MATH 253 Honors
FINAL EXAM
Sections 201-203
Solutions
Spring 1999
P. Yasskin
1. Find the volume of the parallelepiped with
Ÿ3, 2, 0 ,
edges
Ÿ"1, 1, 2 and
Ÿ0, 4, 1 .
a. "23
b. "19
c. 19
correctchoice
d. 21
e. 23
3
2 0
"1 1 2
0
4 1
3
1 2
4 1
"2
Volume must be positive:
2. Find the unit tangent vector
"1 2
0
V
1
3Ÿ1 " 8 " 2Ÿ"1 "19
19
T
to the curve
rŸt Ÿ3t, 2t 2 , 4t 3 at the point
rŸ1 Ÿ3, 2, 4 .
a.
b.
c.
d.
e.
3 , 4 , 12
correctchoice
13 13 13
3 , 4 , 12
29
29
29
3 , 4 , 12
169 169 169
3 , 4 , 12
29 29 29
3 , "4 , 12
169 169 169
v Ÿ3, 4t, 12t 2 vŸ1 Ÿ3, 4, 12 |v|
9 16 144
13
T
v
|v|
3 , 4 , 12
13 13 13
1
3. If a jet flies around the world from East to West, directly above the equator, in what
B
direction does the unit binormal
point?
a. North
b. South
correctchoice
c. East
d. West
e. Down (toward the center of the earth)
T points West,
4. At the point
plane
2
3
4
5
e. 6
points Down
N
and by the right hand rule
where the line
r Ÿt we have
5,
xyz
Ÿx, y, z 2x " y z
Ÿ2 t, 3 " t, t B points South.
intersects the
a.
b.
c.
d.
correctchoice
Plug the line into the plane and solve for t:
2Ÿ2 t " Ÿ3 " t Ÿt 5
4t 1 5
t1
Plug back into the line:
So
xyz
Ÿx, y, z Ÿ2 t, 3 " t, t Ÿ3, 2, 1 6
T 4P
>
the pressure and > is the density. At a certain point
Q
5. The temperature in an ideal gas is given by
where 4 is a constant, P is
Ÿ1, 2, 3 ,
we have
PŸQ 4
PŸQ Ÿ"3, 2, 1 >ŸQ 2
>ŸQ Ÿ3, "1, 2 So at the point Q, the temperature is
T ŸQ 24
and its gradient is
T ŸQ a. 4Ÿ"4.5, 0, 2.5 b. 4Ÿ1.5, 0, 2.5 c. 4Ÿ1.5, 2, "4.5 d. 4Ÿ"4.5, 2, "1.5 correctchoice
e. 4Ÿ"1.5, 2, 2.5 By chain rule:
(Think about each component separately.)
4
T
T
> >
P " 4P2 > 4 Ÿ"3, 2, 1 " 442 Ÿ3, "1, 2 T P 2
P
>
2
>
3 , 1, 1 4Ÿ"3, 1, "2 4 " 9 , 2, " 3
4 "
2
2
2
2
2
6. The saddle surface
may be parametrized as
z xy
the plane tangent to the surface at the point
Ÿ1, 2, 2 .
a. 3x y " z
b. 2x y " z
3
2
R
Ÿu, v Ÿu, v, uv .
Find
correctchoice
c. 3x 2y " z 5
d. 2x " y z 2
e. 3x " y z 3
METHOD I:
f
Plane tangent to a graph:
xy
fŸ1, 2 z
fx
2
fŸ1, 2 f x Ÿ1, 2 Ÿx
y
f x Ÿ1, 2 z
fy
2
fŸx, y x
f y Ÿ1, 2 " 1 f y Ÿ1, 2 Ÿy " 2 xy
1
2 2Ÿx " 1 1Ÿy " 2 METHOD II:
Plane tangent to a parametric surface:
R
Ÿu, v R u • R v Ÿ"v, "u, 1 R
N
R
u Ÿ1, 0, v v Ÿ0, 1, u Ÿ"2, "1, 1 .
At
P Ÿ1, 2, 2 u 1, v 2
So
N
" 2x " y z "2 " 2 2 "2
NX NP
7. Find the minimum value of the function
plane
x 2y 3z
f
x2
y2
z2
2x y " 2
Ÿu, v, uv on the
14.
a. 0
b. 7
4
c. 7
2
d. 14
e. 28
correctchoice
METHOD I:
Lagrange Multipliers
f
g Ÿ1, 2, 3 f Ÿ2x, 2y, 2z 5 2y 25 2z 35
x 5 y5
z 35
2
2
5 25 3 35
Use the constraint:
2
2
x1
y2
z3
fŸ1, 2, 3 2x
g
5
14
75
149
14
5
2
14
METHOD II:
Eliminate a Variable:
x 14 " 2y " 3z
f Ÿ14 " 2y " 3z 2 y 2 z 2
f f , f Ÿ"4Ÿ14 " 2y " 3z 2y, "6Ÿ14 " 2y " 3z 2z y z
10y 12z 56
12y 20z 84
®
y 2, z 3
So
x1
and
fŸ1, 2, 3 1 4 9 14
Ÿ0, 0 3
3
9
0
y
; ; 2 y cosŸx 2 dx dy
8. Compute
a. 1 sin 81
correctchoice
4
b. 1 cos 9 " 1
2
2
c. 9 sin 81 cos 9 " 1
2
d. " 9 sin 81 9 sin y 4
2
2
e. 9 sin 81 " cos 9 1
2
3
2
y
1
0
2
4
6
8
x
Reverse the order of integration:
3
9
; ; 2 y cosŸx 2 dx dy
0
y
; ;
0
9
; 2x cosŸx 2 dx
0
;;; z 2 dV
9. Compute
x
9
0
y cosŸx 2 dy dx
1 sinŸx 2 4
9
x0
;
y2
cosŸx 2 2
9
0
x
y0
dx
1 sin 81
4
x2
over the solid sphere
y2
z 2 t 4.
a. 64=
b.
c.
d.
e.
5
256=
3
48=
5
64=
15
128=
15
correctchoice
In spherical coordinates:
;;; z 2 dV
2=
=
2
0
0
0
x
> sin C cos 2
y
> sin C sin 2
z
> cos C
; ; ; > 2 cos 2 C > 2 sin C d> dC d2
2= " " 1 " " 1
3
3
32
5
128=
15
J
> 2 sin C
2= "
cos 3 C
3
=
C0
>5
5
2
> 0
4
3
for
over the surface of the cube
Ÿx, y , z dS
F
;; F
with outward normal.
0 t y t 1,
0tzt1
10. Compute
0 t x t 1,
a. 1
b. 2
c. 3
correctchoice
d. 4
e. 6
F 1 3y 2 1 2 3y 2
By Gauss’ Theorem:
dS
;; F
F dV
;;; ¡2y y 3 ¢ 10
1
0
1
1
; ; ; 2 3y 2 dx dy dz
0
0
1
1
1
0
0
0
; 1 dx ; 1 dz ; 2 3y 2 dy
3
11. (15 points) Find the area of the diamond shaped region between the curves
1 ex,
y
4
You must use the curvilinear coordinates
y
ex,
y
4
3
y2
1
-1
0
1
x
2
3
-1
J
" 1
2u
v 1/2
2u 1/2
A
;; 1 dA
1
2v
u 1/2
2v 1/2
4
; ;
1
u
1/4
2u
ye "x
x
y
x
u
y
u
x
v
y
v
" 1
2u
v 1/2
2u 1/2
1/2
1/4
y
and
4e "x .
v
ye x .
1tvt4
1 ln v 1 lnŸv " 1 lnŸu u
2
2
2
1/2 1/2
uv u v
"
1
2v
u 1/2
2v 1/2
1
1
"
4u 1/2 v 1/2
4v 1/2 u 1/2
1
2u 1/2 v 1/2
1 u "1/2 v "1/2 du dv 1 ; 4 v "1/2 dv ; 1 u "1/2 du
2 1
1/4 2
1/4
1
1
2¡2 " 1 ¢ 1 "
21
1
2
2
; ;
1
1
4
and
1 tut1
4
x
v ye e 2x
u
ye "x
uv y 2
4
J du dv
1
e "x
1/2
1/2
" 1 u 1/2 " 1 v 1/2
2u 2v
2v 2u
1
1 2v 1/2
2
1
5
12. (10 points) Find the mass of a wire in the shape of the curve
if the density is
> siny x .
0txt =
e
4
Note: The wire may be parametrized as
v
>
M
1, " sin t Ÿ1, " tan t |v|
cos t
sin x sin t sin t tan t
cos t
ey
e lnŸcos t ; > ds
; >Ÿt |v| dt
;
=/4
y
lnŸcos x for
rŸt Ÿt, lnŸcos t .
1 tan 2 t
tan t sec t dt
=/4
sec t
2 "1
0
0ttt =
4
for
sec t
0
around the boundary of the triangle with
> x dx z dy " y dz
and
traversed in this order of the
Ÿ0, 0, 0 ,
Ÿ0, 1, 0 Ÿ0, 0, 1 ,
Hint: The yz-plane may be parametrized as
R
Ÿu, v Ÿ0, u, v .
13. (10 points) Compute
vertices
vertices.
Let
F
Ÿx, z, "y .
By Stokes’ Theorem, the integral is
;; • F dS.
ds
>F
F
î
•F
k
x y z
x
z
Ÿ"2, 0, 0 "y
Using the parametrization of the triangle
î 1, 0, 0
•k
F
R
N
R
Ÿ
u Ÿ0, 1, 0 F
v Ÿ0, 0, 1 k
Since the triangle is traversed counterclockwise as seen from the positive x-axis, the
normal is in the correct direction. So
;; • F N du dv
ds
>F
1
"2 ; Ÿ1 " u du
0
1
; ;
0
1"u
0
Ÿ"2 2
"2 u " u
2
dv du
1
0
"2 ;
"2 1 " 1
2
1
1"u
v
0
v0
du
"1
6
14.
• F dS
;; (15 points) Compute
S
for
F
Ÿx
2
x2
sphere
y, y, z
y2
2
z2
over the piece of the
0tzt4
for
25
with normal pointing away from the z-axis.
Hint: Parametrize the upper and lower edges.
By Stokes’ Theorem
• F dS
;; > F
> F
ds
ds
ds
>F
S
S
lower
upper
By the right hand rule, since the normal points outward, the upper circle must be
traversed clockwise while the lower circle must be traversed counterclockwise as
seen from the positive z-axis. We compute each line integral:
Upper Circle:
x 2 y 2 25 " z 2 25 " 16 9
z4
rŸt Ÿ3 cos t, 3 sin t, 4 v Ÿ"3 sin t, 3 cos t, 0 This is clockwise, so we reverse the velocity:
v Ÿ3 sin t, "3 cos t, 0 2
2
2
F Ÿx y, y, z Ÿ27 cos t sin t, 3 sin t, 16 > F
ds
v dt
> F
upper
;
;
2=
0
81 cos 2 t sin 2 t " 9 sin t cos t
upper
2=
81 sin 2 Ÿ2t " 9 sinŸ2t 2
4
0
dt
dt
81 1 Ÿ2= 9 cosŸ2t 2
4 2
2
81=
4
Lower Circle:
z0
x 2 y 2 25 " z 2 25
rŸt Ÿ5 cos t, 5 sin t, 0 v Ÿ"5 sin t, 5 cos t, 0 This is counterclockwise, so we do not need to reverse the velocity.
2
2
2
Ÿx y, y, z Ÿ125 cos t sin t, 5 sin t, 0 F
> F
ds
v dt
> F
lower
;
lower
2=
;
0
0
"625 cos 2 t sin 2 t 25 sin t cos t
" 625 sin 2 Ÿ2t 25 sinŸ2t 2
4
Total Boundary:
81= " 625=
• F dS ;; 4
4
S
2=
" 544=
4
dt
dt
cosŸ2t " 625 1 Ÿ2= " 25
2
4 2
2
" 625=
4
"136=
7