Name
ID
MATH 311
Final Exam
Section 200
Solutions
1.
Spring 2001
1
/30 4
/20
2
/10 5
/10
3
/15 6
/15
P. Yasskin
30 points Let SŸ2, 2 be the set of 2 • 2 symmetric matrices, i.e. 2 • 2 matrices M satisfying M T M.
Consider the function L : MŸ2, 2 ¯ SŸ2, 2 given by LŸX X X T .
a. Ÿ5 Show that SŸ2, 2 is a subspace of MŸ2, 2 , the vector space of 2 • 2 matrices.
A, B S
b.
®
Ÿ
sA tB T sA T tB T sA tB
®
sA tB S
5 Find a basis for SŸ2, 2 . What is the dimension of SŸ2, 2 ?
Ÿ M
a b
M
a b
S
c d
1 0
0 0
bc
«
1 0
a
b d
Basis is
c.
A T A, B T B
®
0 0
0 1
,
1 0
0 1
b
,
d
1 0
0 0
0 1
0 0
0 1
Dimension 3
.
5 Show L is linear.
Ÿ Recall LŸX X X T . So
LŸsA tB ŸsA tB ŸsA tB T ŸsA tB ŸsA T tB T sŸA A T tŸB B T sLŸA tLŸB d.
Ÿ
15 For the linear function L, identify
S
1 DomŸL MŸ2, 2 Ÿ dim DomŸL 4
S
1 CoDomŸL SŸ2, 2 Ÿ dim CoDomŸL 3
S
3 KerŸL Ÿ M s.t. LŸM 0
M s.t. M M T 0
M s.t. M T "M
antisymmetric matrices
Span
a b
s.t.
c d
0
a c
b d
"a "b
"c "d
0
b
"b 0
1
"1 0
dim KerŸL 1
1
S
a b
3 RanŸL £LŸM ¤ £M M T ¤ Ÿ 1 0
2a
0 0
Span
1 0
,
0 0
c d
0 1
Ÿb c 1 0
0 1
,
1 0
a c
2d
2a
b d
bc
b c 2d
0 0
0 1
0 0
SŸ2, 2 0 1
dim RanŸL 3
S
3 1 " 1?
Circle:
3 onto?
Ÿ No
Yes
No
RanŸL CoDomŸL SŸ2, 2 Why?
S
Yes
dim KerŸL 1 p 0
Why?
S
Circle:
Ÿ 1 Verify the Nullity-Rank Theorem for L.
Ÿ dim KerŸL dim RanŸL 1 3 4 dim MŸ2, 2 dim DomŸL 10 points
2.
Consider the function of two matrices X a b
c d
and Y p q
r s
given by
X, Y ™ trŸXY ˜
where tr means ”trace” which is the sum of the principle diagonal entries, i.e. tr
Explain why
,
y z
is an inner product on SŸ2, 2 , but is not an inner product on MŸ2, 2 .
X, Y ™ trŸXY tr
a b
p q
c d
r s
Y, X ™ trŸYX tr
p q
a b
r s
c d
˜
˜
So
w x
tr
tr
ap br aq bs
cp dr cq ds
ap cq pb qd
ra sc br ds
w z.
ap br cq ds
ap cq br ds ˜X, Y ™
is symmetric.
˜rX Z, Y ™ trŸŸrX Z Y trŸrXY ZY trŸrXY trŸZY r trŸXY trŸZY r ˜X, Y ™ ˜Z, Y ™
So , is linear.
,
X, X ™ trŸXX tr
˜
a b
a b
c d
c d
tr
a 2 bc ab bd
a 2 2bc d 2
ca dc bc d 2
If X SŸ2, 2 then b c and ˜X, X ™ a 2 2b 2 d 2 u 0 and 0 only if a b d 0, so that X 0.
So , is positive definite on SŸ2, 2 .
If X MŸ2, 2 then a 2 2bc d 2 may not be positive, e.g. if X So
,
0
1
"1 0
then ˜X, X ™ "2.
is NOT positive definite on MŸ2, 2 .
2
pŸ"1 3.
15 points
Consider the linear map L : P 2
¯
R given by LŸp 3
pŸ0 .
pŸ 1 a.
5 Find the matrix of L relative to the bases e £e 1 1, e 2 t, e 3 t 2 ¤ for P 2 and
1
0
0
,i 2 ,i 3 i i 1 for R 3 . Call it A.
0
1
0
Ÿ L Ÿe 1 L Ÿe 2 L Ÿe 3 0
0
e 1 Ÿ" 1 1
e 1 Ÿ0 i 1 i 2 i 3
1
e 1 Ÿ1 1
e 2 Ÿ" 1 "1
e 2 Ÿ0 0
e 2 Ÿ1 1
e 3 Ÿ" 1 1
e 3 Ÿ0 1 "1 1
" i 1 i 3
A
i
ªe
1
0 0
1
1 1
i 1 i 3
0
1
e 3 Ÿ1 b.
1
5 Find the matrix of L relative to the bases q £q 1 1 t 2 , q 2 t t 2 , q 3 t 2 ¤ for P 2 and
0
1
2
v 1 for R 3 . Call it B.
, v 2 , v 3 v
0
0
1
Ÿ 1
1
q 1 Ÿ" 1 L Ÿq 1 L Ÿq 2 L Ÿq 3 q 1 Ÿ0 2
1
q 1 Ÿ1 2
q 2 Ÿ" 1 0
q 2 Ÿ0 0
q 2 Ÿ1 2
q 3 Ÿ" 1 1
q 3 Ÿ0 q 3 Ÿ1 2
0
v 3
0 2 0
2v1
B
ª
v q
0 0 1
1 0 0
v 2
1
3
c.
5 Find the matrix B by a second method.
Ÿ B
ª
v q
C
A
C
ª iªe eªq
v i
1 0 0
q 1 1 t e 1 e 3 ,
q 2 t t e 2 e 3 ,
2
v 1 i 3 ,
2
v 2 i 1 i 3 ,
q 3 t e 3
2
C
i
B
ª
v q
4.
C
A
C
ª iªe eªq
v i
ªv
"2 0
1
0 0
0 1 0
1
1
1 1
1 1 1
1
0
0
C
ª
v i
1 1 2
1 0 0
1
1 1 1
0 0 1
1 "1 1
0
0 1 0
0 1 2
v 3 2i1 i 2 2i3
"1
C
ª
e q
"1
0
1
"2 0
0
1
1
0
0 2 0
0 0 1
1 0 0
20 points Consider the helix H parametrized by rŸt Ÿ4 cos t, 4 sin t, 3t between A Ÿ4, 0, 0 and
B Ÿ"4, 0, 3= .
a.
Ÿ
B
ds of the vector field F
Ÿyz, "xz, z along the helix H.
10 Compute the line integral ; F
A
H
Ÿyz, "xz, z Ÿ12t sin t, "12t cos t, 3t v Ÿ"4 sin t, 4 cos t, 3 0ttt=
F
v "48t sin 2 t " 48t cos 2 t 9t "48t 9t "39t
F
B
ds ; = "39t dt "39 t 2 = " 39= 2
; F
2 0
2
A
0
H
b.
Ÿ
10 Find the total mass of the helix H if the linear mass density is > z 2 .
|v| 16 sin 2 t 16 cos 2 t 9 5
> z 2 9t 2
3 =
15= 3
M ; > ds ; > |v|dt ; 9t 2 5 dt 45 t
3 0
0
=
10 points
5.
Compute > x dx z dy " y dz around the boundary of the triangle with vertices Ÿ0, 0, 0 ,
0, 1, 0 and Ÿ0, 0, 1 , traversed in this order of the vertices.
Ÿu, v Ÿ0, u, v .
HINT: The yz-plane may be parametrized as R
Ÿ
•F
dS where T is the triangle and T is its boundary. From the
ds ;; By Stokes therorm ; F
T
T
direction the boundary is traversed, the normal to T must point in the positive x-direction.
e u • e v Ÿ1, 0, 0 e v Ÿ0, 0, 1 e u Ÿ0, 1, 0 N
•F
i
j
k
x y z
x
z
Ÿ"2, 0, 0 •F
N
"2
"y
The triangle satisfies 0 t y t 1 and 0 t z t 1 " y, or 0 t u t 1 and 0 t v t 1 " u. So
•F
dS ; 1 ; 1"u Ÿ"2 dv du ; 1 Ÿ"2 Ÿ1 " u du "2 u " u 2 1 "1
;; 2 0
0 0
0
T
4
6.
15 points
Gauss’ Theorem states
F
dS
dV ;; F
;;; V
V
where V is the total boundary of V with OUTWARD normal.
Let V be the solid cone x 2 y 2 t z t 2.
2
x 2 y 2 for z t 2
Let C be the conical surface z with UPWARD normal.
-2
Let D be the disk x y t 4 with z 2
2
2
0
r
-2
2
2
with UPWARD normal.
dS for F
Ÿxy 2 , yx 2 , z 3 in two ways.
Compute ;; F
C
a.
dS explicitly.
5 Method I: Parametrize C and compute ;; F
Ÿ C
C: Polar coordinates:
C Ÿr, 2 Ÿr cos 2, r sin 2, r R
Ÿxy 2 , yx 2 , z 3 F
r 3 cos 2 sin 2 2, r 3 sin 2 cos 2 2, r 3
e r • e 2 Ÿ"r cos 2, "r sin 2, r e 2 Ÿ"r sin 2, r cos 2, 0 e r Ÿcos 2, sin 2, 1 N
2= 2
dS ;; F
N
dr d2 ; ; "r 4 cos 2 2 sin 2 2 " r 4 sin 2 2 cos 2 2 r 4 dr d2
;; F
0
D
UP, correct
0
D
r 5 2 ; 2= "2 sin 2 2 cos 2 2 1 d2 32 ; 2= 1 " 1 sin 2 Ÿ22 d2
5 0
5 0 0
2
32
48
1
1
3
32
= =
2= "
Ÿ 2= 5 2
5
5
2 2
b.
Ÿ
F
dS and ;;; dS.
dV and solve for ;; F
10 Method II: Parametrize D and V, compute ;; F
D
V
C
Be very careful with the orientation of the surfaces.
D: Polar coordinates:
D Ÿr, 2 Ÿr cos 2, r sin 2, 2 R
Ÿxy 2 , yx 2 , z 3 F
r 3 cos 2 sin 2 2, r 3 sin 2 cos 2 2, 8
e r • e 2 Ÿ0, 0, r e 2 Ÿ"r sin 2, r cos 2, 0 e r Ÿcos 2, sin 2, 0 N
dS ;; F
N
dr d2 ; 2= ; 2 8r dr d2 2=¡4r 2 ¢ 2 32=
;; F
0
D
UP, correct
0
0
D
V: Cylindrical coordinates:
V Ÿr, 2, z Ÿr cos 2, r sin 2, z R
2
2=
2
0
0
r
F
y 2 x 2 3z 2 r 2 3z 2
Jr
F
dV ; ; ; Ÿr 2 3z 2 r dz d2 dr 2= ; ¡r 2 z z 3 ¢ 2 r dr
;;; r
2
0
V
2
r 2 2 8 ¢ " ¡r 3 r 3 ¢ r dr 2= ; Ÿ2r 3 8r " 2r 4 dr 2= 1 r 4 4r 2 " 2 r 5
5
2
0
0
112
64
2= 8 16 "
=
5
5
F
dS " ;; F
dS
dV ;; F
Since C is oriented upward, ;;; 2= ;
So ;;
C
dS F
2
Ÿ¡
V
;;
D
D
2
0
C
F
dS " ;;; dV 32= " 112 = 48 =
F
5
5
V
5