Exam 1 Solutions
Problem 1. (10 points) Consider the transformation T : R3 R3 such that
T (1, 0, 0) = (2, 1, 4), T (0, 1, 0) = (4, 3, 6), T (0, 0, 1) = (0, 1, 2).
1. Determine the null space of T .
2. If A is the plane formed by span(cfw_(2, 5, 3), (1, 1,
MODEL ANSWERS TO HWK #2
(18.022 FALL 2010)
(1) (1.3.27)
(a) First let us x a labeling for the points: W1 , W2 and W3 are the centers of what we will
call the circles 1, 2 and 3, respectively, O is the point where all three circles intersect,
and A, B and
MODEL ANSWERS TO HWK #5
1. Let f : R2 R2 be the function given by f (x, y ) = (x2 + y 2
1, y 2 x2 (x + 1). Then we are looking for solutions to the equation
f (x, y ) = (0, 0).
We compute the derivative of f ,
Df (x, y ) =
2x
2y
.
2x 3x2 2y
The determina
MODEL ANSWERS TO HWK #7
(18.022 FALL 2010)
2
2
(1) (a) F is a gradient eld given by the potential f (x, y ) = x + y2 + C .
2
(b) The ow line r(t) = (x(t), y (t) satises x (t) = x(t), y (t) = y (t) and x(0) = a, y (0) = b.
The solution is x(t) = A1 et and
Practice Exam 2 Solutions
Problem 1. Let f : R2 R be a scalar eld. For each of the following questions,
answer yes or no. If the answer is yes, cite a theorem or give a brief sketch
of a proof. If the answer is no, provide a counterexample.
1. Suppose f (
Exam 2 Solutions
Problem 1. Consider f (x, y ) = (xy + y )10 on the square Q = [0, 1] [0, 1]. Evaluate
f dxdy .
Q
Solution The function
is continous on R2 . Thus,
Fubinis Theorem to get:
Q
10
(xy +y ) d xd y =
0
1
1
f (x, y ) = (xy + y )10
0
f (x, y )dx i
FINAL EXAM PRACTICE MATERIALS
1. Announcements
Oce Hours the week of May 9th:
Vera  Wednesday, 34 pm; Friday, 24 pm
Christine  Tuesday, 4:305:30 pm; Wednesday 910 am; Sunday, 58 pm
Do not expect posted solutions to all of the problems liste
Practice Exam 1 Solutions
Problem 1. Let A be an m n matrix and r be the rank of A.
1. Describe the dimension of the solution space of the equation Ax = 0 in terms
of m, n, r.
2. Suppose there exists c such that Ax = c does not have a solution. What can
y
THIRD MIDTERM
MATH 18.022, MIT, AUTUMN 10
You have 50 minutes. This; host; is Closed book, closed notes, 110 calculators.
Name: If [ODEL NS BEES
Signature:
Recitation Time:
There are 5 problems, and the total number of points is 100. Show
all y
MODEL ANSWERS TO HWK #4
(18.022 FALL 2010)
(1) (i) f is nowhere continuous. Let = 1 . If x is rational, since irrational numbers are dense in
2
reals, for any > 0 there is irrational y such that y x < , and f (y ) f (x) = 1 > .
Therefore f is not cont
SECOND MIDTERM
MATH 18.022, MIT, AUTUMN 10
You have 50 minutes. This test is closed book, closed notes, no calculators.
Name:_\)jl_5§:L.&\Lr2£S
Signature:
Recitation Time:
There are 5 problems, and the total number of points is 100. Show
MODEL ANSWERS TO HWK #1
1.1.20. (a) Suppose that = (a1 , a2 , a3 ). Then
0.
a
0 = (0 a1 , 0 a2 , 0 a3 ) = (0, 0, 0) = .
a
0
The case of a vector in R2 is similar (and easier).
(b) . Suppose that = (a1 , a2 , a3 ). Then
a
a
a
1 = (1 a1 , 1 a2 , 1 a3 ) = (a
MODEL ANSWERS TO HWK #6
(18.022 FALL 2010)
(1) The curve C is given in rectangular coordinates by () = (f () cos(), f () sin(). Then
r
() = (f () cos() f () sin(), f () sin() + f () cos(),
r
and the arc length of C is given by
s() =
( ) d
r
=
(f ( ) c
MODEL ANSWERS TO HWK #3
(18.022 FALL 2010)
(1) In cartesian coordinates D is the region
(x a)2 + (y a)2 a2
1 z 3
This translates to the cylindrical coordinates
r2 2ar(cos + sin ) a2
1 z 3
(2) The region D is
2a
a sin cos a
(3) (i) True. For any c C ther
GREEN'S THEOREM PROBLEM
Theorem 0.1. Let R1 be the region in R2 bounded by the curves x = d, y = c, y = f (x) where f (x) = y is the same curve as x = g(y) (i.e. f is invertible on the interval of interest) and f (x) > c on the region of interest. Prove P
GREEN'S THEOREN AND I T S APPLICATIONS
The d i s c u s s i o n i n 1 1 . 1 9

11.27 o f , Apostol i s n o t complete
We
n o r e n t i r e l y r i g o r o u s , a s t h e a u t h o r himself p o i n t s o u t . give here a rigorous treatment.
nr e e n '
Notes  double integrals.
on (Read 11.111.5 of Apostol.)
Just as for the case of a single integral, we have the
following condition for the existence of a double integral:
Theorem 1 (Riemann condition). Q = [arb] x [c,d] given any s
E
.
Then ther
Derivatives  vector functions.
of Recall that if x is a point of a scalar function of x, R "
and if f
f ( 5 ) is (if it
then the derivative of
exists) is the vector
For some purposes, it will be convenient to denote the derivative
of
f
by a row mat
The 
inverse of a matrix
We now consider the pro5lern of the existence of multiplicatiave
inverses for matrices. A t this point, we must take the noncommutativity of matrix.multiplicationinto account.Fc;ritis perfectly possible, given a matrix A, that
Matrices We have already defined what we mean by a matrix. we introduce algebraic operations into the set of matrices. Definition. If A
k by
In this section,
and B are two matrices of the same size, say n matrix obtained by adding
n,
we define A
t
B to be
Linear Spaces
we have seen (12.112.3 of Apostol) that ntuple space
V ,
has the following properties : Addition:
1. (Commutativity) A + B = B + A.
2. (Associativity) A
+ (B+c) = (A+B) +
for all A.
C.
0

3. (Existence of zero)
There is an element
suc
CONTENTS
1
A,
Linear spaces Linear independence GaussJordan elimination Parametric equations of lines and planes in Vn Parametric equations for kplanes in Vn
A1
A7
A17
A25
A32
> Matrices Systems of linear equations Cartesian equations of kpla
StokesI Theorem
Our text states and proves Stokes' Theorem in 12.11, but ituses the scalar form for writing both the line integral and the surface integral
involved. In the applications, it is the vector form of the theorem that is
most likely to be q
Solutions for PSet 10
1. (E7:1,2) 1 Let l be the arclength of C, and parameterize C by its arclength: (t) = (x(t), y(t). Then (t) = 1 thus n(t) = (y (t), x (t). We have l f n ds = (P (t), Q(t) (y (t), x (t) dt = Q dx + P dy
C 0 C
2 Using Green's Theore