MATH 253 Honors
FINAL EXAM
Fall 1999
Sections 201-202
Solutions
P. Yasskin
Multiple Choice: (7 points each)
1. Consider the line through the point
plane
x 2y 3z
a. Ÿ3, 2, 1 b. Ÿ1, 2, 3 P Ÿ4, 4, 4 Its tangent vector is
7.
which is perpendicular to the
correctchoice
c. Ÿ7, 6, 5 d. Ÿ5, 6, 7 e. Ÿ4, 4, 4 ŸA, B, C . In this
If the equation of a plane is Ax By Cz D then the normal is N
case, N Ÿ1, 2, 3 . Since the line is perpendicular to the plane, then its tangent vector
is the normal to the plane. So, v Ÿ1, 2, 3 .
x " yz 0
at the point
Which of the following points does not lie on this plane?
2. Find the plane tangent to the hyperbolic paraboloid
P
Ÿ6, 3, 2 .
a. Ÿ"6, 0, 0 b. Ÿ0, 3, 0 c. Ÿ0, 0, 2 d. Ÿ1, "1, "1 correctchoice
e. Ÿ"1, 1, 1 The hyperbolic paraboloid is a level surface of the function g
g
g Ÿ1, "z, "y . So the normal to the surface at P is N
tangent plane is N X N P, or x " 2y " 3z
point, we find Ÿ1, "1, "1 is not a solution.
x " yz. Its gradient is
Ÿ1, "2, "3 . So the
Ÿ6,3,2 6"23"32
"6. Plugging in each
3. Duke Skywater is flying the Millenium Eagle through a polaron field. His galactic
measured in lightseconds and his velocity is
measured in lightseconds per second. He measures the strength of
p Ÿ3, 2, 2 the polaron field is
p 274 milliwookies
and its gradient is
milliwookies per lightsecond. Assuming a linear approximation for the polaron field and
that his velocity is constant, how many seconds will Duke need to wait until the polaron
286 milliwookies?
field has grown to
coordinates are
Ÿ2300, 4200, 1600 v Ÿ.2, .3, .4 a. 2
b. 3
c. 4
d. 6
correctchoice
e. 12
1
The derivative along Duke’s path is
lightseconds
Ÿ3, 2, 2 milliwookies
second
lightsecond
milliwookies
.6 .6 .8 2
second
So the polaron field increases 2 milliwookies each second. To increase 12 milliwookies,
it will take 6 seconds.
dp
dt
v
p Ÿ.2, .3, .4 R
Ÿu, v 4. Consider the surface S parametrized by
and 0 t v t 4.
dS
;; F
Compute
Ÿu F
where
v, u " v, uv for
0tut2
Ÿy, x, y .
S
a. "32
b. "16
c. 16
correctchoice
d. 32
e. 64
Ru • Rv
R
N
R
u Ÿ1, 1, v v Ÿ1, "1, u Ÿy, x, y Ÿu " v, u v, u " v F
F N Ÿu " v Ÿu v Ÿu v Ÿv " u " 2Ÿu " v dS
;; F
N du dv
;; F
S
4
0
4
; "4 4v dv
0
0
"4v 2v 2
4
v0
tangent to this surface at the point
points does not lie on this plane?
b.
c.
d.
e.
P
v, v " u, "2 "2u 2v
4
2
0
u0
; "u 2 2vu
"16 32
5. Consider the surface S parametrized by
a. Ÿ3, 0, 0 2
; ; "2u 2v du dv
S
Ÿu dv
16
Find the plane
R
Ÿu, v Ÿu v, u " v, uv .
1, 2
Which of the following
RŸ Ÿ3, "1, 2 .
correctchoice
Ÿ0, 4, 0 Ÿ0, 0, "2 Ÿ1, 1, 0 Ÿ0, 6, 1 R u • R v Ÿu v, v " u, "2 R
N
R
u Ÿ1, 1, v v Ÿ1, "1, u The normal at P is N
P N Ÿ1,2 Ÿ3, 1, "2 and the tangent plane is N X N P, or
3x y " 2z 3Ÿ3 Ÿ"1 " 2Ÿ2 4. Plugging in each point, we find Ÿ3, 0, 0 is not a
solution.
2
>Ÿ"x 2 y 2 dx 2xy 3 dy 6. Compute
0tyt
area
4 " x2
over the complete boundary of the semicircular
traversed counterclockwise.
a. 0
b. 16
c. 4
5
d. 80
5
e. 128
correctchoice
5
By Green’s Theorem:
>Ÿ"x 2 y 2 dx 2xy 3 dy ;; Ÿ2xy 3 " Ÿ"x 2 y 2 dx dy
y
x
;; 2Ÿy 2 x 2 y dx dy
; ; 2r 2 r sin 2 r dr d2
=
2
0
0
;;Ÿ2y 3 2x 2 y dx dy
=
2 " cos 2
3 2
5
y3z2
;; x 3z dy dz 3 dz dx z5 dx dy
7. Compute
2
r5
5
0
0
128
5
over the complete surface of the
S
x2
sphere
a. 512=
c.
d.
e.
z2
with outward normal.
4
correctchoice
21
32= 2
5
128=
5
16=
3
256=
15
b.
y2
Apply Gauss’ Theorem in spherical coordinates:
3 2
x3z2 , y z , z5
F
F x2z2 y2z2 z4
5
3
3
I
F dV
;;; 2=
=
2
0
0
0
Ÿx
; ; ; > 4 cos 2 2 > 2 sin 2 d> dI d2
2
y2
z 2 z 2
3
2= " cos 2
3
8. (15 points) Find the point in the first octant on the surface
z
=
0
32
x4y2
1
zx 4
> 2 > 2 cos 2 2
>7
7
2
0
512=
21
which is
closest to the origin.
Minimize f
x2
f
Ÿ2x, 2y, 2z 2x
54zx 3 y 2
x2
2y 2
g
y
zx 4 y 2
2
y2
g
2y
y2
x
z 2 on the surface g
Ÿ4zx
1 y
2
2
z
2z
1
4 2
y
f
y , 2zx 4 y, x 4 y 2 x
2y
zx 4 y 2
3 2
52zx 4 y
2z 2
5x 4 y 2
g
5
5
1
2zx 2 y 2
2z
x4y2
1 y
2
2y
z
2 3/2 y 7
32
Ÿx, y, z 32.
25
y7
2 7/2
2, 2 , 1
3
15
;; x dA
(10 points) Compute
9.
R
10
over the region R in the first quadrant
y
bounded by the curves
y
x2,
y
x4
and
y
5
16.
0
The left edge is y
;; x dA
R
16
y 1/2
1
y 1/4
; ;
x dx dy
256 " 64
4
3
y"1
4
y 1/4 . The right edge is y x 2 or x
y 1/2
16
16
2
y 1/2
y
x
"
dy ;
dy
;
2 xy 1/4
2
2
1
1
16
y 3/2
y2
"
4
3
x 4 or x
2
x
1 " 1
4
3
"
255 " 21
4
y 1/2 .
171
4
(15 points) Find the mass and center of mass
10.
of the solid below the paraboloid
z
4 " x2 " y2
-
above the xy-plane, if the density is
x2
y2.
(11 points for setting up the integrals and the
t r
final formula.)
In cylindrical coordinates, the paraboloid is z
Jacobian is r.
M
;;; - dV
2
2= ; r 3 z
0
2=
0
0
4"r 2
z0
6
2= r 4 " r
6
z-mom
2
; ; ;
;;; z- dV
4"r 2
0
4 " r 2 , the density is -
x
y
2
dr
2= ; r 3 Ÿ4 " r 2 dr
0
2
0
2= 16 " 32
3
2=
2
4"r 2
0
0
0
; ; ;
4"r 2
0
r 2 and the
r 2 r dz dr d2
32=
3
zr 3 dz dr d2
2
2
2
2
2= ; r 3 z
dr = ; r 3 Ÿ4 " r 2 dr
2 z0
0
0
Let u r 2 . Then du 2r dr and r dr 1 du. So
2
4
4
2
=
=
z-mom ; uŸ4 " u du ; uŸ16 " 8u u 2 du
2 0
2 0
= 128 " 512 64 32=
3
2
3
z z-mom 32= 3 1
M
3 32=
= 8u 2 " 8 u 3
3
2
u4
4
4
0
by symmetry.
4
(15 points) Find the area and centroid
11.
1
of the right leaf of the rose
r
2 cos 2 2.
-2
0
-1
1
2
(12 points for setting up the integrals
and the final formula.)
-1
A
;; 1 dA
x-mom
x
y
;
=/2
"=/2
=/2
;
"=/2
;
2 cos 2 2
0
2 cos 2 2
r2
2
r0
r dr d2
d2
;
=/2
"=/2
2 cos 4 2 d2
;
=/2
2 1 cos 22
2
"=/2
2
d2
1 ; =/2 Ÿ1 2 cos 22 cos 2 22 d2 1 ; =/2 1 2 cos 22 1 cos 42 d2
2 "=/2
2
2 "=/2
=/2
3 2 cos 22 1 cos 42 d2 1 3 2 sin 22 1 sin 42 =/2
1 ;
2
8
2 2
2 "=/2 2
"=/2
3 = " " = 3=
4 2
2
4
;; x dA
=/2
;
=/2
"=/2
2 cos 2 2
;
2 cos 2 2
0
r 2 cos 2 dr d2
8 ; =/2 cos 6 2 cos 2 d2
3 "=/2
"=/2
r0
=/2
1
3
8 ;
1 " sin 2 2 cos 2 d2 8 ; Ÿ1 " u 2 3 du
3 "=/2
3 "1
1
5
7
8 u " u 3 3u " u
16 1 " 1 3 " 1
5
7
5
7 "1
3
3
;
x-mom
A
0
r3
3
cos 2 d2
256 4
105 3=
8 ; 1 Ÿ1 " 3u 2 3u 4 " u 6 du
3 "1
16 16 256
3 35
105
1024
315=
by symmetry.
5