also describe ฮฑ + ฮฒ as the diagonal of theparallelogram defined by ฮฑ and ฮฒ, as shown in the diagram.
โLinear Algebra To be Continuedโ
INTENTIONALLY LEFT BLANK
Chapter 3 INTRODUCTON TO INTEGRATION IN VECTOR FIELDS
CURVES, SURFACES, LINE INTEGRALS, VECTOR FIELDS, GREENโS THEOREM, SURFACE INTEGRALS AND STOKEโS 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 ๐ฆ๐ฆ โ 2๐ฅ๐ฅ = 1 (even though this is not 'curved'!), or a circle, say, ๐ฅ๐ฅ2 + ๐ฆ๐ฆ2 = 1, or perhaps a parabola, say ๐ฆ๐ฆ โ ๐ฅ๐ฅ2 = 0.
๐ฆ๐ฆ โ 2๐ฅ๐ฅ = 1 ๐ฆ๐ฆ โ ๐ฅ๐ฅ2 = 0 ๐ฅ๐ฅ2 + ๐ฆ๐ฆ2 = 1
All of these curves are described by means of their cartesian equation
๐๐(๐ฅ๐ฅ,๐ฆ๐ฆ) = ๐๐
where, ๐๐ is a function of two variable ๐ฅ๐ฅ and ๐ฆ๐ฆ and ๐๐ is a constant. From this point of view, a curve is a set of points, namely
โ = {(๐ฅ๐ฅ,๐ฆ๐ฆ) โ โ2: ๐๐(๐ฅ๐ฅ,๐ฆ๐ฆ) = ๐๐}. (1)
These examples are all curves in the plane โ2, but we can also consider curves in โ3 (3-D space)--- for example, the x-axis in โ3 is the straight line given by
{(๐ฅ๐ฅ,๐ฆ๐ฆ, ๐ง๐ง) โ โ3:๐ฆ๐ฆ = 0 & ๐ง๐ง = 0}
and more generally a curve in โ3 might be defined by a pair of equations
๐๐1(๐ฅ๐ฅ,๐ฆ๐ฆ, ๐ง๐ง) = ๐๐1, ๐๐2(๐ฅ๐ฅ, ๐ฆ๐ฆ, ๐ง๐ง) = ๐๐2
Curves of this kind are called level curves, the idea being that the curve in Eq (1), for example, is the set of points (๐ฅ๐ฅ,๐ฆ๐ฆ) in the plane at which the quantity ๐๐(๐ฅ๐ฅ,๐ฆ๐ฆ) reaches the 'level' ๐๐.
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 ๐๐(t) denotes the position vector of the point at time ๐ก๐ก, the curve is described by a function ๐๐ of a scalar parameter ๐ก๐ก with vector values (in โ2 for a plane curve, in โ3 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 โ๐๐ (we shall be interested only in the cases when ๐๐ = 2 ๐๐๐๐๐๐ ๐๐ = 3, but it is convenient to treat both cases simultaneously):
Definition 3.1.1 A parametrized curve in โ๐๐ is a continuous function or map given by
๐๐ โถ ๐ฐ๐ฐ โ โn , ๐ธ๐ธ = ๐๐(t) = โจฮณ1(t), ฮณ2(t), โฆ , ฮณn(t)โฉ for ๐ก๐ก in some open interval ๐ฐ๐ฐ โ โ
where the continuous functions (usually called maps) ๐พ๐พ1(๐ก๐ก), ๐พ๐พ2(๐ก๐ก), โฆ , ๐พ๐พ๐๐(๐ก๐ก) are called the component functions of ๐ธ๐ธ and ๐ก๐ก is called the parameter.
The symbol ๐ฐ๐ฐ denotes the parameter interval for ๐ธ๐ธ and it is usually (but not always) an open interval of the form ๐ฐ๐ฐ = {๐ก๐ก โ โ โถ ๐ผ๐ผ < ๐ก๐ก < ๐ฝ๐ฝ ๐๐๐๐๐๐ ๐ ๐ ๐๐๐ ๐ ๐ ๐ ๐ผ๐ผ,๐ฝ๐ฝ โ โ}. A parametrized curve whose image, ๐ธ๐ธ(๐ฐ๐ฐ), is contained in a level curve โ is called a parametrization of (part of) โ.
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 ๐๐(t) of the plane curve given as โ= {(๐ฅ๐ฅ,๐ฆ๐ฆ) โ โ2: ๐ฆ๐ฆ โ ๐ฅ๐ฅ2 = 0} (called a parabola). You are perhaps more familiar with it written in functional form as: ๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ) = ๐ฅ๐ฅ2, ๐ฅ๐ฅ โ โ
Suppose ๐๐(๐ก๐ก) = โจ๐พ๐พ1(๐ก๐ก), ๐พ๐พ2(๐ก๐ก)โฉ, then the component functions ๐พ๐พ1 and ๐พ๐พ2 of ๐๐ must satisfy the relation
๐พ๐พ2(๐ก๐ก) = ๐พ๐พ2(๐ก๐ก)2 (2)
for all values of ๐ก๐ก โ ๐ฐ๐ฐ where ๐๐ is defined (yet to be decided), and ideally every point (๐ฅ๐ฅ,๐ฆ๐ฆ) on the parabola โ should be equal to โจ๐พ๐พ1(๐ก๐ก), ๐พ๐พ2(๐ก๐ก)โฉ for some value of ๐ก๐ก โ ๐ฐ๐ฐ. Of course, there is an obvious solution to Eq (2): Take ๐พ๐พ1(๐ก๐ก) =๐ก๐ก, ๐พ๐พ2(๐ก๐ก) = ๐ก๐ก2. To get every point on the parabola we must allow ๐ก๐ก to take every real number value (since the ๐ฅ๐ฅ-coordinate of ๐๐(๐ก๐ก) is just ๐ก๐ก, and the ๐ฅ๐ฅ-coordinate of a point on the parabola can be any real number), so we must take ๐ผ๐ผ to be โ. Thus, the parametrization we seek is:
๐๐ โถ โ โถโ2, ๐๐(๐ก๐ก) = โจ๐ก๐ก, ๐ก๐ก2โฉ.
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 ๐๐ โถ โ โถโ2, ๐๐(๐ก๐ก) = โจ๐ก๐ก3, ๐ก๐ก6โฉ. Yet another is ๐๐ โถ โ โถโ2, ๐๐(๐ก๐ก) =โจ2๐ก๐ก, 4๐ก๐ก2โฉ, and of course there are (infinitely many) others (see if you can think of some others!). So the parametrization of a given level curve ๐๐(๐ฅ๐ฅ,๐ฆ๐ฆ) = ๐๐ is not unique.
Example 3.1.2 Now we try to find a parametrization of the plane curve as โ= {(๐ฅ๐ฅ, ๐ฆ๐ฆ) โ โ2: ๐ฅ๐ฅ2 + ๐ฆ๐ฆ2 = 1} (called the unit circle with center at the origin (0,0)). It is tempting to take ๐พ๐พ1(๐ก๐ก) = ๐ก๐ก , ๐พ๐พ2(๐ก๐ก) = โ1 โ ๐ก๐ก2 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 โ1 โ ๐ก๐ก2 is always nonnegative. Similarly, if we had taken ๐พ๐พ1(๐ก๐ก) = ๐ก๐ก, ๐พ๐พ2(๐ก๐ก) = โโ1โ ๐ก๐ก2, 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 ๐พ๐พ1(๐ก๐ก) and ๐พ๐พ2(๐ก๐ก) such that
๐พ๐พ1(๐ก๐ก)2 + ๐พ๐พ2(๐ก๐ก)2 = 1 (3)
for all ๐ก๐ก โ ๐ผ๐ผ, such that every point (๐ฅ๐ฅ,๐ฆ๐ฆ) on the circle is equal to โจ๐พ๐พ1(๐ก๐ก), ๐พ๐พ2(๐ก๐ก)โฉ for some ๐ก๐ก โ ๐ผ๐ผ. There is an obvious solution to Eq (3): Take ๐พ๐พ1(๐ก๐ก) = cos(๐ก๐ก) ๐๐๐๐๐๐ ๐พ๐พ2(๐ก๐ก) = ๐ ๐ ๐ ๐ ๐๐(๐ก๐ก) (since cos2(๐ก๐ก) + sin2(๐ก๐ก) = 1 for all real values of ๐ก๐ก). We can take ๐ผ๐ผ = โ, although this is overkill: Any open interval ๐ผ๐ผ whose length greater than 2๐๐ will suffice. Thus a desired parametrization of the unit circle is
๐๐ โถ [0, 2๐๐] โถโ2, ๐๐(๐ก๐ก) = โจcos ๐ก๐ก , sin ๐ก๐กโฉ
Note: In our parametrization of the unit circle chose an interval ๐ผ๐ผ 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 ๐๐ of the circle with radius ๐๐ > 0 centered at a point (โ,๐๐) โโ2 (Do not choose all of โ 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
๐๐(๐ก๐ก) = โจcos3 ๐ก๐ก, sin3 ๐ก๐กโฉ.
Since cos2 ๐ก๐ก + sin2 ๐ก๐ก = 1 for all ๐ก๐ก, the coordinates ๐ฅ๐ฅ = cos3 ๐ก๐ก, ๐ฆ๐ฆ = sin3 ๐ก๐ก of the point ๐๐(๐ก๐ก) satisfy
๐ฅ๐ฅ23 + ๐ฆ๐ฆ
23 = 1.
This level curve coincides with the image of the map ๐๐.
In this section, we shall be studying curves (and later, surfaces) using methods of the calculus. To differentiate a vector-valued function such as ๐๐(๐ก๐ก) (as in Definition 3.1.1), we differentiate component-wise: If
๐ธ๐ธ(๐ก๐ก) = โจ๐พ๐พ1(๐ก๐ก),๐พ๐พ2(๐ก๐ก), โฆ , ๐พ๐พ๐๐(๐ก๐ก)โฉ,
then
๐๐๐๐๐๐๐ก๐ก
= โจ ๐๐๐พ๐พ1
๐๐๐ก๐ก,๐๐๐พ๐พ2
๐๐๐ก๐ก, โฆ ,
๐๐๐พ๐พ๐๐๐๐๐ก๐ก
โฉ,
๐๐2๐๐
๐๐๐ก๐ก2 = โจ ๐๐2๐พ๐พ1๐๐๐ก๐ก2 , ๐๐
2๐พ๐พ2๐๐๐ก๐ก2 , โฆ , ๐๐
2๐พ๐พ๐๐๐๐๐ก๐ก2 โฉ, etc.
We say that a parametrized curve ๐๐ is smooth if each of its component functions ๐พ๐พ1,๐พ๐พ2, โฆ , ๐พ๐พ๐๐ is smooth i.e., if
all derivative ๐๐๐ธ๐ธ๐ ๐ ๐๐๐ก๐ก
, ๐๐2๐ธ๐ธ๐ ๐ ๐๐๐ก๐ก2 , ๐๐
3๐ธ๐ธ๐ ๐ ๐๐๐ก๐ก3 , โฆ exists, for ๐ ๐ = 1,2, โฆ ,๐๐. In most cases, all parametrized curves studied here
in this section will be assumed to be smooth.
Definition 3.1.2 If ๐๐ is a parametrized curve, then at some ๐ก๐ก โ ๐ฐ๐ฐ its first derivative ๐๐๐๐/๐๐๐ก๐ก is called the tangent vector of ๐๐ at the point ๐๐(๐ก๐ก).
To see the reason for this terminology, note that the vector
1ฮ๐ก๐ก
[๐๐(๐ก๐ก + ฮ๐ก๐ก) โ ๐๐(๐ก๐ก)]
is parallel to the chord joining the two points ๐๐(๐ก๐ก) ๐๐๐๐๐๐ ๐๐(๐ก๐ก + ฮ๐ก๐ก) of the image โ of ๐๐.
๐๐(๐ก๐ก + ฮ๐ก๐ก)
๐๐(๐ก๐ก)
We expect, as ฮ๐ก๐ก โถ 0 (tends to 0), the chord parallel to the tangent to โ at ๐๐(t). Hence, the tangent should be parallel to
limฮ๐ก๐กโ0
1ฮ๐ก๐ก
[๐๐(๐ก๐ก + ฮ๐ก๐ก) โ ๐๐(๐ก๐ก)] =๐๐๐๐๐๐๐ก๐ก
= ๐๐โฒ(๐ก๐ก).
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 ๐๐๐๐๐๐๐ก๐ก
= a for all ๐ก๐ก โ ๐ผ๐ผ, where ๐๐ is a constant vector, we have, integrating component-wise,
๐พ๐พ(๐ก๐ก) = ๏ฟฝ๐๐๐๐๐๐๐๐
๐๐๐๐๐ก๐ก
๐ก๐ก0
= ๏ฟฝ ๐๐ ๐๐๐ก๐ก๐ก๐ก
๐ก๐ก0
= ๐ก๐ก๐๐ + ๐๐,
where b is another constant vector. If ๐๐ โ ๐๐, this is the parametric equation of the straight line parallel to ๐๐ and passing through the point with position vector ๐๐ = ๐ธ๐ธ(๐๐๐๐):
๐ก๐ก๐๐ ๐ธ๐ธ(๐๐)
๐๐
๐๐
If ๐๐ = ๐๐, the image of ๐พ๐พ is a single point (namely, the point with position vector ๐๐ = ๐๐(๐ก๐ก0) ). Q.E.D
Problems
1. Is the curve ๐๐(๐ก๐ก) = โจ๐ก๐ก2, ๐ก๐ก4โฉ a parametrization of the parabola = ๐ฅ๐ฅ2 ? (Explain).
2. Find parametrizations of the following level curves a. ๐ฆ๐ฆ2 โ ๐ฅ๐ฅ2 = 1 ;
b. ๐ฅ๐ฅ2
4+ ๐ฆ๐ฆ2
9= 1.
3. Find the cartesian equations of the following parametrized curves.
a. ๐ธ๐ธ(๐ก๐ก) = โจcos2 ๐ก๐ก, sin2 ๐ก๐กโฉ; b. ๐๐(๐ก๐ก) = โจ๐ ๐ ๐ก๐ก , ๐ก๐ก2โฉ.
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?
6. Show that ๐๐(๐ก๐ก) = โจcos2 ๐ก๐ก โ 1
2, sin(๐ก๐ก) cos(๐ก๐ก) , sin(๐ก๐ก)โฉ is a parametrization of the curve of intersection of
the circular cylinder of radius 12 and axis the ๐ง๐ง-axis with the sphere of radius 1 and center ๏ฟฝโ 1
2, 0, 0๏ฟฝ. (This
is called Viviani's Curve)
3.2 ARC-LENGTH
If ๐๐ = โจ๐ฃ๐ฃ1, ๐ฃ๐ฃ2, โฆ , ๐ฃ๐ฃ๐๐โฉ is a vector in โ๐๐ , its length or magnitude is
โฅ ๐๐ โฅ= ๏ฟฝ๐ฃ๐ฃ12 + ๐ฃ๐ฃ2
2 + โฏ+ ๐ฃ๐ฃ๐๐2.
If ๐๐ is another vector in โ๐๐ , โฅ ๐๐ โ ๐๐ โฅ is the length of the straight line segment joining the points in โ๐๐ with position vectors ๐๐ and ๐๐.
To find a formula for the length of any parametrized curve, say, ๐ธ๐ธ, note that, if ฮ๐ก๐ก is sufficiently small, the part of the image โ of ๐ธ๐ธ between ๐พ๐พ(๐ก๐ก) and ๐พ๐พ(๐ก๐ก + ฮ๐ก๐ก) is nearly a straight line. so its length is approximately
โฅ ๐ธ๐ธ(๐ก๐ก + ฮ๐ก๐ก) โ ๐ธ๐ธ(๐ก๐ก) โฅ.
Again, since ฮ๐ก๐ก is sufficiently small, the vector 1ฮ๐ก๐ก
[๐ธ๐ธ(๐ก๐ก + ฮ๐ก๐ก) โ ๐ธ๐ธ(๐ก๐ก)] โ ๐ธ๐ธโฒ(๐ก๐ก), so the length L is approximately
โฅ ๐ธ๐ธโฒ(๐ก๐ก) โฅ ฮ๐ก๐ก (4)
If we want to calculate the length of (a not necessarily small) part of โ, we can divide it up into segments, each of which corresponds to a small increment ฮ๐ก๐ก in ๐ก๐ก, calculate the length of each segment using Eq (4), and add up the results.
Letting ฮ๐ก๐ก โ 0 should then give the exact length.
This motivates the following definition:
Definition 3.2.1 The arc-length of a parametrized curve ๐ธ๐ธ starting at the point ๐ธ๐ธ(๐ก๐ก0) is the function ๐ ๐ (๐ก๐ก) given by
๐ ๐ (๐ก๐ก) = ๏ฟฝ โฅ๐๐๐ธ๐ธ๐๐๐๐
โฅ ๐๐๐๐ = ๏ฟฝ โฅ ๐ธ๐ธโฒ(๐๐) โฅ ๐๐๐๐๐ก๐ก
๐ก๐ก0
๐ก๐ก
๐ก๐ก0
Thus, ๐ ๐ (๐ก๐ก0) = 0 and ๐ ๐ (๐ก๐ก) is positive or negative according to whether ๐ก๐ก is larger or smaller than ๐ก๐ก0. If we choose a
different starting point, say, ๐ธ๐ธ(๐๐), the resulting arc-length ๏ฟฝ๏ฟฝ๐ differs from ๐ ๐ by the constant โซ โฅ ๐ธ๐ธโฒ(๐ข๐ข) โฅ ๐๐๐ข๐ข ๐๐๐ก๐ก0
.
Example 3.2.1 For a logarithmic spiral
๐ธ๐ธ(๐ก๐ก) = โจ๐ ๐ 0.2๐ก๐ก cos(๐ก๐ก) , ๐ ๐ 0.2๐ก๐ก sin(๐ก๐ก)โฉ,
we have
๐๐โฒ = โจ๐ ๐ 0.2๐ก๐ก(0.2cos(๐ก๐ก) โ sin(๐ก๐ก)), ๐ ๐ 0.2๐ก๐ก (cos(๐ก๐ก) + 0.2 sin(๐ก๐ก))โฉ
โด โฅ ๐ธ๐ธโฒ โฅ2= ๐ ๐ 0.4๐ก๐ก(0.2 cos(๐ก๐ก) โ sin(๐ก๐ก))2 + ๐ ๐ 0.4๐ก๐ก(0.2 sin(๐ก๐ก) + cos(๐ก๐ก))2 = 1.04๐ ๐ 0.4๐ก๐ก .
Hence, the arc-length of gamma starting at ๐ธ๐ธ(0) = โจ1,0โฉ (which corresponds to the point (1,0) ) is
๐ ๐ = ๐ ๐ (๐ก๐ก) = โซ โ1.04๐ ๐ 0.4๐๐๐๐๐๐ =๐ก๐ก0 โ1.04
0.2 (๐ ๐ 0.2๐ก๐ก โ 1) .
If s is the arc-length of a curve ๐ธ๐ธ starting at the point ๐๐(๐ก๐ก0), we have
๐ ๐ โฒ(๐ก๐ก) = ๐๐๐ ๐ ๐๐๐ก๐ก
= ๐๐๐๐๐ก๐ก โซ โฅ ๐ธ๐ธโฒ(๐๐) โฅ ๐๐๐๐ =โฅ ๐ธ๐ธโฒ(๐ก๐ก) โฅ. ๐ก๐ก
๐ก๐ก0 (5)
Thinking of ๐๐(๐ก๐ก) as the position of a moving point at time ๐ก๐ก, ๐๐๐ ๐ ๐๐๐ก๐ก
is the speed of the point (the rate of
change of distance along the curve). For this reason, we make the following definition
Definition 3.2.2 If ๐๐: ๐ฐ๐ฐ โถ โ๐๐ is a parametrized curve, its speed at the point ๐๐(๐ก๐ก) is โฅ ๐๐โฒ(๐ก๐ก) โฅ, and the curve ๐๐ is called a unit-speed curve if the tangent vector, ๐๐โฒ(๐ก๐ก) , is a unit vector (i.e., โฅ ๐๐โฒ(๐ก๐ก) โฅ = 1) for all ๐ก๐ก โ ๐ฐ๐ฐ.
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.
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Theorem 3.2.1 Let ๐ง๐ง(๐ก๐ก) be a unit vector that is a smooth function of a parameter ๐ก๐ก. Then, the dot product
๐ง๐งโฒ(๐ก๐ก) โ ๐ง๐ง(๐ก๐ก) = 0
for all ๐ก๐ก, i.e., ๐ง๐งโฒ(๐ก๐ก) = 0 or else ๐ง๐งโฒ(t) is perpendicular to ๐ง๐ง(๐ก๐ก) for all ๐ก๐ก. In particular, If ๐ธ๐ธ is a unit-speed curve, then ๐๐2๐ธ๐ธ
๐๐๐ก๐ก2
is either 0 or perpendicular to ๐๐๐ธ๐ธ๐๐๐ก๐ก
.
proof We use the 'dot product' formula' for differentiating dot products of vector-valued functions ๐๐(๐ก๐ก) and ๐๐(๐ก๐ก):
๐๐๐๐๐ก๐ก
(๐๐ โ ๐๐) = ๐๐๐๐๐๐๐ก๐กโ ๐๐ + ๐๐ โ ๐๐๐๐
๐๐๐ก๐ก.
Using this to differentiate both sides of the equation ๐ง๐ง โ ๐ง๐ง = 1 with respect to ๐ก๐ก gives
๐ง๐งโฒ โ ๐ง๐ง + ๐ง๐ง โ ๐ง๐งโฒ = 0,
so 2๐ง๐งโฒ โ ๐ง๐ง = 0. The last part follows by taking ๐ง๐ง = ๐๐โฒ Q.E.D
Problems
1. Calculate the arc-length of the catenary ๐๐(๐ก๐ก) = โจ๐ก๐ก, cosh ๐ก๐กโฉ starting at the point (0,1).
2. Show that the following curves a unit-speed curves
(i) ๐๐(๐ก๐ก) = โจ 13
(1 + ๐ก๐ก)32 , 1
3 (1 โ ๐ก๐ก)
32 , ๐ก๐ก
โ2 โฉ;
(ii) ๐๐(๐ก๐ก) = โจ45
cos ๐ก๐ก , 1 โ sin ๐ก๐ก ,โ 35
cos ๐ก๐กโฉ.
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 ๐ผ๐ผ > 0, the cycloid can be parametrized as
๐ธ๐ธ: โโถ โ2, ๐ธ๐ธ(๐ก๐ก) = ๐ผ๐ผโจ๐ก๐ก โ sin ๐ก๐ก , 1 โ cos ๐ก๐กโฉ 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 ๐ธ๐ธ: ๐ฐ๐ฐ โถ โ3,๐ธ๐ธ = ๐ธ๐ธ(๐ก๐ก) = โจcosฮฑt , sinฮฑt, ฮฒtโฉ, where ๐ผ๐ผ and ๐ฝ๐ฝ are constants. Calculate the arc-length function ๐ ๐ along the circular helix starting at the point ๐ธ๐ธ(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 ๐๐: ๐ฑ๐ฑ โถ โ๐๐ is called a reparametrization of ๐ธ๐ธ: ๐ฐ๐ฐ โถ โ๐๐ if there is a smooth bijective map (i.e. a one-to-one and onto map) ๐๐: ๐ฑ๐ฑ โถ ๐ฐ๐ฐ, ๐ก๐ก = ๐๐(๐ข๐ข) such that the inverse map ๐๐โ1: ๐ฐ๐ฐ โถ ๐ฑ๐ฑ, ๐ข๐ข =๐๐โ1(๐ก๐ก) is also smooth and
๐๐: ๐ฑ๐ฑ โถ โ๐๐ , ๐๐(๐ข๐ข) = (๐ธ๐ธ โ ๐๐)(๐ข๐ข) = ๐ธ๐ธ(๐๐(๐ข๐ข))
Note that, since ๐๐has a smooth inverse, ๐ธ๐ธ is a reparametrization of ๐๐:
๐๐๏ฟฝ๐๐โ1(๐ก๐ก)๏ฟฝ = ๐ธ๐ธ๏ฟฝ๐๐๏ฟฝ๐๐โ1(๐ก๐ก)๏ฟฝ๏ฟฝ = ๐ธ๐ธ(๐ก๐ก) for all ๐ก๐ก โ ๐ผ๐ผ.
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 ๐ธ๐ธ(๐ก๐ก) = โจcos ๐ก๐ก, sin ๐ก๐กโฉ for the circle ๐ฅ๐ฅ2 + ๐ฆ๐ฆ2 = 1. Another parametrization is
๐๐(๐ข๐ข) = โจsin๐ข๐ข, cos๐ข๐ขโฉ
(since sin2 ๐ ๐ + cos2 ๐ ๐ = 1 for all s). To see that ๐๐ is a reparametrization of ๐ธ๐ธ, we have to find a reparametrization map ๐๐ such that
โจcos(๐๐(๐ข๐ข)), sin(๐๐(๐ข๐ข))โฉ = โจsin๐ข๐ข, cos๐ข๐ขโฉ
One solution is ๐ก๐ก = ๐๐(๐ข๐ข) = ๐๐2โ ๐ข๐ข (observe that this is a smooth bijective map! whose inverse is ๐ข๐ข = ๐๐โ1(๐ก๐ก) = ๐๐
2โ ๐ก๐ก
which is also a smooth bijective map. Thus ๐ธ๐ธ is also a reparametrization of ๐๐ 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 ๐ธ๐ธ(๐ก๐ก) of a parametrized curve ๐ธ๐ธ is called a regular point if the tangent vector at ๐ธ๐ธ(๐ก๐ก) does not vanish, i.e., if ๐ธ๐ธโฒ(๐ก๐ก) โ ๐๐ (please keep in mind that ๐๐ = โจ0,0, โฆ ,0โฉ in โ๐๐ and 0 is a real number); otherwise, ๐ธ๐ธ(๐ก๐ก) is a singular point. A curve ๐ธ๐ธ 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 ๐พ๐พ is a regular curve, then any reparametrization ๐๐ of ๐พ๐พ is again regular.
proof We aim to show that the curve ๐๐ is a regular curve (i.e., ๐๐๐๐๐๐๐ ๐
is never zero) Since ๐๐ is a reparametrization of ๐พ๐พ, we
have smooth bijective reparametrization maps ๐ก๐ก = ๐๐(๐ข๐ข) and ๐ข๐ข = ๐๐โ1(๐ก๐ก) having the property ๐๐๏ฟฝ๐ก๐กโ1(๐ก๐ก)๏ฟฝ = ๐ก๐ก. Now differentiating on both sides with respect to ๐ก๐ก and using the chain rule yields
๐๐๐๐๐๐๐ข๐ข
๐๐๐๐โ1
๐๐๐ก๐ก= 1
This shows that ๐๐๐๐/๐๐๐ข๐ข is never zero. Since ๐๐(๐ข๐ข) = ๐ธ๐ธ๏ฟฝ๐๐(๐ข๐ข)๏ฟฝ, and ๐ธ๐ธ is a regular curve, another application of the chain rule gives
๐๐๐๐๐๐๐ข๐ข
= ๐๐๐ธ๐ธ๐๐๐ก๐ก
๐๐๐๐๐๐๐ข๐ขโ ๐๐ (do you see why?) Q.E.D
Theorem 3.3.2 If ๐ธ๐ธ is a regular curve, its arc-length, ๐ ๐ (see Definition 3.1.3), starting at any point ๐ธ๐ธ(๐ก๐ก) of ๐ธ๐ธ, is a smooth function of ๐ก๐ก.
proof We have already seen that (whether or not ๐ธ๐ธ is regular) ๐ ๐ is a differentiable function of ๐ก๐ก and that
๐๐๐ ๐ ๐๐๐ก๐ก
= โฅ ๐ธ๐ธโฒ(๐ก๐ก) โฅ.
To simplify the notation, assume form now on that ๐พ๐พ is a plane curve, say
๐ธ๐ธ(๐ก๐ก) = โจ๐ฅ๐ฅ(๐ก๐ก),๐ฆ๐ฆ(๐ก๐ก)โฉ,
where ๐ฅ๐ฅ and ๐ฆ๐ฆ are smooth functions of ๐ก๐ก. Define ๐๐:โ2 โถโ by
๐๐(๐ฅ๐ฅ,๐ฆ๐ฆ) = ๏ฟฝ๐ฅ๐ฅ2 + ๐ฆ๐ฆ2,
๐๐๐ ๐ ๐๐๐ก๐ก
= ๐๐ ๏ฟฝ๐๐๐ฅ๐ฅ๐๐๐ก๐ก
, ๐๐๐ฆ๐ฆ๐๐๐ก๐ก๏ฟฝ = ๏ฟฝ๏ฟฝ๐๐๐ฅ๐ฅ
๐๐๐ก๐ก๏ฟฝ
2+ ๏ฟฝ๐๐๐ฆ๐ฆ
๐๐๐ก๐ก๏ฟฝ
2 (6)
The crucial point is that ๐๐ is smooth on โ2 without the origin, which means that all the partial derivatives of ๐๐ of all orders exists and are continuous functions except at the origin (0,0). For example,
๐๐๐๐๐๐๐ฅ๐ฅ
(๐ฅ๐ฅ,๐ฆ๐ฆ) =๐ฅ๐ฅ
๏ฟฝ๐ฅ๐ฅ2 + ๐ฆ๐ฆ2,
๐๐๐๐๐๐๐ฃ๐ฃ
(๐ฅ๐ฅ,๐ฆ๐ฆ) =๐ฆ๐ฆ
๏ฟฝ๐ฅ๐ฅ2 + ๐ฆ๐ฆ2,
are well defined and continuous except where ๐ฅ๐ฅ = ๐ฆ๐ฆ = 0, and similarly for higher derivatives. Since ๐ธ๐ธ is regular ๐๐๐ฅ๐ฅ/๐๐๐ก๐ก and ๐๐๐ฆ๐ฆ/๐๐๐ก๐ก are never both zero, so the chain rule and Eq (6) shows that ๐๐๐ ๐ /๐๐๐ก๐ก is smooth. For example,
๐๐2๐ ๐ ๐๐๐ก๐ก2 =
๐๐๐๐๐๐๐ฅ๐ฅ
๐๐2๐ฅ๐ฅ๐๐๐ก๐ก2 +
๐๐๐๐๐๐๐ฆ๐ฆ
๐๐2๐ฆ๐ฆ๐๐๐ก๐ก2 ,
and similarly for the higher derivatives of ๐ ๐ . Q.E.D
The main result we want is
Theorem 3.3.4 A parametrized curve ๐ธ๐ธ: ๐ฐ๐ฐ โถ โ๐๐ has a unit-speed reparametrization if and only if it is regular.
proof Suppose the ๐ธ๐ธ: ๐ฐ๐ฐ โถ โ๐๐ is a parametrized curve with a unit-speed reparametrization ๐๐: ๐ฑ๐ฑ โถ โ๐๐ . Let ๐๐: ๐ฑ๐ฑ โถ ๐ฐ๐ฐ, ๐ก๐ก = ๐๐(๐ข๐ข) be the reparametrization map. Then we have
๐๐(๐ข๐ข) = ๐ธ๐ธ๏ฟฝ๐๐(๐ข๐ข)๏ฟฝ = ๐ธ๐ธ(๐ก๐ก)
โด ๐๐๐๐๐๐๐ข๐ข
=๐๐๐ธ๐ธ๐๐๐ก๐ก
๐๐๐ก๐ก๐๐๐ข๐ข
,
โด โฅ๐๐๐๐๐๐๐ข๐ข
โฅ = โฅ๐๐๐ธ๐ธ๐๐๐ก๐ก
โฅ ๏ฟฝ๐๐๐ก๐ก๐๐๐ข๐ข๏ฟฝ.
Since ๐๐ is unit-speed, โฅ ๐๐๐๐๐๐๐ข๐ขโฅ = 1, so clearly ๐๐๐ธ๐ธ/๐๐๐ก๐ก cannot be zero (i.e. ๐ธ๐ธ is a regular curve).
Conversely, suppose that the tangent vector ๐๐๐ธ๐ธ๐๐๐ก๐กโ ๐๐ (i.e., ๐พ๐พ is a regular parametrized curve). By Eq (5), we
know that ๐๐๐ ๐ ๐๐๐ก๐ก
> 0 for all ๐ก๐ก, where ๐ ๐ is the arc-length of ๐พ๐พ starting at any point of the curve, and by Theorem 3.3.2 we have that ๐ ๐ is a smooth function of ๐ก๐ก. It follows from the inverse function theorem of multivariable calculus that ๐ ๐ : ๐ผ๐ผ โถ โ is injective (i.e. one-to-one), that its image is an open interval ๐ฝ๐ฝ, and that the inverse map ๐ ๐ โ1: ๐ฝ๐ฝ โถ ๐ผ๐ผ 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 ๐๐ = ๐ ๐ โ1 and let ๐๐ be the corresponding reparametrization of ๐พ๐พ, so that
๐๐(๐ ๐ ) = ๐ธ๐ธ(๐ก๐ก).
Then,
๐๐๐๐๐๐๐ ๐
๐๐๐ ๐ ๐๐๐ก๐ก
=๐๐๐ธ๐ธ๐๐๐ก๐ก
,
โด โฅ ๐๐๐๐๐๐๐ ๐ โฅ ๐๐๐ ๐ ๐๐๐ก๐ก
= โฅ ๐๐๐ธ๐ธ๐๐๐ก๐กโฅ = ๐๐๐ ๐
๐๐๐ก๐ก (by Eq (5))
โด โฅ ๐๐๐๐๐๐๐ ๐ โฅ = 1. 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 ๐ธ๐ธ โถ ๐ฐ๐ฐ โถ โ๐๐ be a regular curve and let ๐๐ โถ ๐ฑ๐ฑ โถ โ๐๐ be a unit-speed reparametrization of ๐ธ๐ธ:
๐๐๏ฟฝ๐ข๐ข(๐ก๐ก)๏ฟฝ = ๐ธ๐ธ(๐ก๐ก) for all ๐ก๐ก,
where ๐ข๐ข is a smooth function of ๐ก๐ก. Then, if ๐ ๐ is the arc-length of ๐ธ๐ธ (starting at any point), we have
๐ข๐ข = ยฑ๐ ๐ + ๐๐, (7)
where ๐๐ is a constant. Conversely, if ๐ข๐ข is given by Eq (7) for some value of ๐๐ and with either sign, then ๐๐ is a unit-speed reparametrization of ๐ธ๐ธ
proof The calculation in the first part of the proof of Theorem 3.3.4 shows that ๐ข๐ข gives a unit-speed reparametrization of ๐ธ๐ธ if and only if
๐๐๐ข๐ข๐๐๐ก๐ก
= ยฑ โฅ ๐๐๐ธ๐ธ๐๐๐ก๐กโฅ = ยฑ ๐๐๐ ๐
๐๐๐ก๐ก, (by Eq (5))
Hence, ๐ข๐ข = ยฑ๐ ๐ + ๐๐ for some constant ๐๐. 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
๐ธ๐ธ(๐ก๐ก) = โจ๐ ๐ 0.2๐ก๐ก cos ๐ก๐ก, ๐ ๐ 0.2๐ก๐ก sin ๐ก๐กโฉ,
we found in Example 3.1.4 that
โฅ ๐ธ๐ธโฒ(๐ก๐ก) โฅ2= 1.04๐ ๐ 0.04๐ก๐ก .
This is never zero, so ๐ธ๐ธ is regular. The arc-length of ๐พ๐พ starting at (1,0) was found to be ๐ ๐ = โ1.04(๐ ๐ 0.2๐ก๐ก โ 1). Hence,
๐ก๐ก = 10.2
ln ๏ฟฝ sโ1.04
+ 1๏ฟฝ, so a unit-speed reparametrization of ๐ธ๐ธ is given by the rather unwieldly formula
๐๐(๐ ๐ ) = โจ ๏ฟฝ๐ ๐
โ1.04+ 1๏ฟฝ cos ๏ฟฝ
10.2
ln ๏ฟฝ๐ ๐
โ1.04+ 1๏ฟฝ๏ฟฝ , ๏ฟฝ
๐ ๐ โ1.04
+ 1๏ฟฝ sin๏ฟฝ1
0.2๐๐๐๐ ๏ฟฝ
๐ ๐ โ1.04
+ 1๏ฟฝ๏ฟฝ โฉ
Example 3.3.2 The twisted cubic is the space curve given by
๐ธ๐ธ(๐ก๐ก) = โจ๐ก๐ก, ๐ก๐ก2, ๐ก๐ก3โฉ, โโ < ๐ก๐ก < โ
We have
๐ธ๐ธโฒ(๐ก๐ก) = โจ1, 2๐ก๐ก, 3๐ก๐ก2โฉ,
โด โฅ ๐ธ๐ธโฒ(๐ก๐ก) โฅ = ๏ฟฝ1 + 4๐ก๐ก2 + 9๐ก๐ก4 .
This is never zero, so ๐พ๐พ is regular. The arc-length starting at ๐ธ๐ธ(0) = ๐๐ = โจ0,0,0โฉ (is the position vector associated with origin (0,0,0) ) is
๐ ๐ = ๐ ๐ (๐ก๐ก) = โซ โ๐ธ๐ธโฒ(๐ข๐ข)โ ๐๐๐ข๐ข =๐ก๐ก0 โซ โ1 + 4๐ข๐ข2 + 9๐ข๐ข4 ๐๐๐ข๐ข๐ก๐ก
0 .
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 ๐ธ๐ธ = ๐ธ๐ธ(๐ก๐ก) = โจ๐ก๐ก, ๐ก๐ก2โฉ of the level curve (or the parabola)
๐ฅ๐ฅ2 โ ๐ฆ๐ฆ = 0, we have ๐๐๐ธ๐ธ๐๐๐ก๐ก
= ๐ธ๐ธโฒ(๐ก๐ก) = โจ1, 2๐ก๐กโฉ โ ๐๐ for all ๐ก๐ก , so ๐ธ๐ธ is a regular curve. But the curve ๐๐ given by
๐๐(๐ก๐ก) = ๐พ๐พ(๐ข๐ข(๐ก๐ก)) = ๐พ๐พ(๐ก๐ก3) = โจ๐ก๐ก3, ๐ก๐ก6โฉ
is a reparametrization of the parabola ๐ธ๐ธ (where ๐ข๐ข = ๐ข๐ข(๐ก๐ก) = ๐ก๐ก3 is the reparametrization map) . This time
๐๐โฒ(๐ก๐ก) = ๐ข๐ขโฒ(๐ก๐ก)๐ธ๐ธโฒ๏ฟฝ๐ข๐ข(๐ก๐ก)๏ฟฝ = 3๐ก๐ก2โจ1, 2๐ก๐ก3โฉ = โจ3๐ก๐ก2, 6๐ก๐ก5โฉ, and this is zero when ๐ก๐ก = 0, so the reparametrization ๐๐ 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) ๐พ๐พ(๐ก๐ก) = โจcos2 ๐ก๐ก, sin2 ๐ก๐กโฉ for โโ < ๐ก๐ก < โ; (ii) the same is curve as in (i), but with 0 < ๐ก๐ก < ๐๐
2;
(iii) ๐พ๐พ(๐ก๐ก) = โจ๐ก๐ก, cosh ๐ก๐กโฉ for โโ < ๐ก๐ก < โ.
2. The cissoid of Diocles (shown above) is the plane curve whose equation in terms of polar coordinates
(๐๐,๐๐) (you should what polar coordinates are look it up if you don't!) is
๐๐ = ๐๐(๐๐) = ๐ ๐ ๐ ๐ ๐๐๐๐๐ก๐ก๐๐๐๐๐๐, โ๐๐2
< ๐๐ <๐๐2
.
write down a parametrization of the cissoid using ๐๐ as your parameter and show that the curve given by
๐ธ๐ธ(๐ก๐ก) = โจ๐ก๐ก2,๐ก๐ก3
โ1 โ ๐ก๐ก2โฉ , โ1 < ๐ก๐ก < 1,
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' ๐ฅ๐ฅ2 + ๐ฆ๐ฆ2 = 0 is a single point. The correct conditions to impose on a function ๐๐(๐ฅ๐ฅ,๐ฆ๐ฆ) in order that ๐๐(๐ฅ๐ฅ,๐ฆ๐ฆ) = ๐๐, where ๐๐ 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 ๐๐ = 0 (since we could replace ๐๐ by ๐๐ โ ๐๐).
Theorem 3.4.1 Let ๐๐(๐ฅ๐ฅ,๐ฆ๐ฆ) be a smooth function of two variables x and y. Assume that โ๐๐(๐ฅ๐ฅ,๐ฆ๐ฆ) โ ๐๐ for all points (๐ฅ๐ฅ,๐ฆ๐ฆ) of the level curve given by
โ= {(๐ฅ๐ฅ,๐ฆ๐ฆ) โ โ2:๐๐(๐ฅ๐ฅ,๐ฆ๐ฆ) = 0}.
If ๐๐(๐ฅ๐ฅ0,๐ฆ๐ฆ0) is a point of โ, there is a regular parametrized curved ๐ธ๐ธ = ๐ธ๐ธ(๐ก๐ก), defined on an open interval ๐ฐ๐ฐ containing 0, such that the curve ๐ธ๐ธ passes through ๐๐(๐ฅ๐ฅ0,๐ฆ๐ฆ0) when ๐ก๐ก = 0 and ๐ธ๐ธ(๐ก๐ก) is contained in โ for all ๐ก๐ก.
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 ๐๐๐๐๐๐๐๐(๐๐) = โ ๐๐(๐ฅ๐ฅ,๐ฆ๐ฆ) = โจ๐๐๐๐๐๐๐ฅ๐ฅ
(๐ฅ๐ฅ, ๐ฆ๐ฆ), ๐๐๐๐๐๐๐ฆ๐ฆ
(๐ฅ๐ฅ, ๐ฆ๐ฆ)โฉ โ ๐๐ means
not both ๐๐๐๐๐๐๐ฅ๐ฅ
(๐ฅ๐ฅ,๐ฆ๐ฆ) = 0 and ๐๐๐๐๐๐๐ฆ๐ฆ
(๐ฅ๐ฅ,๐ฆ๐ฆ) = 0 holds (usually one says the 'gradient of ๐๐ never vanishes').
To understand the significance of the conditions placed on f(x, y) in the theorem, we suppose for a moment that ๐๐(๐ฅ๐ฅ0 + ฮ๐ฅ๐ฅ,๐ฆ๐ฆ0 + ฮ๐ฆ๐ฆ) 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 ๐๐(๐ฅ๐ฅ0 + ฮ ๐ฅ๐ฅ,๐ฆ๐ฆ0 + ฮ ๐ฆ๐ฆ) = 0. We al so know that
0 = ๐๐(๐ฅ๐ฅ0 + ฮ๐ฅ๐ฅ,๐ฆ๐ฆ0 + ฮ๐ฆ๐ฆ) โ ๐๐(๐ฅ๐ฅ0,๐ฆ๐ฆ0) โ ๐๐๐๐๐๐๐ฅ๐ฅ
(๐ฅ๐ฅ0,๐ฆ๐ฆ0)ฮ๐ฅ๐ฅ + ๐๐๐๐๐๐๐ฆ๐ฆ
(๐ฅ๐ฅ0,๐ฆ๐ฆ0)ฮ๐ฆ๐ฆ (this you should know!).
If follows that for sufficiently small ฮ๐ฅ๐ฅ and ฮ๐ฆ๐ฆ we get that
๐๐๐๐๐๐๐ฅ๐ฅ
(๐ฅ๐ฅ0,๐ฆ๐ฆ0)ฮ๐ฅ๐ฅ + ๐๐๐๐๐๐๐ฆ๐ฆ
(๐ฅ๐ฅ0,๐ฆ๐ฆ0)ฮ๐ฆ๐ฆ โ 0, (8)
which says that the vector โจฮ๐ฅ๐ฅ,ฮ๐ฆ๐ฆโฉ is nearly (and equal if we let both ฮ๐ฅ๐ฅ and ฮ๐ฆ๐ฆ approach 0) tangent to the level curve at ๐๐ and that the vector โ๐๐ is perpendicular to the level curve at ๐๐ (this statement should not surprise you because you have seen it many times in Calculus III ).
โ
๐ฆ๐ฆ
โ๐๐
โจฮ๐ฅ๐ฅ,ฮ๐ฆ๐ฆโฉ
P
๐ฅ๐ฅ
The hypothesis in the theorem tells us that โ๐๐ never vanishes at every point P of the level curve. Suppose
now for example, that ๐๐๐๐๐๐๐ฆ๐ฆโ 0 at ๐๐. Then,โ๐๐ is not parallel to the ๐ฅ๐ฅ-axis at ๐๐, so the tangent vector to the
tangent line to the level curve at P is not parallel to the ๐ฆ๐ฆ-axis.
โ
Tangent line at ๐๐
๐๐(๐ฅ๐ฅ0,๐ฆ๐ฆ0) Rectangle on which functon exists
๐ฅ๐ฅ = ๐๐2
๐ฅ๐ฅ = ๐๐_1 ๐ฅ๐ฅ = ๐ฅ๐ฅ0
This implies that vertical lines of the form ๐ฅ๐ฅ = ๐๐ near the line ๐ฅ๐ฅ = ๐ฅ๐ฅ0 all intersect the level curve in a unique point (๐ฅ๐ฅ,๐ฆ๐ฆ) near ๐๐. In other words, the equation
๐๐(๐ฅ๐ฅ,๐ฆ๐ฆ) = 0 (9)
has a unique solution ๐ฆ๐ฆ near ๐ฆ๐ฆ0 for every ๐ฅ๐ฅ near ๐ฅ๐ฅ0. Note that this may fail to be the case if the tangent to the level curve at ๐๐ is parallel to the ๐ฆ๐ฆ-axis:
๐๐
In this example, lines ๐ฅ๐ฅ = ๐๐ for ๐๐ < ๐ฅ๐ฅ0 do not meet the level curve near ๐๐, while those vertical lines just to the right of ๐ฅ๐ฅ = ๐ฅ๐ฅ0 and near P meets the level curve in more than one point.
The statement in red about ๐๐ in the last paragraph means that there is a function ๐๐(๐ฅ๐ฅ), defined for ๐ฅ๐ฅ near ๐ฅ๐ฅ0, such that ๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ) is the unique solution of Eq (9) near ๐ฅ๐ฅ0. We can now define a parametrization \gamma of the part of the level curve near ๐๐ by
๐ธ๐ธ(๐ก๐ก) = โจ๐ก๐ก,๐๐(๐ก๐ก)โฉ.
If we accept that ๐๐ is smooth (which follows from the inverse function theorem), then ๐ธ๐ธ is certainly regular since
๐พ๐พโฒ(๐ก๐ก) = โจ1,๐๐โฒ(๐ก๐ก)โฉ
is obviously never zero. This 'proves' the theorem. Q.E.D
3.5. LINE INTEGRALS, VECTOR FIELDS, GREENโS THEOREM, SURFACEINTEGRALS, STOKEโS 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 ฮณ be any smooth parametrized space curve given by
ฮณ(s) = ใx(s), y(s), z(s)ใ) 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 ฮณ with respect to arc length from a to b to be:
In space
โซ
ฮณ
f(x, y, z) ds = limโPโโ0
nโ
i=1
f(xโi , y
โi , zโi ) โsi (provided this limit exists)
or
In the Plane
โซ
ฮณ
f(x, y) ds = limโPโโ0
nโ
i=1
f(xโi , y
โi ) โsi (provided this limit exists)
Theorem 3.5.1. [Evaluation Theorem for Line Integrals with respect to Arc-length] If the planecurve ฮณ is given parametrically by x = x(s), y = y(s), a 6 s 6 b or the space curve ฮณ by x = x(s), y =y(s), z = z(s), a 6 s 6 b, then
(In the plane)
โซ
ฮณ
f(x, y) ds =
โซ b
a
f(x(s), y(s)) ds
or
(In Space)
โซ
ฮณ
f(x, y, z) ds =
โซ b
a
f(x(s), y(s), z(s)) ds
In practice, the arc-length parameter s is usually difficult to obtain, therefore, the above Evaluation Theorem becomesimpractical. A more practical Evaluation Theorem for line integrals with respect to arc-length is given in the nexttheorem.
Theorem 3.5.2. [Evaluation Theorem for Line Integrals with respect to Arc-length] If the curve ฮณis 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,
thenโซ
ฮณ
f(x, y) ds =
โซ b
a
f(x(t), y(t))โ
[xโฒ(t)]2 + [yโฒ(t)]2 dt =
โซ b
a
f(x(t), y(t)) โ ฮณโฒ(t) โ dt
or
โซ
ฮณ
f(x, y, z) ds =
โซ b
a
f(x(t), y(t), z(t))โ
[xโฒ(t)]2 + [yโฒ(t)]2 + [zโฒ(t)]2 dt =
โซ b
a
f(x(t), y(t), z(t)) โ ฮณโฒ(t) โ dt
Example 3.5.1. How to find the mass of a Helical Spring: Find the mass of a spring in the shape of theHelix defined parametrically by ฮณ(t) = ใ2 cos t, t, 2 sin tใ, for 0 6 t 6 6ฯ, with density ฯ(x, y, z) = 2y.
Solution. First we assume that the Mass, m is approximated as follows: m โ โni=1 ฯ(xi, yi, zi)โsi so that
m = limโpโโ0
โni=1 ฯ(xi, yi, zi)โsi =
โซ
ฮณ ฯ(x, y, z) ds. Since the density of the spring at point (x, y, z) on the spring
(the curve ฮณ) is given as ฯ(x, y, z) = 2y = 2t and the arc-length element ds is given by
ds = โ ฮณโฒ(t) โ dt =โ
(โ2 sin t)2 + (1)2 + (2 cos t)2 dt =โ
5 dt
Thus,
m = Mass =
โซ
ฮณ
ฯ(x, y, z) ds =
โซ 6ฯ
0
2โ
5 t dt = 36ฯ2โ
5 mass-units
โโ
PROBLEMS YOU SHOULD TO DO
Exercise 3.5.1. Evaluate (if possible)โซ
ฮณ x2y ds, where oriented curve ฮณ is determined by the parametric equations
x = 3 cos(t), y = 3 sin(t), 0 6 t 6 ฯ/2. Also show that the parametrization x =โ
9 โ y2, y = y, 0 6 y 6 3,gives the same value. Draw a picture of ฮณ too! and if the above integral is impossible to evaluate by hand, thendonโt worry about it!
Exercise 3.5.2. Evaluate (if possible)โซ
ฮณ(2x2 โ 3yz) ds, where the parametrized curve ฮณ is given by ฮณ(t) =
ใcos t, sin t, cos tใ, for 0 6 t 6 2ฯ. Do not Draw a picture of ฮณ. If the above line integral is impossible to evaluateby hand, then donโt worry about it!
Theorem 3.5.3. Let f(x, y, z) be a continuous function defined on some region D containing the parametrizedcurve ฮณ. Then, if ฮณ is a piece-wise smooth curve, with ฮณ = ฮณ1 โช ฮณ2 โช ยท ยท ยท โช ฮณn, where ฮณ1, ฮณ2, . . . , ฮณn are all smoothand where the terminal point of ฮณi is the same as the initial point of ฮณi+1, for i = 1, 2, . . . , n โ 1, we have
(i)โซ
โฮณ f(x, y, z) ds =โซ
ฮณ f(x, y, z) ds and
(ii)โซ
ฮณ f(x, y, z) ds =โn
i=1
โซ
ฮณif(x, y, z) ds.
Interpretation of Line Integrals of f(x, y) with respect to Arc-Length
As before, if y = f(x) โฅ 0 for all x โ [a, b] thenโซ b
a f(x) dx measures the area of the plane region bounded by thevertical lines x = a, y = b and the interval [a, b] and the graph of y = f(x). In a similar way, if z = f(x, y) โฅ 0 forall (x, y) โ D โ R
2, thenโซ
ฮณf(x, y) ds measures the surface area of the vertical cylinder (usually called a โcurtainโ)
bounded above by the graph of z = f(x, y) and the vertical lines (in space parallel to the z-axis) passing throughthe initial and terminal points of the curve ฮณ (with the arc-length parameter s) and also bounded below by the curveฮณ itself.
Example 3.5.2. Evaluating a line integral over a Piece-Wise smooth Curve ฮณ: Evaluate the line integralโซ
ฮณ(3x โ y) ds, where the plane cure ฮณ is the line segment from (1, 2) to (3, 3), followed by the portion of the circle
x2 + y2 = 18 traversed from (3, 3) clockwise around to (3,โ3).
Solution. Since ฮณ is piece-wise smooth, we must find smooth parametrization of the smooth portions of ฮณ whichare ฮณ1 (the line segment from (1, 2) to (3, 3)) and ฮณ2 the circular part of the circle x2 + y2 = 18 traversed clockwisefrom (3, 3) to (3,โ3).
(A) For the line Segment portion: Use the parametrization formula for any line segment given as:
ฯ(t) = (1 โ t)ฯ0 + tฯ1 for 0 6 t 6 1.
Now treat the initial point of the line segment as position vector ฮณ10= ฯ0 = ใ1, 2ใ and the terminal point (3, 3) as
the position vector ฮณ11= ฯ1 = ใ3, 3ใ so that
ฮณ1(t) = ใ1 + 2t, 2 + tใ for 0 6 t 6 1.
Thus ds = โ ฮณโฒ1(t) โ dt = โ ใ2, 1ใ โ dt =
โ5 dt and f(ฮณ1(t)) = f(1 + 2t, 2 + t) = 3(1 + 2t) โ (2 + t) = 1 + 5t and
so we have
โซ
ฮณ1
f(x, y) ds =
โซ
ฮณ1
(3x โ y) ds =
โซ 1
0
[3(1 + 2t) โ (2 + t)]โ
22 + 12 dt =
โซ 1
0
(5t + 1)โ
5 dt =7
2
โ5
(B) For the curved portion of ฮณ: This is not a line segment so our approach has to be different from that inpart (A). The usual parametrization of a circle with counterclockwise orientation is given as:
ฮณ(t) = ใR cos t, R sin tใ for ฮฑ 6 t 6 ฮฒ.
We want a parametrization with clockwise orientation of a circle of radius R = 3โ
2. So we want to start whenโt = ฯ/4 and end when โt = โฯ/4. This means replacing t in the counterclockwise parametrization of a circlewith โt will gives the clockwise orientation we seek: Thus using the trigonometric identities cos(โt) = cos t andsin(โt) = โ sin t we arrive at
ฮณ2(t) = ใ3โ
2 cos t, โ3โ
2 sin tใ for โ ฯ/4 6 t 6 ฯ/4.
Now f(ฮณ2(t)) = f(3โ
2 cos t,โ3โ
2 sin t) = 3(3โ
2 cos t) โ (โ3โ
2 sin t) = 3โ
2(3 cos t + sin t) and
ds = โ ฮณโฒ2(t) โ dt = โ ใโ3
โ2 sin t, โ3
โ2 cos tใ โ dt =
โ
(โ3โ
2 sin t)2 + (โ3โ
2 cos t)2 dt = 3โ
2 dt
andโซ
ฮณ2
f(x, y) ds =
โซ ฯ/4
โฯ/4
(3 cos t + sin t) 18 dt = 54โ
2.
Finally, combining the results of (A) and (B) we have that
โซ
ฮณ
f(x, y) ds =
โซ
ฮณ1
f(x, y) ds +
โซ
ฮณ2
f(x, y) ds =
โซ
ฮณ
(3x โ y) ds =7
2
โ5 + 54
โ2.
โโ
Exercise 3.5.3. Evaluate (if possible)โซ
ฮณ(x2 + y2) ds, where ฮณ is the piecewise smooth plane curve given by the
circle x2 + y2 = 4 traversed clockwise from (0, 2) to (0,โ2) and the line segment from (0,โ2) to (โ2,โ2) and theline segment from (โ2,โ2) to (โ2, 2) and finally the line segment from (โ2, 2) to (0, 2). Draw a picture of ฮณ. If theabove line integral is impossible to evaluate by hand, then donโt worry about it!
Definition 3.5.2. The line integral of f(x, y, z) with respect to the parameter x along the smoothparametrized space-curve ฮณ(t) = ใx(t), y(t), z(t)ใ for ฮฑ 6 t 6 ฮฒ is written as:
โซ
ฮณ
f(x, y, z) dx = limโ P โโ0
nโ
i=1
f(xi, y1, zi) โxi (provided that is limit exists )
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 parametrizedspace-curve ฮณ(t) = ใx(t), y(t), z(t)ใ for ฮฑ 6 t 6 ฮฒ as:
โซ
ฮณ
f(x, y, z) dy = limโ P โโ0
nโ
i=1
f(xi, y1, zi) โyi
and the line integral of f(x, y, z) with respect to the parameter z along the smooth parametrized space-curveฮณ(t) = ใx(t), y(t), z(t)ใ for ฮฑ 6 t 6 ฮฒ as:
โซ
ฮณ
f(x, y, z) dz = limโ P โโ0
nโ
i=1
f(xi, y1, zi) โzi.
In each case, the line integral is defined whenever the corresponding limit exists and is independent of how we choosethe points (xi, yi, zi) on the curve.
Theorem 3.5.4. [Evaluation Theorem of Line Integrals with respect to the coordinate axes] LetP (x, y, z), Q(x, y, z) and R(x, y, z) be a continuous functions defined on a Path-connected region E โ R
3 containingthe smooth parametrized space-curve ฮณ(t) = ใx(t), y(t), z(t)ใ for ฮฑ 6 t 6 ฮฒ. Then
โซ
ฮณ
P (x, y, z) dx =
โซ ฮฒ
ฮฑ
P (x(t), y(t), z(t)) xโฒ(t) dt
โซ
ฮณ
Q(x, y, z) dy =
โซ ฮฒ
ฮฑ
Q(x(t), y(t), z(t)) yโฒ(t) dt
โซ
ฮณ
R(x, y, z) dz =
โซ ฮฒ
ฮฑ
R(x(t), y(t), z(t)) zโฒ(t) dt.
Notation:
IN THE PLANE
โซ
ฮณ
P (x, y) dx + Q(x, y) dy =
โซ
ฮณ
P (x, y) dx +
โซ
ฮณ
Q(x, y) dy
IN SPACE
โซ
ฮณ
P (x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz =
โซ
ฮณ
P (x, y, z) dx +
โซ
ฮณ
Q(x, y, z) dy +
โซ
ฮณ
R(x, y, z) dz.
PROBLEM 4 Evaluating A Line Integral in Space: You Try these Now!
Evaluate the line integralโซ
ฮณ(4xz + 2y) dx, where the piecewise smooth space-curve ฮณ 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).
Calculateโซ
ฮณ4x dx + 2y dz, where ฮณ 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
โซ
ฮณ
P (x, y) dx + Q(x, y) dy =
โซ
ฮณ
y
x2 + y2dx +
โx
x2 + y2dy,
where ฮณ(t) = ใcos3t, sin3 tใ; 0 6 t 6 ฯ/2. [This hint might be helpful for evaluating the definite integral: Letu = 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 ฮณ, then
(i) If ฮณ is a piecewise smooth curve, then
โซ
โฮณ
f(x, y, z) dx = โโซ
ฮณ
f(x, y, z) dx.
(ii) If ฮณ = ฮณ1 โช ฮณ1 โช ยท ยท ยท โช ฮณn where all the ฮณ1, ฮณ2, . . . , ฮณn are all smooth and the terminal point of ฮณi is the sameas the initial point of ฮณi+1 for all i = 1, 2, . . . , n โ 1, then
โซ
ฮณ
f(x, y, z) dx =
nโ
i=1
โซ
ฮณi
f(x, y, z) dx.
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 ฮณ, for ฮฑ 6 t 6 ฮฒ and if f(x) is continuous on D . Then
โซ
ฮณ
f(x) ds =
โซ ฮฒ
ฮฑ
g(t) dt = G(ฮฒ) โ G(ฮฑ)
where G(t) is an anti-derivative of g(t) = f(ฮณ(t))โ ฮณโฒ(t) โ
Problem 5 Consider the function f(x, y) = x2 + y2 which is continuous on D = R2. Let ฮณ(t) = ใR cos t, R sin tใ
be a circle of radius R > 0 in D with counterclockwise orientation i.e., 0 6 t 6 2ฯ. Provide an interpretation ofthe line integral
โซ
ฮณf(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 wantsto 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 coordinategrid similar to R
2. At special points in your grid you would probably draw arrows (some long and some short pointin different directions) to represent the velocity of the river at point (x, y). In fact, from a mathematical point ofview, you actually have a graph called a vector field. Note this type of graph is different from the graphs thatyouโve 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 โ R
2 whose range consists only of vectors inR
2(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 โ R
2 (a mapping from theplane to a vector space) defined by
F(x, y) = ใP (x, y), Q(x, y)ใ for (x, y) โ 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 โ R
3 (a mapping from the space to a 3-D vectorspace) defined by
F(x, y, z) = ใP (x, y, z), Q(x, y, z), R(x, y, z)ใ for (x, y, z) โ 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 throughouta region in space (it might be the force of gravity or an electromagnetic force of some kind) and that
ฮณ(t) = ใg(t), h(t), k(t)ใ, a 6 t โค b,
is a smooth curve in the region. Then the integral of F ยทT, the scalar component of F in the direction of the curveโsunit 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 ฮณ from t = a to t = b is
W =
โซ t=b
t=a
F ยท T ds. (1)
where the scalar coordinate functions (or component functions) P (x, y, z), Q(x, y, z) and R(x, y, z) are contin-uous (and possibly have continuous first partial derivatives) on an open and path-connected domain E โ R
3.
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) ฮณ with arc-length parametrization contained in E .
In order to accomplish this (calculate work done), we let ฮณ(s) = x(s) i + y(s) j + z(s) k (or ฮณ(s) = ใx(s), y(s), z(s)ใ)be the position vector for a point P0(x, y, z) on the curve ฮณ. If T = T(s) = xโฒ(s) i + yโฒ(s) j + zโฒ(s) k is the unittangent vector at P0, then FT = F ยท T (a scalar function) is the tangential component of the vector field Fat the point P0. The work done by the vector field F in moving the particle from point P0 a short distance โsalong the curve is approximately FT โs = F ยท T โs, and consequently the work done in moving the particle fromarbitrary point A to B along ฮณ is defined to be the work integral given by
W =
โซ
ฮณ
FT ds =
โซ
ฮณ
F ยท T ds
Different ways of writing the work integral
Work =
โซ ฮฒ
ฮฑ
(F ยท T) ds =
โซ
ฮณ
FT ds (The Definition of the Work Integral)
=
โซ
ฮณ
(
F ยท d~ฮณ
ds
)
ds
(Expanded to include ds; emphasizes the
arc-length parameter s and the velocity vector d~ฮณ/ds
)
=
โซ
ฮณ
F ยท d~ฮณ (Compact differential form)
=
โซ
ฮณ
(
F ยท d~ฮณ
dt
)
dt
(Expanded to include dt; emphasizes the parameter t
and the velocity vector d~ฮณ/dt
)
=
โซ
ฮณ
(
Pdx
dt+ Q
dy
dt+ R
dz
dt
)
dt (Abbreviates the components of ฮณ)
=
โซ
ฮณ
P dx + Q dy + R dz (dtโฒs canceled; the most commonly used form).
and T = dxds i + dy
ds j + dzds k = ใdx
ds , dyds , dz
ds ใ = d~ฮณds and โTโ = 1
HOW TO EVALUATE A WORK INTEGRAL
To evaluate the work integral, take these steps.
Step 1. Find a Parametrization of the curve ฮณ with parameter t (if this has not been given)
Step 2. Evaluate F on the curve ฮณ as a function of the parameter t, i.e., Calculate g(t) = F(ฮณ(t)).
Step 3. Find ~ฮณโฒ(t) = d~ฮณ/dt i.e., Take the first derivative with respect to the parameter t of each component ofthe parameterize curve ฮณ.
Step 4. Dot the two vector functions F(ฮณ(t)) with ~ฮณโฒ(t), i.e., calculate F(~ฮณ(t)) ยท ~ฮณโฒ(t)
Step 5. Integrate the function of a single variable computed in Step 4 with respect to the parameter t fromt = ฮฑ to t = ฮฒ (the parameter interval you have from Step 1); i.e., Evaluate the definite integral
โซ t=ฮฒ
t=ฮฑ
F(~ฮณ(t)) ยท ~ฮณโฒ(t) dt.
Hereโs a Quick Example
Example 3.5.4. [Finding work done by a Variable Force Over a Space Curve] Find the work done bythe variable force F given by
F(x, y, z) = ใP (x, y, z), Q(x, y, z), R(x, y, z)ใ = ใy โ x2, z โ y2, x โ z2ใ on E = R3
in moving a particle along the path (or smooth curve) given by ฮณ(t) = ใt, t2, t3ใ; 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).
Solution. To Calculate the work integralโซ
ฮณF ยท d~ฮณ using the 5 Steps above to accomplish this Task.
Step 1. A parametrization of the path or curve ฮณ has been given as
ฮณ = ฮณ(t) = ใt, t2, t3ใ; ฮฑ = 0 6 t 6 1 = ฮฒ.
Step 2. Calculate the force at each point on the path, i.e., F(ฮณ(t)).
F(ฮณ(t)) = F(t, t2, t3) = ใt2 โ t2, t3 โ t4, t โ t6ใ = ใ0, t3 โ t4, t โ t6ใ; 0 6 t 6 1.
Step 3. Find ~ฮณโฒ(t).
~ฮณโฒ(t) = ใxโฒ(t), yโฒ(t), zโฒ(t)ใ = ใ1, 2t, 3t2ใ; 0 6 t 6 1.
Step 4. Dot the two vector functions F(ฮณ(t)) and ~ฮณโฒ(t).
F(ฮณ(t)) ยท d~ฮณ
dt= ใ0, t3 โ t4, t โ t6ใ ยท ใ1, 2t, 3t2ใ
= (t3 โ t4)(2t) + (t โ t6)(3t2)
= 2t4 โ 2t5 + 3t3 โ 3t8; 0 6 t 6 1.
Step 5. Integrate the expression 2t4 โ 2t5 + 3t3 โ 3t8 over the parameter interval 0 6 t 6 1.
WORK =
โซ 1
0
(2t4 โ 2t5 + 3t3 โ 3t8) dt
=[2/5 t5 โ 1/3 t6 + 3/4 t4 โ 1/3 t9
]t=1
t=0=
29
60(Work Units).
โโ
Exercise 3.5.5. Find the work done by the Gravitational vector field
F = F(x, y, z) = โGmM
โจx
(x2 + y2 + z2)3/2,
y
(x2 + y2 + z2)3/2,
z
(x2 + y2 + z2)3/2
โฉ
in moving a particle along the straight-line curve ฮณ in space from A(0, 3, 0) to B(4, 3, 0). Here F is measure inNewtons and length in meters
Exercise 3.5.6. Write out the details for the work integral along a piecewise smooth curve ฮณ defined in an openand connected region D of R
2 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) = ใP (x, y, z), Q(x, y, z), R(x, y, z)ใ and suppose that instead of calling Fa force field, we now view F as the velocity field of a fluid (liquid, air, molten lava, etc) flowing through a region inspace (a tidal basin or the turbine chamber of a hydroelectric generator, for example). Under these circumstances,the integral of the quantity F ยท T along a curve ฮณ in the region gives us the fluidโs flow along ฮณ.
Definition 3.5.5. If ฮณ(t) is a smooth parametrized curve in the domain of a continuous velocity field F, the flowalong the curve from t = ฮฑ to t = ฮฒ is
Flow of F along ฮณ =
โซ
ฮณ
FT ds =
โซ
ฮณ
F ยทT ds (2).
The integral in this case is called a flow integral.
If the curve ฮณ is a simple closed curve, the flow integral is called the circulation of F around the curve ฮณ.
Circulation of F around the closed path ฮณ =
โฎ
ฮณ
FT ds =
โฎ
ฮณ
F ยทT ds.
(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. Ourinterpretation 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 fluidโs velocity field F(x, y, z) = ใx, z, yใ. Find the flowalong the helix
ฮณ(t) = ใcos(t), sin(t), tใ, 0 6 t โค ฯ/2.
Solution. To Calculate the Flow integralโซ
ฮณ F ยท d~ฮณ using the same 5-step procedure outlined above for workintegrals.
Step 1. A parametrization of the path or curve ฮณ has been given as
ฮณ = ฮณ(t) = ใcos(t), sin(t), tใ; ฮฑ = 0 6 t 6 ฯ/2 = ฮฒ.
Step 2. Calculate the velocity at each point alonge the path, i.e., F(ฮณ(t)).
F(ฮณ(t)) = F(cos(t), sin(t), t) = ใcos(t) t, sin(t)ใ, 0 6 t 6 ฯ/2.
Step 3. Find ~ฮณโฒ(t).
~ฮณโฒ(t) = ใxโฒ(t), yโฒ(t), zโฒ(t)ใ = ใโ sin(t), cos(t), 1ใ; 0 6 t 6 ฯ/2.
Step 4. Dot the two vector functions F(ฮณ(t)) and ~ฮณโฒ(t).
F(ฮณ(t)) ยท ~ฮณโฒ(t) = ใcos(t), t, sin(t)ใ ยท ใโ sin(t), cos(t), 1ใ= โ sin(t) cos(t) + t cos(t) + sin(t).
Step 5. Integrate the expression โ sin(t) cos(t) + t cos(t) + sin(t). over the given parameter interval 0 6 t 6
ฯ/2.
Flow =
โซ ฯ/2
0
(โ sin(t) cos(t) + t cos(t) + sin(t)) dt
=
[cos2(t)
2+ t sin(t)
]t=ฯ/2
t=0
=ฯ
2โ 1
2.
โโ
Example 3.5.6. [Finding Circulation Around a circle] Find the circulation of the field F(x, y) = ใx โ y, xใaround the unit circle
ฮณ(t) = ใcos t, sin tใ, 0 โค t โค 2ฯ.
Solution. To Calculate the Circulation integralโฎ
ฮณF ยท d~ฮณ using the same 5-step procedure for work integrals
Step 1. A parametrization of the unit circle ฮณ has been given as
ฮณ = ฮณ(t) = ใcos(t), sin(t)ใ; 0 = ฮฑ 6 t 6 ฮฒ = 2ฯ.
Step 2. Calculate the velocity at each point along the circle:
F(ฮณ(t)) = F(cos(t), sin(t)) = ใcos(t) โ sin(t), cos(t)ใ, 0 6 t 6 2ฯ.
Step 3. Find ~ฮณโฒ(t).~ฮณโฒ(t) = ใxโฒ(t), yโฒ(t)ใ = ใโ sin(t), cos(t)ใ; 0 6 t 6 2ฯ.
Step 4. Dot the two vector functions F(ฮณ(t)) and ~ฮณโฒ(t).
F(ฮณ(t)) ยท ~ฮณโฒ(t) = ใcos(t) โ sin(t), cos(t)ใ ยท ใโ sin(t), cos(t)ใ= โ sin(t) cos(t) + sin2 t + cos2(t)
= 1 โ sin t cos t.
Step 5. Integrate the expression 1 โ sin(t) cos(t) over the given parameter interval 0 6 t 6 2ฯ.
Circulation =
โซ 2ฯ
0
(1 โ sin(t) cos(t)) dt
=
[
t โ sin2 t
2
]t=2ฯ
t=0
= 2ฯ.
โโ
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 ฮณ in the xy-plane, wecalculate the line integral over ฮณ of F ยท n, the scalar component of the fluidโs velocity field in the direction of thecurveโs outward pointing normal vector. The value of this integral is the flux of F across ฮณ. 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, theintegral of F ยท n would still be called the flux of the field across ฮณ.
Definition 3.5.6. [Flux Across a Closed Curve in the Plane] If ฮณ is a smooth closed curve in the domainof a continuous vector field F(x, y) = ใP (x, y), Q(x, y)ใ in the plane and if n is the outward-pointing unit normalvector on ฮณ, the flux of F across ฮณ is
Flux of F across ฮณ =
โฎ
ฮณ
F ยท n ds (3)
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 toanother depends only on the objectโs initial and final positions and not on the path taken in between. Now wediscuss the notion of path independence of work integrals and describe some properties of vector fields in which thework integrals are path independent.
PATH INDEPENDENCE AND CONSERVATIVE VECTOR FIELDS
If A and B are two points in an open region E in space, the workโซ
F ยท d~ฮณ done in moving a particle from A to Bby a field F defined on E usually depends on the path taken. For some special fields, however, the integralโs valueis 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
โซF ยท d~ฮณ 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
E in space and suppose that for any two points A and B in E the workโซ B
AF ยท d~ฮณ done in moving a particle from A
to B is the same value over all paths ฮณ from A to B. Then the integralโซ
F ยท d~ฮณ is said to be path independent inE 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 ascalar function ฯ; that is, if and only if F = โฯ for some ฯ. The function ฯ is then called a potential functionfor F.
Definition 3.5.8. [Potential Function] If F is a vector field defined on E and F = โฯ for some scalar functionฯ defined on an open region E in space, then ฯ 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 scalarfunction whose gradient field is a gravitational field, and so on. As we will see, once we have found a potentialfunction ฯ for a vector field F, we can evaluate all the work integrals in the domain of F with the formula
โซ B
A
F ยท d~ฮณ =
โซ B
A
โฯ ยท d~ฮณ = ฯ(B) โ ฯ(A).
If you think of โF for functions F of several variables as being something like the derivative f โฒ for functions of thesingle variable, then you see that the above equation is the vector calculus analogue of the Fundamental Theorem ofCalculus formula
โซ b
a
f โฒ(x) dx = f(b) โ f(a).
Conservative vector fields have other remarkable properties we as we go along. For example, saying that a vectorfield 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 aboveequation 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 manysmooth 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 = โฯ, this continuity requirement guarantees that the mixed second derivatives of thepotential function ฯ are equal (Clairautโs Theorem), a result we will find revealing in studying conservativevector fields.
3. E is an open region in space. This means essentially that every point in E is the center of a sphere that liesentirely in E .
4. E is connected (all in one piece) which means in an open region every point of E can be connected to everyother 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 connectedset D (subset of R
n). Then the line integralโซ
ฮณ
F ยท d~ฮณ
is independent of path in D if and only if F = โฯ for some scalar function ฯ : Rn โ R (ฯ 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 openconnected subset D of R
n. Then the following conditions are equivalent:
A) The vector field F is conservative on D (i.e., F = โฯ for some scalar function ฯ : Rn โ R)
B) The line integralโซ
ฮณ F ยท d~ฮณ is independent of path in D .
C)โซ
ฮณ F ยท d~ฮณ = 0 for every closed path ฮณ in D .
Theorem 3.5.9. [Test for Path Independence in Space] Let F(x, y, z) = ใP (x, y, z), Q(x, y, z), R(x, y, z)ใ,where the scalar functions P (x, y, z), Q(x, y, z) and R(x, y, z) are continuous together with their first-order partialderivatives in an open and connected subset E of R
3. Then the vector field F is conservative on E if and only ifโร F = 0, that is, if and only if
โP
โy=
โQ
โx,
โP
โz=
โR
โx,
โQ
โz=
โR
โy
In the two variable case, where F(x, y) = ใP (x, y), Q(x, y)ใ. F is conservative on D if and only if
โP
โy=
โQ
โx
Theorem 3.5.10. [The Fundamental Theorem for Line Integrals] Let ฮณ be a piecewise smooth curve givenparametrically by ฮณ = ฮณ(t), a 6 t 6 b, which begins at a = ฮณ(a) and ends at b = ฮณ(b). If the scalar function ฯ iscontinuously differentiable on an open set containing the curve ฮณ, then
โซ
ฮณ
โฯ(ฮณ) ยท d~ฮณ = ฯ(b) โ ฯ(a)
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) = โฮบฮณ
โ ฮณ โ2= โฮบ
โจx
x2 + y2 + z2,
y
x2 + y2 + z2,
z
x2 + y2 + z2
โฉ
where ฮณ = ใx, y, zใ. Show that vector field F is conservative on itโs domain E of definition by finding a potentialfunction for F.
3.6. GREENโS THEOREM IN THE PLANE
In the preceding section, we learned how to evaluate flow integrals for conservative fields. We found a potentialfunction for the field, evaluated it at the path endpoints, and calculated the integral as the appropriate difference ofthose values.
In this section, we see how to evaluate flow and flux integrals across closed plane curves when the vector field is notconservative. The means for doing so is a theorem known as Greenโs Theorem which converts line integrals to doubleintegrals.
Greenโs 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 Greenโs 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, thatGreenโs Theorem applies to any vector field satisfying certain mathematical conditions. It does not depend for itsvalidity on the fieldโs having a particular physical interpretation.
Flux Density at a Point: Divergence
We need two new ideas for Greenโs 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 velocityfield of a fluid flow in the plane and that the first partial derivatives of P and Q are continuous at eachpoint of a region D . We let (x, y) be a point in the region and let R denote a small rectangle withone of its vertices at (x, y) that, along with its interior, lies entirely in the region R. The sides of therectangle, parallel to the coordinate axes, have lengths โx and โy. The rate at which fluid leaves therectangle across the bottom edge is approximately
F(x, y) ยท (โj)โx = โQ(x, y) โx.
This is the scalar component of the velocity at (x, y) in the direction of the outward normal times thelength of the segment. If the velocity is in meeters per second, for example, the exit rate will be inmeters per second times meters or square meters per second. The rates at which the fluid crosses theother 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 + โy) ยท j โx = Q(x, y + โ) โx
Bottom: F(x, y) ยท (โj) โx = โQ(x, y) โx
Right: F(x + โx, y) ยท i โy = P (x + โx) โy
Left: F(x, y) ยท (โi) โy = โP (x, y) โy.
Combining opposite pairs gives us:
Top and Bottom: (Q(x, y + โy) โ Q(x, y)) โx โ(
โQโy โy
)
โx
Right and Left: (P (x + โx, y) โ P (x, y)) โy โ(
โPโx โx
)โy.
Adding these last two equations gives
Flux across rectangle boundary โ(
โP
โx+
โQ
โy
)
โx โy.
We now divide by โx โy to estimate the total flux per unit area or flux density for the rectangle:
Flux across rectangle boundary
rectangle areaโ(
โP
โx+
โQ
โy
)
.
Finally, we let โx and โy 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 โdivergenceof Fโ or โdiv F.โ
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 = โ ยท F =โP
โx+
โQ
โy. (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 divergenceof F is the scalar
div F = โ ยทF =โP
โx+
โQ
โy+
โR
โz
Intuitively, if water were flowing in to a region through a small hole at the point (x0, y0), the lines of flow would divergethere (hence the name) and, since water would be flowing out of a small rectangle about (x0, y0), the divergence ofF at (x0, y0) would be positive. If water were draining out of the hold instead of flowing in, the divergence would benegative.
Exercise 3.6.1. Find the divergence of F(x, y) = (x2 โ y) i + (xy โ y2) j.
Circulation Density at a Point: The k-component of Curl
The second of the two ideas we need for Greenโs Theorem is the idea of circulation density of a vector field F at apoint.
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 offlow rates along the sides. For the bottom edge, the flow rate is approximately
F(x, y) ยท i โx = P (x, y) โx.
This is the scalar component of the velocity F(x, y) in the direction of the tangent vector i times thelength of the segment. The rates of flow along the other sides in the counterclockwise direction areexpressed in a similar way. In all, we have
Top: F(x, y + โy) ยท โi โx = โP (x, y + โ) โx
Bottom: F(x, y) ยท (i) โx = P (x, y) โx
Right: F(x + โx, y) ยท j โy = Q(x + โx) โy
Left: F(x, y) ยท (โj) โy = โQ(x, y) โy.
We add opposite pairs to get:
Top and Bottom: โ(P (x, y + โy) โ P (x, y)) โx โ โ(
โPโy โy
)
โx
Right and Left: (Q(x + โx, y) โ Q(x, y)) โy โ(
โQโx โx
)
โy.
Adding these last two equations gives
Circulation around rectangleโs boundary โ(
โQ
โxโ โP
โy
)
โx โy.
We now divide by โx โy to estimate the total flux per unit area or flux density for the rectangle:
Circulation around the rectangleโs boundary
rectangle areaโ(
โP
โx+
โQ
โy
)
.
Finally, we let โx and โy 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 verticalaxis, looking downward on the xy-plane from the tip of the (vertical) unit vector k. The circulation value is actuallythe k-component of a more general circulation vector we define later on, called the curl of the vector field F. ForGreenโs Theorem, we need only the k-component.
Definition 3.6.2. [k-Component of Circulation Density or Curl] The k-component of the circulationdensity 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 ยท k = โร F ยท k =โQ
โxโ โP
โy. (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 putinto 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 โ y) i + (xy โ y2) j.
GREENโS THEOREM
Theorem 3.6.1. [Greenโs Theorem in the Plane] Let R be a domain (i.e., an open and connected subset) ofthe xy-plane and let ฮณ be a piecewise smooth simple closed curve in R whose interior is also in R. Let P (x, y) andQ(x, y) be functions defined and continuous and having continuous first partial derivatives in R. Then
โฎ
ฮณ
P (x, y) dx + Q(x, y) dy =
โซโซ
D
(โQ
โxโ โP
โy
)
dA
where D is the closed region (i.e., it consists of a domain together with its boundary curve) bounded by ฮณ.
Two forms for Greenโs Theorem
In one form, Greenโs Theorem says that under suitable conditions the outward flux of a vector field across a simpleclosed curve in the plane equals the double integral of the divergence of the field over the region enclosed by thecurve.
Theorem 3.6.2. [Outward Flux-Divergence or Normal Form of Greenโs Theorem] The Outward fluxof a vector field F(x, y) = P (x, y) i + Q(x, y) j across a positively oriented simple close curve ฮณ (with arc-lengthparametrization) in the plane equals the double integral of div F over the region D enclosed by the curve ฮณ.
The outward flux of F across ฮณ =
โฎ
ฮณ
F ยท n ds =
โฎ
ฮณ
โQ(x, y) dx + P (x, y) dy =
โซโซ
D
(โP
โx+
โQ
โy
)
dA. (3)
Exercise 3.6.3. Calculate the outward Flux of the vector field F(x, y) = x i + y2 j across the square boundedby the lines x = ยฑ 1 and y = ยฑ 1 by direct methods and also by using greenโs Theorem. Also, draw the positivelyoriented piecewise smooth parametrized simple closed curve ฮณ enclosing the region as stated in Greenโs Theorem.
In another form, Greenโs Theorem says that the counterclockwise circulation of a vector field around a simple closedcurve 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 Greenโs Theorem] Thecounterclockwise circulation of a vector field F(x, y) = P (x, y) i + Q(x, y) j around a positively oriented simpleclosed curve ฮณ (with arc-length parametrization) in the plane equals the double integral of the k-component of thecurl of the field (i.e., the double integral of curl F ยท k) over the region D enclosed by the curve ฮณ.
Counterclockwise Circulation of F around ฮณ =
โฎ
ฮณ
F ยทT ds =
โฎ
ฮณ
P (x, y) dx + Q(x, y) dy =
โซโซ
D
(โQ
โxโ โP
โy
)
dA
Exercise 3.6.4. Calculate the counterclockwise circulation of the vector field F(x, y) = tanโ1(y/x) i + ln(x2 + y2) jaround the curve ฮณ where ฮณ is the boundary of the region defined by the polar coordinate inequalities 1 6 r 6 2,0 6 ฮธ 6 ฯ using only Greenโs Theorem. Also draw a picture of the region together with its positively orientedboundary curve ฮณ. What is the length of ฮณ? Use your arc-length formula to verify your answer.
3.7. SURFACE INTEGRALS, STOKEโS 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. Choosesample point (xโ
i , yโi ) โ Ri, and let (xโ
i , yโi , zโi ) = (xโ
i , yโi , f(xโ
i , yโi )) 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
โซโซ
S
G(x, y, z) dS = limโPโโ0
nโ
i=1
G(xโi , y
โi , zโi ) โSi (provided this limit exists)
where โSi is the area of the ith surface patch Gi. This definition extends in a natural way to non-Rectangularregions in R
2 (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 attentionto the โsi (arc-length of curve) and โSi (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 value1, 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, theintegral 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 theirintegrals and so on. The domain additivity property takes the form
โซโซ
S
G(x, y, z) dS =
โซโซ
S1
G(x, y, z) dS1 +
โซโซ
S2
G(x, y, z) dS2 + ยท ยท ยท +
โซโซ
Sn
G(x, y, z) dSn.
The idea is that if S is partitioned by smooth curve into a finite number of non-overlapping patches (i.e., if S ispiecewise smooth), then the integral over S is the sum of the integrals over the patches. Thus, the integral of afunction 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 givenby 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 partialderivatives and G(x, y, z) = G(x, y, f(x, y)) is continuous on Dxy, then
โซโซ
S
G(x, y, z) dS =
โซโซ
Dxy
G(x, y, f(x, y))โ
fx(x, y)2 + fy(x, y)2 + 1 dA. (4)
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. Ify = 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
โซโซ
S
G(x, y, z) dS =
โซโซ
Dxz
G(x, h(x, z), z)โ
hx(x, z)2 + hz(x, z)2 + 1 dA. (5)
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. Ifx = 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
โซโซ
S
G(x, y, z) dS =
โซโซ
Dyz
G(u(y, z), y, z)โ
uy(y, z)2 + uz(y, z)2 + 1 dA. (6)
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 atthe points of the surface S and D is the shadow region of S, then we can write
โซโซ
S
G(x, y, z) dS =
โซโซ
D
G(x, y, z)โโH(x, y, z)โ
|โH(x, y, z) ยท p | dA. (7)
where p is a unit vector normal to the shadow region D and โH ยท 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 โ f(x, y) =0 = c or H(x, y, z) = f(x, y) โ z = 0 = c and the unit vector normal to the shadow region D of S is p = k andso 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) toevaluate the surface integral โซโซ
S
y dS,
where the surface S is defined by the equation z = f(x, y) = x + y2 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 thatโ
fx(x, y)2 + fy(x, y)2 + 1 =โ
2โ
2y2 + 1.
Thus, we have
โซโซ
S
G(x, y, z) dS =
โซโซ
D
G(x, y, f(x, y))โ
fx(x, y)2 + fy(x, y)2 + 1 dA
=
โซโซ
D
yโ
2โ
2y2 + 1 dA =โ
2
โซโซ
D
yโ
2y2 + 1 dA (All about Calc II&III)
= 2โ3/2
โซ 1
0
[โซ 2
0
โ
2y2 + 1 4y dy
]
dx
= 2โ3/2
โซ 9
1
โu du =
13โ
2
3.
Using formula (7), we would view the given surface S whose equation z = f(x, y) = x + y2 as a Level surfaceH(x, y, z) = z โ f(x, y) = 0 and so H(x, y, z) = z โ x โ y2. Now
โH(x, y, z) =
โจโH
โx,
โH
โy,
โH
โz
โฉ
= ใโ1, โ2y, โ 1ใ and โโH(x, y, z) โ = โ โ1 i โ2y j โ 1 k โ =โ
2โ
2y2 + 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
we have that |โH(x, y, z) ยท p| = |โH(x, y, z) ยท k| = | โ 1| = 1 and so โ โH(x,y,z) โ|โH(x,y,z)ยทp| =
โ2โ
2y2 + 1. Thus we can
evaluate the surface integral as follows:
โซโซ
S
G(x, y, z) dS =
โซโซ
D
G(x, y, z)โ โH(x, y, z) โ|โH(x, y, z) ยท p| dA =
โซโซ
D
yโ
2โ
2y2 + 1 dA =13
โ2
3
which agrees with the result obtained using formula (4). โโ
Exercise 3.7.1. Use formula (7) to Evaluate the surface integral
โซโซ
S
xyz dS
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
โซโซ
S
xyz dS
where S is the portion of the cone z2 = x2 + y2 between the planes z = 1 and z = 4. Include a sketch of S and asketch of D (the shadow region of S) on separate coordinate axes. Note this problem asks for two evaluation of thegiven 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 unitnormals vectors on S that varies continuously with position. Any patch or sub-portion of an orientable surface is stillorientable. 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 calln(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 itsnormal vector field an oriented surface. The vector n at any point on the surface is called the positive directionat 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 ยท nover S; the outward flux of F across S. Thus, flux is the integral over the surface S of the scalar component ofF 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 unitnormal n is given as
Flux of F across S in the direction of n =
โซโซ
S
F ยท dS =
โซโซ
S
(F ยท n) dS where dS = n dS
If we view the vector field F as the velocity field of a three-dimensional fluid flow(such as air), the flux of F acrossS 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) = +โH(x, y, z)
โโH(x, y, z)โ or n = n(x, y, z) = โ โH(x, y, z)
โโH(x, y, z)โ
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
Outward Flux of F across S =
โซโซ
S
F ยท dS
=
โซโซ
S
(F ยท n) dS (Please note the difference between dS and dS)
=
โซโซ
D
(
F ยท ยฑ โH(x, y, z)
โโH(x, y, z)โ
) โ โH(x, y, z) โ|โH(x, y, z) ยท p | dA
=
โซโซ
D
1
| โH(x, y, z) ยท p | (F ยท ยฑโH(x, y, z)) dA
= ยฑโซโซ
D
1
| โH(x, y, z) ยท p | (F ยท โH(x, y, z)) dA
(6)
Exercise 3.7.3. Find the outward Flux of the vector field F(x, y, z) = yz j + z2 k across S where S is the portioncut from the cylinder y2 + z2 = 1, z > 0, by the planes x = 0 and x = 1. Include a sketch of S and a sketch ofthe 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 =โซโซ
S
ฯ dS (ฯ = ฯ(x, y, z) = density at (x, y, z), mass per unit area)
First moments about the coordinate planes:
Myz =
โซโซ
S
xฯ dS, Mxz =
โซโซ
S
yฯ dS, Mxy =
โซโซ
S
zฯ dS
Coordinates of Center of mass:
x =Myz
M, y =
Mxz
M, z =
Mxy
M
Moments of inertia about coordinate axes:
Ix =
โซโซ
S
(y2 + z2)ฯ dS, Iy =
โซโซ
S
(x2 + z2)ฯ dS, Iz =
โซโซ
S
(x2 + y2)ฯ dS
IL =
โซโซ
S
r2ฯ dS
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 =
โ
IL
M
Example 3.7.2. Letโs find the coordinates of the center of mass of a thin hemispherical shell of radius R > 0whose mass per unit area at each point (x, y, z) on the thin hemispherical shell is the constant ฮบ (i.e., the densityฯ = ฯ(x, y, z) = ฮบ 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 + y2 + z2 = R2, z > 0
The symmetry of the surface about the z-axis tells us that x = y = 0. It remains only to calculate z from theformula z =
Mxy
M . The mass M of the shell is given by
M =
โซโซ
S
ฯ dS = ฮบ
โซโซ
S
dS = ฮบA(S) =1
2(4ฯR2)ฮบ = 2ฯR2ฮบ.
To evaluate the surface integral Mxy we take p to be k so that
โ โH(x, y, z) โ = โ 2ใx, y, zใ โ = 2โ
x2 + y2 + z2 = 2R
and | โH(x, y, z) ยท p | = | โH(x, y, z) ยท k | = 2 | z | = 2z. Recall that
dS =โ โH(x, y, z) โ| โH(x, y, z) ยท p | dA =
2R
2zdA =
R
zdA.
Then
Mxy =
โซโซ
S
zฯ dS =
โซโซ
D
zฮบR
zdA = RฮบA(D) = Rฮบ(ฯR2) = ฯฮบR3;
and now we have the z-coordinate of the center of mass as
z =Mxy
M=
ฯฮบR3
2ฯฮบR2=
R
2.
Hence the center of mass of the thin hemispherical shell of radius R > 0 with constant density is (0, 0, R2 ). โโ
Exercise 3.7.4. In Example 3.7.2 we avoided direct computation of the x and y coordinates of the center of massof the thin shell by use of the phrase โby symmetry of the surface about the z-axis tells us that x = 0 and y = 0.โShow 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) andy = u(x, z) and the Implicit form: F (x, y, z) = c = 0 (Also known as a level surface for the function F ). There isalso 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 โ 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) โ 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 ofD . We call the range of r the surface S defined or traced out by r. Equation (9) together with the domainR constitutes a parametrization of the surface S. The variable u and v are the parameters, and D is theparameter domain. To simplify our discussion, we take D to be the rectangle R defined by inequalities of theform 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 ensuresthat 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) โ R.
PARAMETRIZATIONS OF SOME POPULAR SURFACES
Example 3.7.3. [Parametrizing the Cone] Find a parametrization of the cone
z2
c2=
x2
a2+
y2
b2.
Solution. Cylindrical coordinates provide everything we need. A typical point (x, y, z) on the cone has x = ar cos ฮธ,y = br sin ฮธ, and z = cr, with โH
c 6 r 6Hc and 0 6 ฮธ 6 2ฯ. Taking u = r and v = ฮธ in Equation (9) give the
parametrization
r(r, ฮธ) = a(r cos ฮธ) i + b(r sin ฮธ) j + cr k, โH
c6 r 6
H
c, 0 6 ฮธ 6 2ฯ.
Note: If a = b = c, then we have the familiar RIGHT CIRCULAR CONE. โโ
Example 3.7.4. [Parametrizing an Ellipsoid] Find a parametrization of the Ellipsoid
x2
a2+
y2
b2+
z2
c2= 1.
Solution. Spherical coordinates provide what we need. A typical point (x, y, z) on the Ellipsoid has x =a sinฯ cos ฮธ, y = b sinฯ sin ฮธ, and z = c cosฯ, 0 6 ฯ 6 ฯ, 0 6 ฮธ 6 2ฯ. Taking u = ฯ and v = ฮธ inEquation (9) gives the parametrization
r(ฯ, ฮธ) = (a sin ฯ cos ฮธ) i + (b sinฯ sin ฮธ) j + c cosฯ k, 0 6 ฯ 6 ฯ, 0 6 ฮธ 6 2ฯ.
Note: If a = b = c, then we the familiar SPHERE. โโ
Example 3.7.5. [Parametrizing an Elliptic Cylinder] Find a parametrization of the Cylinder
x2
a2+
y2
b2= 1 (It may look like an ellipse; but it is not!).
Solution. Cylindrical coordinates provide what we need. A typical point (x, y, z) on the elliptic cylinder hasx = a cos ฮธ, y = b sin ฮธ, and z = z, 0 6 ฮธ 6 2ฯ, โH 6 z 6 H . Taking u = ฮธ and v = z in Equation (9) givesthe parametrization
r(ฮธ, z) = a cos ฮธ i + b sin ฮธ j + z k, 0 6 ฮธ 6 2ฯ, โH 6 z 6 H, H > 0.
Note: If a = b, then we have the familiar RIGHT CIRCULAR CYLINDER OF HEIGHT 2H. โโ
Example 3.7.6. [Parametrizing an Elliptic Paraboloid] Find a parametrization of the Elliptic Paraboloid
z
c=
x2
a2+
y2
b2(It may look like an ellipse; but it is not!).
Solution. Cylindrical coordinates provide what we need. A typical point (x, y, z) on the Elliptic Paraboloid has
x = ar cos ฮธ, y = br sin ฮธ, and z = cr2, 0 6 ฮธ 6 2ฯ, 0 6 r 6
โHc . Taking u = r and v = ฮธ in Equation (9)
gives the parametrization
r(r, ฮธ) = ar cos ฮธ i + br sin ฮธ j + cr2 k, 0 6 ฮธ 6 2ฯ, 0 6 r 6
โ
H
c, H > 0.
Note: If a = b, then we have the familiar CIRCULAR PARABOLOID. โโ
Example 3.7.7. [Parametrizing a Hyperbolic Paraboloid] Find a parametrization of the Hyperbolic Paraboloid
z
c=
x2
a2โ y2
b2(It may look like an ellipse; but it is not!).
Solution. Cylindrical coordinates provide what we need. A typical point (x, y, z) on the Elliptic Paraboloid has
x = ar cosh ฮธ, y = br sinh ฮธ, and z = cr2, 0 6 ฮธ 6 2ฯ, 0 6 r 6
โHc . Taking u = r and v = ฮธ in Equation (9)
gives the parametrization
r(r, ฮธ) = ar cosh ฮธ i + br sinh ฮธ j + cr2 k, 0 6 ฮธ 6 2ฯ, 0 6 r 6
โ
H
c, H > 0.
Note: If a = b, then we have the familiar CIRCULAR PARABOLOID. โโ
Here is what I called the trivial Parametrization: Suppose a smooth surface S is given as the level surfaceof 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) โ Domain (f) and z = f(x, y).
r(x, z) = x i + g(x, z) j + z k; for (x, z) โ Domain (g) and y = g(x, z).
r(y, z) = h(y, z) i + y j + z k; for (y, z) โ 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 ร rv is never zero on theparameter 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) =
โซ d
c
โซ b
a
โ ru ร rv โ du dv (10)
Exercise 3.7.5. Find the area of the closed surface S consisting of the upper-half circular cone z = HR
โ
x2 + y2
and the intersecting plane z = H where H and R are positive constants denoting the height and radius of the conerespectively.
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 alevel surface H(x, y, z) = c (non-parametrized), then the Area of S is given by
A(S) =
โซโซ
S
dS =
โซโซ
D
1
| โH(x, y, z) ยท p | โ โH(x, y, z) โ dA (11)
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 notrecognize 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 isโซโซ
S
G(x, y, z) dS =
โซโซ
R
G(r(u, v)) โru ร rvโ du dv
=
โซ d
c
โซ b
a
G(f(u, v), g(u, v), h(u, v)) โ ru ร rv โ du dv
(12)
where R is the uv-parameter domain.
Exercise 3.7.7. Integrate G(x, y, z) = x2 over the upper-half circular cone z =โ
x2 + y2, 0 6 z 6 1.
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 positivelyorientable surface S, then the outward Flux of F across S is given by the surface integral
outward flux of F across S =
โซโซ
S
F ยท dS =
โซโซ
S
F ยท n dS = +
โซโซ
R
F(r(u, v)) ยท ru ร rv du dv (13)
Example 3.7.8. Find the outward flux of the vector field F(x, y, z) = ใx, y, zใ through the
Sphere: r(ฯ, ฮธ) = ใR sin ฯ cos ฮธ, R sin ฯ sin ฮธ, R cosฯใ; 0 โค ฯ โค ฯ, 0 โค ฮธ โค 2ฯ
of radius R > 0.
Solution. First, F(r(ฯ, ฮธ)) = R ใsin ฯ cos ฮธ, sin ฯ sin ฮธ, cosฯใ. Next, rฯ = R ใcos ฯ cos ฮธ, cosฯ sin ฮธ, โ sinฯใand rฮธ = R ใโ sin ฯ sin ฮธ, sinฯ cos ฮธ, 0ใ so that rฯ ร rฮธ = R2 ใsin2 ฯ cos ฮธ, sin2 ฯ sin ฮธ, sinฯ cos ฯใ. Therefore,
F(r(ฯ, ฮธ)) ยท rฯ ร rฮธ = R ใsin ฯ cos ฮธ, sin ฯ sin ฮธ, cosฯใ ยท R2 ใsin2 ฯ cos ฮธ, sin2 ฯ sin ฮธ, sin ฯ cosฯใ= R3(sin3 ฯ cos2 ฮธ + sin3 ฯ sin2 ฮธ + cos2 ฯ sin ฯ)
= R3(sin3 ฯ + cos2 ฯ sin ฯ)
= R3 sin ฯ(sin2 ฯ + cos2 ฯ)
= R3 sin ฯ
Now outward flux is
โซโซ
D
F(r(ฯ, ฮธ)) ยท rฯ ร rฮธ dฯ dฮธ = R3
โซ 2ฯ
0
โซ ฯ
0
sinฯ dฯ dฮธ = 4ฯR3 = 3 (volume of the sphere)
โโ
Example 3.7.9.
Evaluate the surface integralโซโซ
S
F ยท dS for the vector field F(x, y, z) = ใx, โz, yใ where S is the
part of the sphere x2 + y2 + z2 = 4 in the first octant. In other words, find the flux of F acrossS (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 โ z j + y k and the open smooth surface S whichis part of the closed surface (a sphere ) x2 + y2 + z2 = 4 in the first octant, with orientation towards the origin.
Solution.Step 1. Use formulas
Outward Flux of F across S =
โซโซ
S
F ยท dS =
โซโซ
S
F ยท n dS = ยฑโซโซ
D
1
| โH(x, y, z) ยท p | F ยท โH(x, y, z) dA
in Definition 3.7.1
Step 2. Choosing the required n: We view the sphere x2 + y2 + z2 = 4 as the closed non-parametrized levelsurface of H(x, y, z) = c = 0 where H(x, y, z) = x2 + y2 + z2 โ 4; so that it is enough to choose n bycalculating ยฑ โH(x, y, z) = ยฑ2 (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 โH(0, 0, 2) = 4 k (which points away from the origin). Since the problemrequires the opposite orientation we can choose n as a unit vector pointing in the direction of โโH(x, y, z).Thus, we would choose n in the direction of
โ โH(x, y, z) = โ (2x i + 2y j + 2z k).
Based on this choice of n our formula in step 1 above becomesโซโซ
S
F ยท dS =
โซโซ
S
F ยท n dS = โโซโซ
D
1
| โH(x, y, z) ยท p | (F ยท โH(x, y, z)) dA
Step 3. Since portion of the sphere x2 + y2 + z2 = 4 that we are interested in lies in the first octant, then weknow that z > 0 and the projection of S in the xy-plane is the quarter disk
D = {(x, y) | 0 6 x 6 2, 0 6 y 6
โ
4 โ x2}.
Thus a unit vector normal to the shadow region D is k (i.e. p = k.) Hence, | โH(x, y, z) ยทp | = | โH(x, y, z) ยทk | = 2 | z | = 2 z (Recall that z =
โ
4 โ (x2 + y2) > 0).
Step 4. We calculate
โ 1
| โH(x, y, z) ยท p | (F ยท โH(x, y, z)) dA.
This gives us
โ 1
| โH(x, y, z) ยท p | (FยทโH(x, y, z)) dA = โ 1
2zใx, โz, yใยทใ2x, 2y, 2zใ dA = โ x2
zdA = โ x2
โ
4 โ (x2 + y2)dA
Step 5. We can now integrate as followsโซโซ
S
F ยท dS = โโซโซ
D
1
| โH(x, y, z) ยท p | (F ยท โH(x, y, z)) dA.
= โโซโซ
D
x2
โ
4 โ (x2 + y2)dA (where D = {(x, y)| x > 0, y > 0 and 0 6 x2 + y2
6 4})
= โโซ ฯ/2
0
โซ 2
0
r2 cos2 ฮธโ4 โ r2
r dr dฮธ (In polar coordinates)
= โ(โซ ฯ/2
0
cos2 ฮธ dฮธ
) (โซ 2
0
r3(4 โ r2)โ1/2 dr
)
Step 6. To get a result, we must now calculate the definite integrals
โซ ฯ/2
0
cos2 ฮธ dฮธ =1
2
โซ ฯ/2
0
dฮธ +1
4
โซ ฯ/2
0
cos 2ฮธ 2 dฮธ =ฯ
4+ 0 =
ฯ
4
and โซ 2
0
r2
โ4 โ r2
r dr (We must be careful, because the integral is an improper integral !!)
Try letting ฯ = 4 โ r2 so that dฯ = โ2r dr and r2 = 4 โ ฯ . So the integral becomes
โ 1
2
โซ 0
4
4 โ ฯโฯ
dฯ =1
2
โซ 4
0
4 โ ฯโฯ
dฯ
=1
2lim
ฮปโ0+
[โซ 4
ฮป
(4ฯโ1/2 โ ฯ1/2) dฯ
]
=1
2lim
ฮปโ0+
(
16 โ 16
3โ 8
โฮป +
2
3ฮป3/2
)
=16
3
Step 7. Finally, we arrive at an answer (you check my calculations for errors !) Thus the Flux of F across Sin the direction of n (towards the origin) is given as
โซโซ
S
F ยท dS = โโซโซ
D
1
| โH(x, y, z) ยท p | (F ยท โH(x, y, z)) dA
= โโซโซ
D
x2
โ
4 โ (x2 + y2)dA (where D = {(x, y) | 0 6 x 6 2, 0 6 y 6
โ
4 โ x2})
= โโซ ฯ/2
0
โซ 2
0
r2 cos2 ฮธโ4 โ r2
r dr dฮธ (In polar coordinates)
= โ(โซ ฯ/2
0
cos2 ฮธ dฮธ
) (โซ 2
0
r3(4 โ r2)โ1/2 dr
)
= โ ฯ
4ยท 16
3= โ 4 ฯ
3.
โโ
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 aparametrization of S
r(u, v) = 2 sinu cos v i + 2 sinu sin v j + 2 cosu k, 0 6 u 6 ฯ/2, 0 6 v 6 ฯ/2.
An outward unit normal n points in the direction of the vector ru ร rv. The problem requires a unit normal in theopposite direction (i.e., in the direction of rv ร ru = โ(ru ร rv)). Thus the flux of F across S is given by
โซโซ
S
F ยท dS =
โซโซ
S
F ยท n dS
= โโซ ฯ/2
0
โซ ฯ/2
0
F(r) ยท ru ร rv du; dv
Step 1. F(r) = 2 sinu cos v i โ 2 cosu j + 2 sinu sin v k
Step 2. ru = 2 cosu cos v i + 2 cosu sin v j โ 2 sinu k and rv = โ2 sinu sin v i + 2 sinu cos v j + 0 k
Step 3. ru ร rv = 4 sin2 u cos v i + 4 sin2 u sin v j + 4 sin u cosu k
Step 4. F(r) ยท ru ร rv = 8 sin3 u cos2 v
Step 5.โซโซ
S
F ยท n dS = โโซ ฯ/2
0
โซ ฯ/2
0
F(r) ยท ru ร rv du; dv
= โ8
(โซ ฯ/2
0
sin3 u cosu du
)
ยท(โซ ฯ/2
0
cos2 v dv
)
= โ8 ยท 2
3ยท ฯ
4= โ 4ฯ
3โโ
Example 3.7.11. Find the flux of the vector field F(x, y, z) = ใxy, yz, zxใ across the surface S which is part ofthe paraboloid z = 4 โ x2 โ y2 that lies above the square D = {(x, y)| 0 6 x 6 1, 0 6 y 6 1}, and has upwardorientation.
Solution. We use formula (6) (I will deal with the orientation of S later): First, view S as the level surfaceH(x, y, z) = x2 + y2 + z โ 4 = 0 so that โH(x, y, z) = ใ2x, 2y, 1ใ and also that
F ยท โH(x, y, z) = ใxy, yz, zxใ ยท ใ2x, 2y, 1ใ= 2x2y + 2y2z + zx
= 2x2y + 2y2(4 โ x2 โ y2) + x(4 โ x2 โ y2)
= 2x2y + 8y2 โ 2x2y2 โ 2y4 + 4x โ x3 โ xy2.
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 ใ0, 0, 1ใ = k. Thus | โH(x, y, z) ยทk | = | โ 1 | = 1. Now the doubleintegral โซโซ
S
F ยท n dS = ยฑโซโซ
D
1
| โH(x, y, z) ยท p | F ยท โH(x, y, z) dA
= ยฑโซโซ
D
(2x2y + 8y2 โ 2x2y2 โ 2y4 + 4x โ x3 โ xy2) dA
= ยฑโซ 1
0
โซ 1
0
(2x2y + 8y2 โ 2x2y2 โ 2y4 + 4x โ x3 โ xy2) dx dy
= ยฑโซ 1
0
(2
3y + 8y2 โ 2
3y2 โ 2y4 +
7
4โ 1
2y2
)
dy = ยฑ 713
180.
Now there is the issue of n: We want the upward (very bad wording here!) orientation (i.e., we want S orientedso that its gradient field โH(x, y, z) is outward). To choose the proper n, I choose a convenient point on S, letโssay P (0, 0, 4). Now calculate โH(0, 0, 4) = ใ0, 0, 1ใ = k; which points outward away from S. Thus we want n to
point in the same direction as โH(x, y, z). This tells us that the required n = + โH(x,y,z)โ โH(x,y,z) โ = ใ2x, 2y, 1ใโ
4x2 +4y2 +1and
so finally we have that the flux of F across S is 713/180. โโ
Exercise 3.7.8. Calculate the outward Flux of the vector field F(x, y, z) = โx i โ y j + z2 k across a smooth
parametrized surface S where S is the portion of the cone z =โ
x2 + y2 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!).
STOKEโS THEOREM
Stokeโs Theorem says that, under conditions normally met in practice, the circulation of a vector field around theboundary of an oriented surface in space in the direction counterclockwise with respect to the surfaceโs unit normalvector field n equals the integral of the normal component of the circulation density of the field over the surface.
Theorem 3.7.6. [Stokeโs Theorem] Let S be an orientable surface (i.e., S is a two-sided surface) with acontinuously varying unit normal vector field n = n(x, y, z). Let the boundary of S (which we denote by โS) be apiecewise smooth, simple closed curve, oriented consistently with n (i.e., ฮณ = โS is a positively oriented piecewisesmooth 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 withP (x, y, z), Q(x, y, z), and R(x, y, z) having continuous first-order partial derivatives on S and its boundary curveโS. If T denotes the unit tangent vector to ฮณ = โS, then the Circulation of F around the boundary curveฮณ = โS is given by the surface integral
circulation of F around ฮณ =
โซโซ
S
curl F ยท dS =
โซโซ
S
curl F ยท n dS =
โฎ
ฮณ
F ยทT ds =
โฎ
ฮณ
F ยท d~ฮณ
Recall that curl F = โร F =
โฃโฃโฃโฃโฃโฃโฃโฃโฃ
i j k
โโx
โโy
โโz
P Q R
โฃโฃโฃโฃโฃโฃโฃโฃโฃ
=(
โRโy โ โQ
โz
)
i +(
โPโz โ โR
โx
)j +
(โQโx โ โP
โy
)
k.
Example 3.7.12. Calculate the work done by the โforce fieldโ given by
F(x, y, z) = ใxx + z2, yy + x2, zz + y2ใ
when a particle moves under its influence around the edge of the part of the sphere x2 + y2 + z2 = 32 that lies inthe first octant, in a counterclockwise direction as viewed from above.
Solution. Observe that we are being asked to calculate a work integral in space. that is we must calculate a lineintegral
โซ
โS F ยท dโS or if you preferโซ
ฮณ F ยท d~ฮณ where ฮณ = โS. Doing this calculation directly might not be a wise
idea (but you can certainly try!). So we will use Stokeโs Theorem (equivalently Greenโs Theorem in Space). So wecalculate curl F as
curl F(x, y, z) = โร F(x, y, z) =
โฃโฃโฃโฃโฃโฃโฃโฃโฃโฃโฃ
i j k
โโx
โโy
โโz
xx + z2 yy + x2 zz + y2
โฃโฃโฃโฃโฃโฃโฃโฃโฃโฃโฃ
= ใ2y, 2z, 2xใ or 2ใy, z, xใ.
Observe that curl F 6= 0 so that F is a non-conservative force field. Next, we must decide on whether we want toparametrize S or not. I choose to parametrize S as follows:
Parametrized S := r(u, v) = ใ3 sinu cos v, 3 sinu sin v, 3 cosuใ; 0 6 u 6 ฯ/2, 0 6 v 6 ฯ/2;
which naturally gives us the outer unit normal and the positively oriented boundary curve โS or ฮณNow ru =ใ3 cosu cos v, 3 cosu sin v, โ3 sinuใ and rv = ใโ3 sinu sin v, 3 sinu cos v, 0ใ and so
ru ร rv = 9 ใsin2 u cos v, sin2 u sin v, sinu cosuใ
and curl F(r(u, v)) = 6 ใsin u sin v, cosu, sin u cos vใ. Now we have that
curl F(r(u, v)) ยท ru ร rv = 54 (sin3 u sin v cos v + sin2 u cosu sin v + sin2 u cosu cos v).
Therefore,
Work done by F =
โฎ
ฮณ
F ยท d~ฮณ =
โฎ
ฮณ
F ยท T ds =
โซโซ
S
curl F ยท dS
=
โซโซ
D
curl F(r(u, v)) ยท ru ร rv du dv
= 54
โซ ฯ/2
0
โซ ฯ/2
0
(sin3 u sin v cos v + sin2 u cosu(sin v + cos v)) du dv
= 54
(โซ ฯ/2
0
sin3 u du
)(โซ ฯ/2
0
sin v cos v dv
)
+ 54
(โซ ฯ/2
0
sin2 u cosu du
)(โซ ฯ/2
0
sin v + cos v dv
)
= 54 ยท 1
2
[
โ cosu +1
3cos3 u
]u=ฯ/2
u=0
+ 54 ยท 1
3(1 + 1)
= 54 ยท 1
2
[
1 โ 1
3
]
+ 54 ยท 1
3(1 + 1)
= 54 work-energy units
โโ
Exercise 3.7.9. Use Stokeโs theorem to evaluateโฎ
ฮณ F ยทT ds, where F = 2z i + (8x โ 3y) j + (3x + y) k where ฮณ
is the triangular positively oriented curve with vertices at (0, 0, 2), (0, 1, 0), and (1, 0, 0). Include a sketch of S andits boundary curve in your work.
DIVERGENCE THEOREM AND A UNIFIED THEORY
The divergence form of Greenโs Theorem in the plane states that the net outward flux of a vector field across asimple 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 ofa vector field across a closed surface in space can be calculated by integrating the divergence of the field over theregion enclosed by the surface.
Divergence Theorem
The Divergence Theorem says that: Under suitable conditions, the outward flux of a vector field across a closedsurface (oriented outward) equals the triple integral of the divergence of the field over the solid region enclosed bythe surface.
Theorem 3.7.7. [Gaussโs Divergence Theorem] Suppose F(x, y, z) = P (x, y, z); i + Q(x, y, z) j + R(x, y, z) kis a vector field with P , Q, and R having continuous first-order partial derivatives on a solid E with boundary surfaceS (E is the solid region in space enclosed by S). If n denotes the outward unit normal to the surface S = โE , then
Outward flux of F across closed surface S =
โซโซ
S
F ยท dS =
โซโซ
S
F ยท n dS =
โซโซโซ
E
div F dV
Example 3.7.13. [Verifying the divergence theorem] Evaluate both sides of the equation in the DivergenceTheorem for the field F = ใx, y, zใ over the sphere x2 + y2 + z2 = R2.
Solution. Calculating Outward flux using the surface integralโซโซ
S
F ยท n dS:
The outward unit normal field is n =2 ใx, y, zใ
โ
4 (x2 + y2 + z2)=
1
Rใx, y, zใ
and so
F ยท n dS =x2 + y2 + z2
RdS =
R2
RdS = R dS
since x2 + y2 + z2 = R2 on the surface. Therefore,
โซโซ
S
F ยท n dS =
โซโซ
S
R dS = R
โซโซ
S
dS = R(4ฯR2) = 4ฯR3 = 3 (Volume of E ).
Next, we calculateโซโซโซ
E
div F dV . Since F = ใx, y, zใ, then the divergence of F is
div F =โ
โx(x) +
โ
โy(y) +
โ
โz(z) = 1 + 1 + 1 = 3.
Therefore,โซโซโซ
E
div F dV = 3
โซโซโซ
E
dV = 3 (volume of E ) = 3
(4
3ฯR3
)
= 4ฯR3
โโProblem 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
โซโซ
S
(2x + 2y + z2)dS
where S is the sphere x2 + y2 + z2 = a2, a > 0.
MISCELLANEOUS EXERCISES
Problem 1. Let S denote the portion of the parabolic cylinder z = 1 โ x2; 0 6 y 6 2 which lies in the firstoctant. Let ฮณ be piecewise smooth curve denoting the boundary of S, which is positively oriented when viewedfrom above. Consider now the continuous vector field F(x, y, z) = ใ1, 0, y2ใ defined every where. Calculate thecirculation of F around ฮณ 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 + y2 and the circular disk x2 + y2 = 4at z = 4. Let F(x, y, z) = ใx, y, 1ใ be a continuous vector field and the surface S is the boundary of E . Verify theDivergence Theorem.
Problem 3. Let D denote the plane region bounded x2 + y2 โ y = 0, y = 0 and y = โx. Let ฮณ denotethe boundary of D . Calculate
โฎ
ฮณP (x, y) dx + Q(x, y) dy where P (x, y) = โy and Q(x, y) = x by verifying the
counterclockwise circulation form of Greenโs theorem.
Problem 4. Calculate the area of the portion of the sphere x2 + y2 + z2 = 22 that lies between the planes z = 0and z =
โ2.
Problem 5. Consider the continuous vector field F(x, y, z) = ใ2z, ez, 2x + yezใ.(a) Does the vector field have a potential? (Explain by finding it or say why it does not).
(b) Evaluateโซ
ฮณ F ยท d~ฮณ where ฮณ is the space curve given by y = 1 โ x2, z = 0 with 0 6 x 6 1 and oriented sothat x is increasing.
Problem 6. Let E be the solid contained between the surfaces z = 4 โ x2 โ y2 and z = 0. Let S denote theboundary of E positively oriented (i.e., oriented with the outward unit normal field). Consider the continuous vectorfield F(x, y, z) = ใx, y, 1ใ. Evaluate the surface integral
โซโซ
S
F ยท dS as a double and Triple integral.
Problem 7. Let E be the solid inside the cylinder x2 + y2 = 1 bounded by the planes z = 0 and z = 2. LetF(x, y, z) = ใzy, zx, y2 + z2ใ. Setup; but do not Evaluate the surface integral
โซโซ
S
F ยท dS both as a double and
a Triple integral. Clearly state the limits of integration for both integrals.
Problem 8. Let ฮณ be the curve of intersection of the plane z = 3x โ 7 and the right circular cylinder x2 + y2 = 1and has the clockwise orientation as viewed from above. Let F be the continuous vector field given by F(x, y, z) =ใ4z โ 1, 2x, 5y + 1ใ. Calculate
โฎ
ฮณF ยท d~ฮณ (a) directly as a line integral (impossible for your level of understanding
thus far) and (b) calculate using (Stokeโs Theorem) (Of course, I could be wrong about this! At any rate, I will justassume 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 difficultywith the problem lies totally in its wording. The inventor of the problem did not intend for us to do both the lineintegral directly (without the use of Stokeโs Theorem) and the surface integral directly (even though the problemstated 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 analyzesome of the questions, you will find that they contain โlogical flawsโ in the sense that they asked you to do the samething twice! even though the intent is for you to do two different things altogether (this is usually the testerโs faultnot yours). Anyway, real intent was that you do just of the two integrals (one of them is easy and the other is ratherdifficult (but not impossible!) as you would have discovered, so take your pick of only one!). The easy integral isthe surface integral
โซโซ
S
(โรF ยท n) dS (from Stokeโs Theorem). The surface S is the plane z = f(x, y) = 3x โ 7 or
H(x, y, z) = โ3x + z + 7 = 0 (Here we view S as a level surface of H). So letโs calculate this surface integral bythe numbers:
curl F(x, y, z) = โร F(x, y, z) =
โฃโฃโฃโฃโฃโฃโฃโฃโฃ
i j k
โโx
โโy
โโz
4z โ 1 2x 5y + 1
โฃโฃโฃโฃโฃโฃโฃโฃโฃ
= ใ5, 4, 2ใ or 5i + 4j + 2k.
Next, since we are traversing ฮณ clockwise as viewed from above, it means that the surface S is to our right as we
travel around ฮณ, in other words, we negatively orient the surface and so n = โH(x,y,z)โ โH(x,y,z) โ = 1โ
10ใโ3, 0, 1ใ.
However our orientation is negative is we use โn as our unit normal. Thus, by our formula for surface integrals wehave โฎ
ฮณ
F ยท T ds =
โซโซ
D
1
| โH(x, y, z) ยท k | (curl F(x, y, z) ยท โn) dS = โโซโซ
D
โ 13โ10
โ10 dA = 13ฯ
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 unitnormal to the shadow region.
Hereโs the correct wording: Evaluateโซ
ฮณ F ยท d~ฮณ, for the vector field F(x, y, z) = ใ4z โ 1, 2x, 5y + 1ใ, where
the curve ฮณ is the intersection of the circular cylinder x2 + y2 = 1 and the plane z = 3x โ 7, oriented so that it istraversed clockwise when viewed from high up on the positive z-axis. โโ
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
Name (Printed)_______________________________________________
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. (a) Find the inverse of the matrix ๐ด๐ด = ๏ฟฝ1 4 12 7 31 7 5
๏ฟฝ.
(b) Use the matrix ๐ด๐ดโ1 that you found in (a) to solve the system ๏ฟฝ๐ฅ๐ฅ + 4๐ด๐ด + ๐ง๐ง = 1
2๐ฅ๐ฅ + 7๐ด๐ด + 3๐ง๐ง = โ2๐ฅ๐ฅ + 7๐ด๐ด + 5๐ง๐ง = โ1
๏ฟฝ
Antony L Foster
Antony Foster
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
No credit for any other method! Show your work on this page and on the page to the left.
MY ANSWER:
(a) (๐จ๐จ|๐ฐ๐ฐ๐๐) = ๏ฟฝ1 4 12 7 31 7 5
1 0 00 1 00 0 1
๏ฟฝ reduce to reduced row echelon form using just a few elementary row
operations to get. I did not state my operations here!
โ
โโ1 0 0
0 1 00 0 1
โ๐๐ ๐๐๐๐๐๐
โ ๐๐๐๐
๐๐ โ ๐๐๐๐
๐๐๐๐
โ๐๐ ๐๐๐๐
๐๐๐๐โ
โโ
from which we have the desire inverse of ๐ด๐ด as:
๐จ๐จโ๐๐ =
โ
โโโ๐๐ ๐๐๐๐
๐๐โ ๐๐
๐๐
๐๐ โ ๐๐๐๐
๐๐๐๐
โ๐๐ ๐๐๐๐
๐๐๐๐โ
โโ
.
(b) Since the coefficient matrix ๐ด๐ด is nonsingular then the solution to the matrix equation ๐ด๐ด๐๐ = ๐ต๐ต is therefore
๐๐ = ๏ฟฝ๐ฅ๐ฅ๐ด๐ด๐ง๐ง๏ฟฝ = ๐ด๐ดโ1๐ต๐ต =
โ
โโโ2 13
7โ 5
7
1 โ 47
17
โ1 37
17โ
โโ๏ฟฝ
1โ2โ1
๏ฟฝ = ๏ฟฝโ5 2โ2
๏ฟฝ.
Therefore, the unique solution to the given linear non-homogeneous system ๐ด๐ด๐๐ = ๐ต๐ต is:
๐๐ = โ๐๐, ๐๐ = ๐๐, ๐๐ = โ๐๐
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
2 (a) Find all eigenvalues and eigenvectors of the matrix ๏ฟฝ7 โ13 3 ๏ฟฝ
(b) Use your answer to (a) to solve ๏ฟฝ๐ด๐ด1โฒ (๐ด๐ด) = 7๐ด๐ด1(๐ด๐ด) โ ๐ด๐ด2(๐ด๐ด)
๐ด๐ด2โฒ (๐ด๐ด) = 3๐ด๐ด1(๐ด๐ด) + 3๐ด๐ด2(๐ด๐ด)
๏ฟฝ for ๐ด๐ด1(๐ด๐ด) and ๐ด๐ด2(๐ด๐ด) subject to initial conditions
๐ด๐ด1(0) = 2 and ๐ด๐ด2(0) = 1.
MY ANSWER
(a) We seek real values ๐๐ and non zero vectors ๐๐ such that ๐ด๐ด๐๐ = ๐๐๐๐ holds.
๐๐๐๐ โ ๐ด๐ด = ๏ฟฝ๐๐ โ 7 1โ3 ๐๐ โ 3๏ฟฝ and so ๐๐(๐๐) = ๏ฟฝ๐๐ โ 7 1
โ3 ๐๐ โ 3๏ฟฝ = ๐๐2 โ 10๐๐ + 24 = (๐๐ โ 6)(๐๐ โ 4) = 0.
Therefore the eigenvalues of the matrix ๐ด๐ด are ๐๐ = ๐๐ ๐๐๐ด๐ด๐๐ ๐๐ = ๐๐. For the eigenvalue ๐๐ = 6, we solve the homogeneous linear system (6๐๐ โ ๐ด๐ด)๐๐ = 0 for ๐๐ which tells us
that the vector ๏ฟฝ๐๐๐๐๏ฟฝ is the eigenvector corresponding to the eigenvalue ๐๐ = 6.
For the eigenvalue ๐๐ = 4, we solve the homogeneous linear system (4๐๐ โ ๐ด๐ด)๐๐ = 0 for ๐๐ which tells us
that ๏ฟฝ๐๐๐๐๏ฟฝ is the eigenvector corresponding to the eigenvalue ๐๐ = 4.
(b) The given system of first-order linear differential equations has general solution (in matrix form)
๐๐(๐๐) = ๏ฟฝ๐๐๐๐(๐๐)๐๐๐๐(๐๐)๏ฟฝ = ๐๐๐๐ ๏ฟฝ
๐๐๐๐๐๐๐๐๐๐๐๐๏ฟฝ ๐๐
๐๐๐๐๐๐ + ๐๐๐๐ ๏ฟฝ๐๐๐๐๐๐๐๐๐๐๐๐๏ฟฝ ๐๐
๐๐๐๐๐๐ = ๐๐๐๐ ๏ฟฝ๐๐๐๐๏ฟฝ๐๐
๐๐๐๐ + ๐๐๐๐ ๏ฟฝ๐๐๐๐๏ฟฝ ๐๐
๐๐๐๐
where the constants ๐๐1 ๐๐๐ด๐ด๐๐ ๐๐2 are to be determined. Now imposing the initial conditions ๐ด๐ด1(0) = 2 ๐๐๐ด๐ด๐๐ ๐ด๐ด2(0) = 1 we get that
๐๐1 + ๐๐2 = 2 ๐๐๐ด๐ด๐๐ ๐๐1 + 3๐๐2 = 1 (๐๐๐ด๐ด๐ด๐ด๐ด๐ดโ๐น๐น๐น๐น ๐น๐น๐ ๐ ๐๐๐ ๐ ๐ ๐ ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐ ๐ ๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐ ๐ ๐ ๐ ๐น๐น)
from which it follows that ๐๐1 = 52
๐๐๐ด๐ด๐๐ ๐๐2 = โ 12 and so the solution to the initial value system is
๐๐(๐ด๐ด) = ๏ฟฝ๐ด๐ด1(๐ด๐ด)๐ด๐ด2(๐ด๐ด)๏ฟฝ =
52๏ฟฝ1
1๏ฟฝ ๐น๐น6๐ด๐ด โ
12๏ฟฝ1
3๏ฟฝ ๐น๐น4๐ด๐ด
๐๐๐๐(๐๐) =๐๐๐๐๐๐
๐๐๐๐ โ๐๐๐๐๐๐
๐๐๐๐ ๐๐๐๐๐๐ ๐๐๐๐(๐๐) =๐๐๐๐๐๐
๐๐๐๐ โ๐๐๐๐๐๐
๐๐๐๐
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
3. (a) Find the rank of the matrix ๐ด๐ด = ๏ฟฝ 1 โ4 9โ1 2 โ4 5 โ6 10
โ7 1 7๏ฟฝ
(b) Solve the linear system ๏ฟฝ ๐ฅ๐ฅ โ 4๐ด๐ด + 9๐ง๐ง โ 7๐ค๐ค = 0โ๐ฅ๐ฅ + 2๐ด๐ด โ 4๐ง๐ง + ๐ค๐ค = 0
5๐ฅ๐ฅ โ 6๐ด๐ด + 10๐ง๐ง + 7๐ค๐ค = 0๏ฟฝ. Indicate clearly all row operations that you use.
(c) Find the determinant of the matrix ๏ฟฝ3 4 0 601
1000
1 2 12 3 4
2000 3000 4001
๏ฟฝ
MY ANSWER
(a) Bring the matrix to reduced-row echelon form and then count the number of rows with leading oneโs.
๏ฟฝ 1 โ4 9โ1 2 โ4 5 โ6 10
โ7 1 7๏ฟฝ ๐น๐น1 + ๐น๐น2 โ ๏ฟฝ
1 โ4 9 ๐๐ โ๐๐ ๐๐ 5 โ6 10
โ7โ๐๐ 7๏ฟฝ โ 5๐น๐น1 + ๐น๐น3 โ ๏ฟฝ
1 โ4 9 0 โ2 5 ๐๐ ๐๐๐๐ โ๐๐๐๐
โ7โ6 ๐๐๐๐
๏ฟฝ
7๐น๐น2 + ๐น๐น3 โ ๏ฟฝ 1 โ4 9 0 โ2 5 ๐๐ ๐๐ ๐๐
โ7โ6 ๐๐๏ฟฝ โ 2๐น๐น2 + ๐น๐น1 โ ๏ฟฝ
๐๐ ๐๐ โ๐๐ 0 โ2 5 0 0 0
๐๐โ6 0๏ฟฝ โ
12๐น๐น2 โ ๏ฟฝ
๐๐ 0 โ1 0 ๐๐ โ๐๐/๐๐ 0 0 0
5 ๐๐ 0๏ฟฝ
From the last matrix (which is in reduced row echelon form) we can see that the Rank of the original matrix ๐ด๐ด is 2, we have two pivots. Furthermore, the row of zeroes tells us that the homogeneous linear system has infinitely many solutions. So letโs 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
the equivalent system ๏ฟฝ
๐๐ โ ๐๐ + ๐๐๐๐ = ๐๐๐๐ โ ๐๐
๐๐๐๐ + ๐๐๐๐ = ๐๐๐๐ = ๐๐๐๐ = ๐๐
๏ฟฝ from which we let ๐ง๐ง = ๐น๐น ๐๐๐ด๐ด๐๐ ๐ค๐ค = ๐ด๐ด and then determine that the
system has infinitely many solutions which we write in matrix form as
๏ฟฝ๐๐๐๐๐๐๐๐๏ฟฝ = ๏ฟฝ
๐๐๐๐๐๐๐๐๐๐
๏ฟฝ๐๐ + ๏ฟฝโ๐๐โ๐๐ ๐๐ ๐๐๏ฟฝ๐๐ where ๐๐, ๐๐ โ โ.
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
By the way the solutions of the system ๐ด๐ด๐๐ = ๐๐ which we express as linear combination of two
independent vectors ๐๐1 = ๏ฟฝ
๐๐๐๐๐๐๐๐๐๐
๏ฟฝ ๐๐๐ด๐ด๐๐ ๐๐2 = ๏ฟฝโ๐๐โ๐๐ ๐๐ ๐๐
๏ฟฝ spans a fundamental 2-dimensional subspace of โ4
called the Nullspace of the matrix ๐ด๐ด. An important theorem in linear algebra says that the rank of ๐จ๐จ plus the nullity of ๐จ๐จ equals the dimension of ๐จ๐จ. The rank of ๐จ๐จ is the dimension of the row-space of A, this is the number of leading oneโs in the row echelon form of ๐ด๐ด. In our case, the rank of ๐ด๐ด = 2. The null-space of ๐จ๐จ is the solution set to the homogeneous system ๐ด๐ด๐๐ = ๐๐. The dimension of the nullspace is called the nullity of ๐จ๐จ. In our case, the nullity of ๐ด๐ด = 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):
๏ฟฝ๐๐ ๐๐ ๐๐ ๐๐๐๐๐๐
๐๐๐๐๐๐๐๐
๐๐ ๐๐ ๐๐๐๐ ๐๐ ๐๐
๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐
๏ฟฝ = โ๏ฟฝ๐๐ ๐๐ ๐๐ ๐๐๐๐๐๐
๐๐๐๐๐๐๐๐
๐๐ ๐๐ ๐๐๐๐ ๐๐ ๐๐
๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐
๏ฟฝ =
โ๏ฟฝ๐๐ ๐๐ ๐๐ ๐๐๐๐๐๐
๐๐๐๐๐๐๐๐
๐๐ ๐๐ ๐๐โ๐๐ โ๐๐ โ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐
๏ฟฝ = โ ๏ฟฝ๐๐ ๐๐ ๐๐ ๐๐๐๐๐๐๐๐
๐๐ ๐๐ ๐๐โ๐๐ โ๐๐ โ๐๐๐๐ ๐๐ ๐๐
๏ฟฝ = โ๏ฟฝ ๐๐ ๐๐ ๐๐ ๐๐๐๐๐๐๐๐
๐๐ ๐๐ ๐๐๐๐ โ๐๐ โ๐๐๐๐ ๐๐ ๐๐
๏ฟฝ =
๏ฟฝ ๐๐ ๐๐ ๐๐ ๐๐ ๐๐ ๐๐ ๐๐
๐๐ ๐๐ ๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐ ๐๐
๏ฟฝ = (๐๐)(๐๐)(๐๐)(๐๐) = ๐๐ Thus we get that
det๏ฟฝ3 4 0 601
1000
1 2 12 3 4
2000 3000 4001
๏ฟฝ = det๏ฟฝ ๐๐ ๐๐ ๐๐ ๐๐ ๐๐ ๐๐ ๐๐
๐๐ ๐๐ ๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐ ๐๐
๏ฟฝ = 5
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
4. (a) Find the length of the parametrized curve with position vector ๐ธ๐ธ(๐ด๐ด) = โจcos ๐ด๐ด, cos t,โโ2 sin tโฉ; 0 โค t โค 2ฯ.
(b) Let ๐ธ๐ธ be the intersection curve of the surfaces ๐ง๐ง = ๐ฅ๐ฅ2 + ๐ด๐ด2 and ๐ฅ๐ฅ2 + ๐ด๐ด2 + ๐ง๐ง2 = 2, oriented
counterclockwise as seen from above. Find. โซ ๐ด๐ด ๐๐๐ฅ๐ฅ โ ๐ฅ๐ฅ ๐๐๐ด๐ด + ๐ฅ๐ฅ2๐ด๐ด3๐ง๐ง5 ๐๐๐ง๐ง๐ธ๐ธ
MY ANSWER
(a) Let ๐ฟ๐ฟ denote the length of the curve ๐พ๐พ then we calculate ๐ฟ๐ฟ as follows:
๐ณ๐ณ = ๏ฟฝ โ๐ธ๐ธโฒ(๐๐)๐๐๐๐
๐๐โ ๐๐๐๐ = ๏ฟฝ โ๐๐ ๐๐๐๐ = ๐๐๐๐โ๐๐
๐๐๐๐
๐๐ ๐๐๐๐๐๐๐๐๐๐.
And the arc-length parameter ๐น๐น = โซ โ๐ธ๐ธโฒ(๐๐)โ๐ด๐ด0 ๐๐๐๐ = โซ โ2 ๐๐๐๐ = โ2 ๐ด๐ด๐ด๐ด
0 or that ๐ด๐ด = 1โ2๐น๐น. This means
could parameterize the curve as
๐พ๐พ(๐น๐น) = โจcos ๏ฟฝ1โ2
๐น๐น๏ฟฝ , sin ๏ฟฝ1โ2
๐น๐น๏ฟฝ ,โโ2 sin ๏ฟฝ1โ2
๐น๐น๏ฟฝโฉ , 0 โค ๐น๐น โค 2โ2 ๐๐
(b) The curve ๐ธ๐ธ of intersection of the two given surfaces is ๐ฅ๐ฅ2 + ๐ด๐ด2 = 1; ๐ง๐ง = 1 whose usual parametrization is ๐ธ๐ธ(๐ด๐ด) = โจcos ๐ด๐ด, sin ๐ด๐ด, 1โฉ; 0 โค ๐ด๐ด โค 2๐๐ (This is a simple closed space curve in the plane ๐ง๐ง = 1). Define ๐ญ๐ญ(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = โจ๐ด๐ด, โ๐ฅ๐ฅ, ๐ฅ๐ฅ2๐ด๐ด3๐ง๐ง5โฉ as the vector field (though it is not necessary!). We are being asked to calculate the line integral โฎ ๐ญ๐ญ โ ๐๐๐ธ๐ธ๐ธ๐ธ so as usual
๏ฟฝ ๐ญ๐ญ โ ๐๐๐ธ๐ธ = ๏ฟฝ ๐ญ๐ญ๏ฟฝ๐ธ๐ธ(๐๐)๏ฟฝ โ ๐ธ๐ธโฒ(๐๐) ๐๐๐๐ ๐๐๐๐
๐๐๐ธ๐ธ= ๏ฟฝ โ(๐ฌ๐ฌ๐ฌ๐ฌ๐ฌ๐ฌ๐๐ ๐๐ + ๐๐๐๐๐ฌ๐ฌ๐๐ ๐๐) ๐๐๐๐ = โ๐๐๐๐
๐๐๐๐
๐๐
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
5. Let ๐๐ be the surface ๐ง๐ง = ๏ฟฝ๐ฅ๐ฅ2 + ๐ด๐ด2; ๐ง๐ง โค 9, and let ๐๐ be the point (3,4,5) on ๐๐. (a) Find an equation of the tangent plane to the surface ๐๐ at point ๐๐.
(b) Find the surface area of ๐๐.
MY ANSWER
(a) Let ๐ป๐ป(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = ๏ฟฝ๐ฅ๐ฅ2 + ๐ด๐ด2 โ ๐ง๐ง = 0 where (๐ฅ๐ฅ,๐ด๐ด) โ ๐ท๐ท = {(๐ฅ๐ฅ,๐ด๐ด):๐ฅ๐ฅ2 + ๐ด๐ด2 โค 81} (A circular disk of radius 9). Now โ๐ป๐ป(๐ฅ๐ฅ, ๐ด๐ด, ๐ง๐ง) = โจ๐ป๐ป๐ฅ๐ฅ ,๐ป๐ป๐ด๐ด ,๐ป๐ป๐ง๐งโฉ = โจ ๐ฅ๐ฅ
๏ฟฝ๐ฅ๐ฅ2+๐ด๐ด2 , ๐๐๏ฟฝ๐ฅ๐ฅ2+๐ด๐ด2 ,โ1โฉ. The equation of the tangent plane to the given
surface ๐๐ at the point (3,4,5) is ๐๐๐ฏ๐ฏ(๐๐,๐๐,๐๐) โ โจ๐๐ โ ๐๐,๐๐ โ ๐๐, ๐๐ โ ๐๐โฉ = ๐๐
๐๐(๐๐ โ ๐๐) + ๐๐
๐๐(๐๐ โ ๐๐) โ (๐๐ โ ๐๐) = ๐๐
๐๐๐จ๐จ ๐๐๐๐ + ๐๐๐๐ โ ๐๐๐๐ = ๐๐
(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)
๏ฟฝ โ๐๐๐ป๐ป(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง)โ ๐๐๐จ๐จ = โ๐๐ ๏ฟฝ๐๐๐จ๐จ = โ๐๐ ๐จ๐จ(๐ซ๐ซ) = ๐๐๐๐โ๐๐ ๐๐๐ซ๐ซ
๐๐๐๐๐๐๐๐๐๐๐๐๐ซ๐ซ
Just in case you parametrized the circular cone ๐๐:
๐๐(๐ข๐ข, ๐ ๐ ) = โจ๐ข๐ข๐๐๐ด๐ด๐น๐น ๐ ๐ ,๐ข๐ข sin ๐ ๐ ,๐ข๐ขโฉ; 0 โค ๐ข๐ข โค 9, 0 โค ๐ ๐ โค 2๐๐.
Then ๐๐๐๐(๐ข๐ข, ๐ ๐ ) = โจcos๐ ๐ , sin๐ ๐ , 1โฉ ๐๐๐ด๐ด๐๐ ๐๐๐๐(๐ข๐ข, ๐ ๐ ) = โจโ๐ข๐ข sin ๐ ๐ ,๐ข๐ข cos ๐ ๐ , 0โฉ ๐๐๐ด๐ด๐๐ ๐๐๐๐ ร ๐๐๐๐ = โจโ๐ข๐ข cos ๐ ๐ ,โ๐ข๐ข sin๐ ๐ ,๐ข๐ขโฉ
โ๐น๐น๐ข๐ข ร ๐น๐น๐ ๐ โ = โ2 ๐ข๐ข. Now surface area is
๏ฟฝ๐๐๐ ๐ = ๏ฟฝโ๐๐๐๐ ร ๐๐๐๐โ ๐๐๐๐ ๐๐๐๐ =๐ซ๐ซ
๏ฟฝ ๏ฟฝ โ๐๐๐๐ ร ๐๐๐๐โ๐๐๐๐ ๐๐๐๐ = โ๐๐ ๏ฟฝ ๏ฟฝ ๐๐ ๐๐๐๐ ๐๐๐๐ = ๐๐๐๐โ๐๐ ๐๐ ๐๐๐๐๐๐๐๐๐๐๐๐. ๐๐
๐๐
๐๐๐๐
๐๐
๐๐
๐๐
๐๐๐๐
๐๐๐ ๐
There is a nice formula for the surface area of a right circular cone of radius ๐ ๐ and height ๐ป๐ป is given as:
๐๐๐ ๐ โ๐ ๐ 2 + ๐ป๐ป2 .
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 = ๐ ๐ = ๐ป๐ป.
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
6. (a) Let ๐ญ๐ญ be the vector field defined by ๐ญ๐ญ(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = โจ2๐ฅ๐ฅ๐ง๐ง, 3๐ด๐ด, ๐ฅ๐ฅ2โฉ. Explain why the value of the line integral โซ ๐ญ๐ญ โ ๐๐๐ธ๐ธ๐พ๐พ is the same for all curves ๐ธ๐ธ from ๐๐(โ1,1,0) to ๐๐(1,2,2) and find the value of this
integral.
b) Let ๐๐ be the part of the surface ๐ฅ๐ฅ2 + ๐ด๐ด2 + ๐ง๐ง2 = 4 with ๐ฅ๐ฅ โค 0 and ๐ง๐ง โฅ 0. Evaluate the surface integral โฌ ๐ง๐ง3 ๐๐๐๐๐๐
MY ANSWER
(a) If the line integral โซ ๐ญ๐ญ โ ๐๐๐ธ๐ธ๐ธ๐ธ has the same value for all paths or curves from ๐๐(โ1,1,0) ๐ด๐ด๐ด๐ด ๐๐(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 ๐๐(๐๐,๐๐, ๐๐) = ๐ช๐ช, ๐๐. ๐๐. ,๐๐๐๐(๐๐,๐๐, ๐๐) = ๐ญ๐ญ which in turn means that ๐๐ ร ๐๐๐๐(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) =๐๐๐ข๐ข๐น๐น๐ ๐ ๐ญ๐ญ = ๐๐. But this needs to be checked in order to receive credit as shown below.
๐๐๐ข๐ข๐น๐น๐ ๐ F = โ ร ๐ญ๐ญ = ๏ฟฝ
๐๐ ๐๐ ๐๐๐๐๐๐๐ฅ๐ฅ
๐๐๐๐๐ด๐ด
๐๐๐๐๐ง๐ง
2๐ฅ๐ฅ๐ง๐ง 3๐ด๐ด ๐ฅ๐ฅ2
๏ฟฝ = โจ0,0,0โฉ = ๐๐
Now we find the potential for ๐ญ๐ญ as follows:
๐๐(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = โซ (2๐ฅ๐ฅ๐ง๐ง)๐๐๐ฅ๐ฅ = ๐ฅ๐ฅ2๐ง๐ง + โ(๐ด๐ด, ๐ง๐ง) and ๐๐๐๐ (๐ฅ๐ฅ ,๐ด๐ด ,๐ง๐ง)๐๐๐ด๐ด
= ๐๐โ(๐ด๐ด ,๐ง๐ง)๐๐๐ด๐ด
= 3๐ด๐ด. This tells us
immediately that โ(๐ด๐ด, ๐ง๐ง) = 32๐ด๐ด2 + ๐๐(๐ง๐ง) and so
๐๐(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = ๐ฅ๐ฅ2๐ง๐ง +32๐ด๐ด2 + ๐๐(๐ง๐ง)
๐๐๐๐ (๐ฅ๐ฅ ,๐ด๐ด ,๐ง๐ง)๐๐๐ง๐ง
= ๐ฅ๐ฅ2 + ๐๐โฒ(๐ง๐ง) = ๐ฅ๐ฅ2 which implies that ๐๐(๐ง๐ง) ๐๐๐น๐น ๐๐ ๐๐๐ด๐ด๐ด๐ด๐น๐น๐ด๐ด๐๐๐ด๐ด๐ด๐ด. Finally, we have
๐๐ = ๐๐(๐๐,๐๐, ๐๐) = ๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐ as the potential function. Now we use the potential function to calculate the line
integral โซ ๐ญ๐ญ โ ๐๐๐ธ๐ธ๐ธ๐ธ as follows (That was the point of part (a), to spare you the laboring task of calculating
the line integral directly!)
๏ฟฝ ๐ญ๐ญ โ ๐๐๐ธ๐ธ = ๐ธ๐ธ(๐๐,๐๐,๐๐) โ ๐ธ๐ธ(โ๐๐,๐๐,๐๐) = ๐๐ โ๐๐๐๐
=๐๐๐๐๐๐๐ธ๐ธ
(b) Perhaps it is best to parameterize the given portion of the sphere ๐ฅ๐ฅ2 + ๐ด๐ด2 + ๐ง๐ง2 = 4; ๐ฅ๐ฅ โค 0, ๐ง๐ง โฅ 0 as: ๐๐(๐ข๐ข, ๐ ๐ ) = โจ2 sin๐ข๐ข cos ๐ ๐ , 2 sin๐ข๐ข sin๐ ๐ , 2 cos๐ข๐ขโฉ; 0 โค ๐ข๐ข โค ๐๐
2, ๐๐
2โค ๐ ๐ โค 3๐๐
2.
So that ๐๐๐ข๐ข = โจ2 cos๐ข๐ข cos ๐ ๐ , 2 cos๐ข๐ข sin๐ ๐ ,โ2 sin๐ข๐ขโฉ ๐๐๐ด๐ด๐๐ ๐๐๐ ๐ = โจโ2 sin๐ข๐ข sin๐ ๐ , 2 sin๐ข๐ข cos๐ ๐ , 0โฉ
and ๐๐๐ข๐ข ร ๐๐๐ ๐ = โจ4 cos ๐ ๐ sin2 ๐ข๐ข, 4 sin ๐ ๐ sin2 ๐ข๐ข, 4 sin๐ข๐ข cos๐ข๐ขโฉ ๐๐๐ด๐ด๐๐ โ๐น๐น๐ข๐ข ร ๐น๐น๐ ๐ โ = 4 sin๐ข๐ข
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
Now let ๐บ๐บ(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = ๐ง๐ง3 so that ๐บ๐บ๏ฟฝ๐๐(๐ข๐ข, ๐ ๐ )๏ฟฝ = ๐บ๐บ(2 sin๐ข๐ข cos ๐ ๐ , 2 sin๐ข๐ข sin๐ ๐ , 2 cos๐ข๐ข) = 8 cos3 ๐ข๐ข and then ๐บ๐บ๏ฟฝ๐๐(๐ข๐ข, ๐ ๐ )๏ฟฝ โ๐น๐น๐ข๐ข ร ๐น๐น๐ ๐ โ = โ32 cos3 ๐ข๐ข (โ sin๐ข๐ข) . Hence we can evaluate the surface integral over a scalar field in the usual way:
๏ฟฝ๐ฎ๐ฎ(๐๐,๐๐, ๐๐)๐๐๐ ๐ = ๏ฟฝ๐๐๐๐๐๐๐ ๐ = ๏ฟฝ ๐ฎ๐ฎ๏ฟฝ๐๐(๐๐,๐๐)๏ฟฝโ๐๐๐๐ ร ๐๐๐๐โ๐๐๐๐ ๐๐๐๐ = ๐๐๐๐๏ฟฝ ๏ฟฝ ๐๐๐๐๐๐๐๐ ๐๐๐๐ = ๐๐๐๐๐๐
๐๐
๐๐๐๐๐๐
๐๐๐๐๐น๐น๐ ๐ ๐ ๐
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
7. Let ๐ ๐ be the region in the ๐ฅ๐ฅ๐ด๐ด-plane contained between the curves ๐ด๐ด = ๐ฅ๐ฅ + 2 and ๐ด๐ด = ๐ฅ๐ฅ2 โ 2๐ฅ๐ฅ + 4. Let ๐ธ๐ธ be the boundary curve of ๐ ๐ , oriented counterclockwise. Find โซ ๐ฅ๐ฅ๐ด๐ด ๐๐๐ฅ๐ฅ โ (๐ฅ๐ฅ + ๐ด๐ด) ๐๐๐ด๐ด๐พ๐พ
(a) directly, as a line integral AND
(b) as a double integral, by using Greenโs Theorem
MY ANSWER
(a) Observe that the boundary curve is a piecewise smooth simple closed curve made up of two smooth parametrized curves ๐ธ๐ธ๐๐(๐ด๐ด) = โจ๐ด๐ด + 1, ๐ด๐ด2 + 3โฉ; 0 โค ๐ด๐ด โค 1 ๐๐๐ด๐ด๐๐ ๐ธ๐ธ๐๐(๐ด๐ด) = โจโ๐ด๐ด + 2,โ๐ด๐ด + 4โฉ; 0 โค ๐ด๐ด โค 1.
๏ฟฝ ๐ฅ๐ฅ๐ด๐ด ๐๐๐ฅ๐ฅ โ (๐ฅ๐ฅ + ๐ด๐ด) ๐๐๐ด๐ด = ๏ฟฝ (โ๐ด๐ด3 โ ๐ด๐ด2 โ 5๐ด๐ด + 3)๐๐๐ด๐ด = โ1
12
1
0๐ธ๐ธ๐๐
๏ฟฝ ๐ฅ๐ฅ๐ด๐ด ๐๐๐ฅ๐ฅ โ (๐ฅ๐ฅ + ๐ด๐ด)๐๐๐ด๐ด = ๏ฟฝ (โ๐ด๐ด2 + 4๐ด๐ด โ 2)dt = โ1
3
1
0๐ธ๐ธ๐๐
๏ฟฝ ๐๐๐๐ ๐๐๐๐ โ (๐๐ + ๐๐)๐๐๐๐ = โ๐๐๐๐๐๐ โ
๐๐๐๐ = โ
๐๐๐๐๐๐๐ธ๐ธ
(b) Evaluation using Greenโs Theorem (Circulation or Curl form) straight from the notes:
๐ช๐ช๐ช๐ช๐๐๐๐๐๐๐๐๐๐๐๐๐ช๐ช๐ช๐ช๐๐๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐ช๐ช๐๐๐๐๐๐๐ช๐ช๐๐ ๐๐๐๐๐ช๐ช๐๐๐๐๐๐ ๐ธ๐ธ = ๏ฟฝ๐ท๐ท(๐๐,๐๐)๐๐๐๐ + ๐ธ๐ธ(๐๐,๐๐)๐๐๐๐ =๐ธ๐ธ
๏ฟฝ ๏ฟฝ๐๐๐ธ๐ธ๐๐๐๐
โ๐๐๐ท๐ท๐๐๐๐
๏ฟฝ๐๐๐จ๐จ๐ซ๐ซ
= ๏ฟฝ ๏ฟฝ ๏ฟฝ๐๐๐๐๐๐
(โ๐๐ โ ๐๐) โ๐๐๐๐๐๐
(๐๐๐๐)๏ฟฝ๐๐๐๐ ๐๐๐๐ = โ๐๐๐๐๐๐
๐๐+๐๐
๐๐๐๐โ๐๐๐๐+๐๐
๐๐
๐๐
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
8. Let ๐๐ be the triangle with vertices ๐๐(0,0,0), ๐๐(1,2,3), ๐ ๐ (2,1,2). Let ๐ธ๐ธ be the boundary curve of ๐๐, oriented clockwise as seen from above. Calculate โซ ๐ด๐ด ๐๐๐ฅ๐ฅ โ ๐ฅ๐ฅ ๐๐๐ด๐ด + ๐ง๐ง ๐๐๐ง๐ง๐ธ๐ธ
(a) Directly as a line integral AND
(b) as a double integral, by using Stokeโs Theorem.
MY ANSWER
(a) We parametrize the three line segments as follows:
โฉโชโจ
โชโง
๐ธ๐ธ๐๐(๐ด๐ด) = (1 โ ๐ด๐ด)โจ0,0,0โฉ + ๐ด๐ดโจ1,2,3โฉ = โจ๐ด๐ด, 2๐ด๐ด, 3๐ด๐ดโฉ; 0 โค ๐ด๐ด โค 1.
๐ธ๐ธ๐๐(๐ด๐ด) = (1 โ ๐ด๐ด)โจ1,2,3โฉ + ๐ด๐ดโจ2,1,2โฉ = โจ1 + ๐ด๐ด, 2 โ ๐ด๐ด, 3 โ ๐ด๐ดโฉ; 0 โค ๐ด๐ด โค 1.
๐ธ๐ธ๐๐(๐ด๐ด) = (1 โ ๐ด๐ด)โจ2,1,2โฉ + ๐ด๐ดโจ0,0,0โฉ = โจ2 โ 2๐ด๐ด, 1 โ ๐ด๐ด, 2 โ 2๐ด๐ดโฉ; 0 โค ๐ด๐ด โค 1.
๏ฟฝ
๏ฟฝ ๐๐ ๐๐๐๐ โ ๐๐ ๐๐๐๐ + ๐๐ ๐๐๐๐ = ๏ฟฝ ๐๐๐๐ ๐๐๐๐ + ๏ฟฝ ๐๐ ๐๐๐๐ + ๏ฟฝ ๐๐๐๐ โ ๐๐ ๐๐๐๐ =๐๐
๐๐
๐๐
๐๐
๐๐
๐๐๏ฟฝ ๐๐๐๐๐๐ โ ๐๐ ๐๐๐๐ = ๐๐๐๐๐๐ โ ๐๐๐๐๏ฟฝ|๐๐๐๐ = ๐๐๐๐
๐๐๐ธ๐ธ
(b) โซ ๐ด๐ด ๐๐๐ฅ๐ฅ โ ๐ฅ๐ฅ ๐๐๐ด๐ด + ๐ง๐ง ๐๐๐ง๐ง = โฌ ๐๐ ร ๐ญ๐ญ โ ๐๐ ๐๐๐๐ = โฌ 1|โ๐ป๐ป(๐ฅ๐ฅ ,๐ด๐ด ,๐ง๐ง)โ ๐๐|๐ท๐ท๐๐๐ธ๐ธ ๏ฟฝ๐๐ ร ๐ญ๐ญ โ โ๐๐๐ป๐ป(๐ฅ๐ฅ, ๐ด๐ด, ๐ง๐ง)๏ฟฝ๐๐๐ด๐ด
where ๐ป๐ป(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = โ๐ฅ๐ฅ โ 4๐ด๐ด + 3๐ง๐ง = 0 and ๐ญ๐ญ(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = โจ๐ด๐ด,โ๐ฅ๐ฅ, ๐ง๐งโฉ. Now โโ๐ป๐ป(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = โจ1,4,โ3โฉ and ๐๐ = ๐๐ and ๐ท๐ท is the triangle in the ๐ฅ๐ฅ๐ด๐ด-plane with vertices at (0,0), (1,2)๐๐๐ด๐ด๐๐ (2,1) so that |โ๐ป๐ป(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) โ ๐๐| = 3 and ๐๐ ร ๐ญ๐ญ = โจ0,0,โ2โฉ and ๐๐ ร ๐ญ๐ญ โ โ๐๐๐ป๐ป(๐ฅ๐ฅ, ๐ด๐ด, ๐ง๐ง) = 6. So our formula above becomes
๏ฟฝ๐๐ ร ๐ญ๐ญ โ ๐๐๐ ๐ ๐ ๐
= โ๏ฟฝ๐๐๐๐
(โ๐๐) ๐๐๐จ๐จ = ๐๐๏ฟฝ๐๐๐จ๐จ = ๐๐(๐จ๐จ๐๐๐๐๐๐ ๐ช๐ช๐๐ ๐ซ๐ซ) = โโจ๐๐,๐๐,๐๐โฉ ร โจ๐๐,๐๐,๐๐โฉโ๐ซ๐ซ๐ซ๐ซ
= โโจ๐๐,๐๐,โ๐๐โฉโ = ๐๐
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
9. Let ๐๐ be the solid contained between the surfaces ๐ง๐ง = 4 โ ๐ฅ๐ฅ2 โ ๐ด๐ด2 and ๐ง๐ง = ๐ฅ๐ฅ2 + ๐ด๐ด2 โ 4. Let ๐ญ๐ญ be the vector field defined by ๐ญ๐ญ(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = โจ๐ฅ๐ฅ,๐ด๐ด, 1โฉ. Use the outward pointing unit normal vector to calculate โฌ ๐ญ๐ญ โ ๐๐ ๐๐๐๐๐๐ (a) directly as a surface integral AND
(b) as a triple integral, by using the Divergence Theorem.
MY SOLUTION
(a) Let ๐๐1 denote the surface ๐ง๐ง = 4 โ (๐ฅ๐ฅ2 + ๐ด๐ด2) or ๐ป๐ป1(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = ๐ฅ๐ฅ2 + ๐ด๐ด2 + ๐ง๐ง = 4. Then โ๐ป๐ป1(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = โจ2๐ฅ๐ฅ, 2๐ด๐ด, 1โฉ and |โ๐ป๐ป1(๐ฅ๐ฅ, ๐ด๐ด, ๐ง๐ง) โ ๐๐| = 1 and โ๐น๐น(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) โ โ๐ป๐ป1(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = 2๐ฅ๐ฅ2 + 2๐ด๐ด2 + 1 where ๐ท๐ท = {(๐ฅ๐ฅ,๐ด๐ด): ๐ฅ๐ฅ2 + ๐ด๐ด2 โค 4} So since the surface ๐๐1 is positively oriented, then we use
๏ฟฝ ๐ญ๐ญ โ ๐๐ ๐๐๐๐1 =๐๐1
๏ฟฝ1
|โ๐ป๐ป1(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) โ ๐๐| ๏ฟฝโ๐น๐น(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) โ โ๐ป๐ป1(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง)๏ฟฝ๐๐๐ด๐ด๐ท๐ท
= ๏ฟฝ ๏ฟฝ (2๐น๐น2 + 1)๐น๐น ๐๐๐น๐น ๐๐๐๐ = 20๐๐2
0
2๐๐
0
Let ๐๐2 denote the surface ๐ง๐ง = ๐ฅ๐ฅ2 + ๐ด๐ด2 โ 4 or ๐ป๐ป2(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = โ๐ฅ๐ฅ2 โ ๐ด๐ด2 + ๐ง๐ง = โ4. Then โโ๐ป๐ป2(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = โจ2๐ฅ๐ฅ, 2๐ด๐ด,โ1โฉ and |โ๐ป๐ป2(๐ฅ๐ฅ, ๐ด๐ด, ๐ง๐ง) โ โ๐๐| = 1 and โ๐น๐น(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) โ โโ๐ป๐ป2(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) = 2๐ฅ๐ฅ2 + 2๐ด๐ด2 + 1 where ๐ท๐ท = {(๐ฅ๐ฅ,๐ด๐ด):๐ฅ๐ฅ2 + ๐ด๐ด2 โค 4} So since the surface ๐๐2 is positively oriented, then we use
๏ฟฝ ๐ญ๐ญ โ ๐๐ ๐๐๐๐2 =๐๐2
โ๏ฟฝ1
|โ๐ป๐ป2(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) โ ๐๐| ๏ฟฝโ๐น๐น(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง) โ โ๐ป๐ป2(๐ฅ๐ฅ,๐ด๐ด, ๐ง๐ง)๏ฟฝ๐๐๐ด๐ด๐ท๐ท
= โ๏ฟฝ ๏ฟฝ (1 โ 2๐น๐น2)๐น๐น ๐๐๐น๐น ๐๐๐๐ = 12๐๐2
0
2๐๐
0
Finally we get โฌ ๐ญ๐ญ โ ๐๐ ๐๐๐ ๐ = โฌ ๐ญ๐ญ โ ๐๐ ๐๐๐ ๐ ๐๐ + โฌ ๐ญ๐ญ โ ๐๐ ๐๐๐ ๐ ๐๐ = ๐๐๐๐๐๐ + ๐๐๐๐๐๐ = ๐๐๐๐๐๐๐ ๐ ๐๐๐ ๐ ๐๐๐ ๐
(b) By the Divergence Theorem we have:
๏ฟฝ๐ญ๐ญ โ ๐๐ ๐๐ = ๏ฟฝ๐๐๐๐๐๐ ๐ญ๐ญ ๐๐๐ ๐ = ๐๐๏ฟฝ๐๐๐ ๐ = ๐๐ (๐ ๐ ๐ช๐ช๐ช๐ช๐๐๐ฝ๐ฝ๐๐ ๐ช๐ช๐๐ ๐ป๐ป) = ๐๐๏ฟฝ ๏ฟฝ ๏ฟฝ ๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐๐ ๐ = ๐๐๐๐๐๐๐๐โ๐๐๐๐
๐๐๐๐โ๐๐
๐๐
๐๐
๐๐๐๐
๐๐๐ป๐ป๐ป๐ป๐ ๐
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)
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 donโt 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).
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. ยฉ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐น๐น๐ด๐ด๐น๐น๐ด๐ด๐น๐น๐น๐น 2008
HOW TO SCORE 120 OUT OF 100 (YOU MAKE IT A THING OF PERFECTION!)