Name
ID
MATH 311
Final Exam
Fall 2000
Section 502
Solutions
P. Yasskin
>Ÿ4x " 3y dx Ÿ3x " 2y dy
with vertices
and
Ÿ0, 0 ,
Ÿ0, 3 Ÿ2, 0 .
(HINT: Use Green’s Theorem.)
a. 3
b. 6
c. 9
d. 12
e. 18
correctchoice
1. (5 points) Compute
P
4x " 3y
> Pdx Q dy
3x " 2y
Q
" P
;;
y
x
Q
1-5
/25 7
/11
6
/45 8
/24
counterclockwise around the edge of the triangle
dx dy
;;Ÿ3 " Ÿ"3 dx dy
6 Area
6 1
2
23
18
2. (5 points) Let MŸ2, 2 be the vector space of 2 • 2 matrices. Consider the linear map
L : MŸ2, 2 ¯ MŸ2, 2 given by LŸX X X T
where X T is the transpose of X. Which of the following is FALSE or are they all TRUE?
a. DomainŸL MŸ2, 2 b. CodomainŸL MŸ2, 2 c. KernelŸL antisymmetric 2 • 2 matrices
d. ImageŸL symmetric 2 • 2 matrices
e. All of the above are TRUE
correctchoice
The domain and range are before and after the arrow in L : MŸ2, 2 ¯ MŸ2, 2 .
The kernel is the set of solutions of LŸX X X T 0. But then X T "X and X is antisymmetric.
The image is the set of matrices of the form X X T which is symmetric. But is every symmetric matrix
of this form? Yes because if S is symmetric, then S T S and L 1 S 1 S 1 S T S.
2
2
2
So (a-d) are true.
3. (5 points) If a jet flies around the world from East to West, directly above the equator, in what
direction does the unit binormal
B
point?
(HINT: B T • N )
a. North
b. South
correctchoice
c. East
d. Up (away from the center of the earth)
e. Down (toward the center of the earth)
T points West
N points Down
So B points South by the right hand rule.
1
4. (5 points) Let P 2 be the vector space of polynomials of degree at most 2. Consider the inner product
1
; 3x pŸx qŸx dx
˜p, q ™ Find the angle between the polynomials rŸx 4 2
a. cos "1
correctchoice
5 3
b. cos "1 16
15
5 3
"1
c. cos
4 2
"1 15
d. cos
16
4 3
e. cos "1
5 2
|r|2
1
; 3x rŸx 2 dx
1
0
|s|
2
1
; 3x sŸx dx
2
2
; 3x Ÿ1 " x 2 dx
0
0
;
1
3x 3
cos 2
dx
0
1
˜r, s ™ "
3x 4
4
1
0
Ÿ1
0
" x2 3
2
1
Ÿx
3
x.
1
0
0" "1
2
1
2
3
4
; 3x rŸx sŸx dx ; 3x Ÿ1 " x 2 Ÿx dx
0
0
2
˜r, s ™
5
2 8 4 2
5 3
|r||s|
1 3
5 3
2 4
5. (5 points) Compute ;; F
dS for F
1 " x 2 and sŸx 1
; Ÿ3x 2 " 3x 4 dx
0
5
x 3 " 3x
5
1
0
1" 3
5
2
5
, y 3 , x 2 z y 2 z over the complete surface of the cylinder
C
x 2 y 2 t 4 and 0 t z t 3.
(HINT: Use Gauss’ Theorem.)
a. 24=
b. 48=
correctchoice
c. 96=
d. 144=
e. 324=
dS
;; F
C
;;; FdV
;;;Ÿ3x 2 3y 2 x 2 y 2 dx dy dz
C
C
3
2=
2
0
0
0
; ; ;
4r 2
r dr d2 dz
3 2= ¡r 4 ¢ 20
96=
2
6. (45 points) Consider the vector space V spanned by
e1
cosh 2 x,
e2
sinh 2 x and e 3
cosh x sinh x.
USEFUL FACTS:
sinh 0
0
cosh 2 x " sinh 2 x
cosh 0 1
d sinh x cosh x
dx
d cosh x sinh x
dx
d sinh 2 x 2 cosh x sinh x
dx
d cosh 2 x 2 cosh x sinh x
dx
d cosh x sinh x cosh 2 x sinh 2 x
dx
1
sinhŸ2x 2 sinh x cosh x
coshŸ2x cosh 2 x sinh 2 x
a. (5 pts Extra Credit) Show e 1 , e 2 and e 3 are linearly independent.
Assume
ae 1 be 2 ce 3 0.
Then
a cosh 2 x b sinh 2 x c cosh x sinh x 0 for all x.
Plug x 0 into (1)
aŸ1 bŸ0 cŸ0 0
®
a0
2
2
With a 0 differentiate (1)
b2 cosh x sinh x c cosh x sinh x 0 for all x.
Plug x 0 into (2)
bŸ0 cŸ1 0
®
c0
2
2
With c 0 differentiate (2)
b2 cosh x sinh x 0 for all x.
Plug x 0 into (3)
b2Ÿ1 0
®
b0
Therefore a b c 0 and so e 1 , e 2 and e 3 are linearly independent.
(1)
(2)
(3)
Find the change
f 1 1, f 2 coshŸ2x and f 3 sinhŸ2x .
of basis matrices from the f-basis to the e-basis, C and from the e-basis to the f-basis, C .
b. (10 pts) Another basisfor V is
f se
esf
Since
cosh 2 x " sinh 2 x
e1 " e2
f1
1
f2
coshŸ2x cosh 2 x sinh 2 x
f3
sinhŸ2x 2 sinh x cosh x
e1
e2
2e 3
1 1 0
the change of basis matrix is C
esf
"1 1 0
.
0 0 2
Invert to find C
f se
1 1 0 1 0 0
"1 1 0 0 1 0
®
®
0 1 0
0 0 1
0 1 0
®
0 2 0 1 1 0
0 0 2 0 0 1
0 0 2 0 0 1
1 0 0
1 1 0
1 1 0 1 0 0
1
2
1
2
0
"1
2
1
2
0
0
0
1
2
®
C
f se
0 0 1
1
2
1
2
0
"1
2
1
2
0
1
1
2
0
0
1
2
0
0
0
1
2
0
0
.
1
2
3
c. (5 pts) Consider the linear operator of differentiation
D:V
¯
V given by DŸg dg
dx
Since
D Ÿf 1 D Ÿ1 D Ÿf 2 DŸcoshŸ2x 2 sinhŸ2x 2e 3
D Ÿf 3 DŸsinhŸ2x 2 coshŸ2x 2e 2
0
0 0 0
the matrix of D relative to the f-basis is B
f sf
.
0 0 2
0 2 0
Find the matrix of D relative to the e-basis. Call it A .
ese
A
ese
C
esf
B
f sf
C
f se
1 1 0
0 0 0
"1 1 0
0 0 2
0 0 2
0 2 0
1 1 0
0 0 0
"1 1 0
0 0 1
0 0 2
1 1 0
1
2
1
2
0
"1
2
1
2
0
0
0
1
2
0 0 1
0 0 1
2 2 0
d. (5 pts) Find the matrix of D relative to the e-basis by a second method.
Since
D Ÿe 1 D cosh 2 x
2 cosh x sinh x
2e 3
D Ÿe 2 D sinh 2 x
2 cosh x sinh x
2e 3
D Ÿe 3 DŸcosh x sinh x cosh 2 x sinh 2 x
e1
e2
0 0 1
the matrix of D relative to the e-basis is A
ese
0 0 1
.
2 2 0
4
0 0 0
e. (16 pts) Find the eigenvalues and eigenvectors of the matrix B
.
0 0 2
0 2 0
"5
detŸB " 51 det
0
0
0 "5
2
" 5 Ÿ5 2 " 4 "5Ÿ5 " 2 Ÿ5 2 ŸB
0:
" 51 v 1
0
0
v1 0 0 2
0
0 2 0
0 0 0 0
®
x
®
0 0 1 0
0 2 0 0
2:
ŸB
" 51 v 2
0
0 0
0 "2
2 0
0
0
0
0 "2
2
"2:
ŸB
0
®
®
0 2 2 0
0
y
0 0
z
2 0 0
0
v3 0
0
®
t
v2 1
t
1
0
0
x
®
0 1 1 0
0
0
x
0 2 2
v1 0
1 0 0 0
0 2 2 0
®
0
0
v2 0 2 2
2 0 0 0
1
0
0 0
0 0
" 51 v 3
2 "2
0 1 "1 0
2 "2 0
0
For 5 3
1 0
®
t
z
0
"2
y
0 0 0 0
"2
0
0 1 0 0
0 0 2 0
For 5 2
0
0, 2, "2
0 0 0
For 5 1
2 "5
0
Eigenvalues: 5
y
0 0 0 0
0
z
0
"t
®
t
v3 "1
1
f. (4 pts) Find the eigenvalues and eigenfunctions of the operator D.
1
For 5 1
0:
v1 0
g1
1f 1
1
g2
1f 2
1f 3
g3
0
0
For 5 2
2:
v2 1
cosh 2x sinh 2x
e 2x
1
0
For 5 3
"2:
v3 "1
"1f 2
1f 3
" cosh 2x sinh 2x
e "2x
1
5
7.
(11 points) Compute ;; 1x e
xy
10
dx dy over the diamond shaped
region between the curves
1
and
x
You must use the curvilinear coordinates
and
x uv
y uv.
y
x,
y
9x,
y
y
4.
x
5
0
1
2
3
a. (3 pts) Find the Jacobian:
J
det
"v 1
u2 u
v
u
"2v
u
2v
u
b. (4 pts) Express each boundary curve in terms of u and v:
y
x
®
uv
y
®
uv
y
®
uv
y
9x
1
x
4
x
®
uv
v
u
9 uv
u
v
4 uv
®
u
1
®
u
3
®
v
1
®
v
2
c. (2 pts) Express the integrand in terms of u and v:
xy
v2
xy
v,
1
x
u,
v
1e
x
xy
u ev
v
d. (2 pts) Compute the integral:
;;
1e
x
xy
dx dy
2 3
; ;
1 1
u e v 2v du dv
v
u
2;
2
1
;
3
1
e v du dv
2 u
2
1
ev
3
1
2Ÿe 3 " e 6
8.
(24 points) Use two methods to compute
• F dS
;; S
for
F
Ÿy, "x, z 2
x2
sphere
y2
over the piece of the
z2
0tzt4
for
25
with normal pointing away from the z-axis.
• F and compute the double integral ;; • F dS
a. (12 pts) Parametrize the surface, compute S
directly.
R
ŸI, 2 Ÿ5 sin I cos 2, 5 sin I sin 2, 5 cos I R I Ÿ5 cos I cos 2, 5 cos I sin 2, "5 sin I R
2 Ÿ"5 sin I sin 2, 5 sin I cos 2, 0 N 25 sin 2 I cos 2, 25 sin 2 I sin 2, 25 sin I cos I which points away from the z-axis.
§
§
i
•F
§
j
k
Ÿ0, 0, "2 x y z
"x z 2
y
On the upper circle, z
• F dS
;; 4, x 2
• F N dI d2
;; S
S
9, tan I
"50 ;
2=
0
;
=/2
sin 2 I
2
"50¡2= ¢
y2
tan "1 Ÿ3/4 • F dS
b. (12 pts) By Stokes’ Theorem ;; =/2
3 , I tan "1 3 . On the lower circle, I
4
4
sin I cos I dI d2
=
2
tan "1 Ÿ3/4 "50= 1 "
3
5
2
"=¡50 " 18 ¢
"32=
where S is the boundary of S.
ds
>F
S
S
for each circle.
Parametrize the upper and lower circles and compute > F
ds
Be sure to discuss the orientation of the circles when you add up the integrals.
On the upper circle, z 4, x 2 y 2
2
Ÿy, "x, z Ÿ3 sin t, "3 cos t, 16 ,
F
;
ds
> F
2=
0
Ÿ"9 dt 9, rŸt Ÿ3 cos t, 3 sin t, 4 , v
2
F
v "9 sin t " 9 cos 2 t "9
Ÿ"3 sin t, 3 cos t, 0 "18=
upper
On the lower circle, z 0, x 2 y 2
2
Ÿy, "x, z Ÿ5 sin t, "5 cos t, 0 ,
F
;
ds
> F
lower
2=
0
Ÿ"25 dt 25, rŸt Ÿ5 cos t, 5 sin t, 0 , v Ÿ"5 sin t, 5 cos t, 0 F v "25 sin 2 t " 16 cos 2 t "25
"50=
Both the upper and lower integrals were computed counterclockwise. We need the upper
integral clockwise. So it gets a minus sign.
> F
" > F
"50= 18=
ds
ds
ds
>F
S
lower
"32=
upper
7