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Unformatted text preview: Name: +—__K~€ Lab Section: MATH 215 — Fall 2005 FINAL EXAM READ THIS This exam contains nine problems, worth a total of 100 points. The ﬁrst six questions are multiple
choice. Your answers for these six questions are to be entered in the table below. Problem zero has
been done as an example. No partial credit will be given for the ﬁrst six problems, so double check
your work. Please do not cheat. The use of books, calculators, cell phones, computers, notes, cheat
sheets, and all similar aids is strictly prohibited. Good luck. SHOW YOUR WORK. —
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“—m Points Score TOTAL 100  Problem 0. (0 points) Which of the following statements will, in Stephen DeBacker’s opinion,
be most useful for this exam.  A. sin2($) +cm2(x) = 1 and 008(23) = 032(3) — sin2(m)
1 — 005(23) B. sin(2m) = 2sin(x) cos(:1:) and sin2(a:) = 2 C. 0052(3) = 292(3). D. The formulae for spherical coordinates are
3 = 9608(9) sinﬁp) y = psin(9) sin(<p) z = 1060609)
@All of the above. Problem 1. (20 = 5 + + 10 points)
(a) (5 points) If 1?: (2, ——1,3) and 6= (3, 2, —4), then {71 if is equal to A. 22
B —22 (SL8 E. none of the above. u 3: {.52 Lav<12 +>
'2: {9“}. [7. +2? (b) (5 pomts) If 11' = (2, —1, 3) and 5= (3, 2, —4), then if x if is equal to A. (2,17,—7)
®(_231717)
C. (—2,—17,7)
D (2,—17,7
E. none of the above.
.4. A LJ k
UKV": 2! 3 '5 <4'b,‘("?"1),4"('35>
5 1. 4! a (2, I4, 47 (c) (5 points) This problem is (potentially) more djﬂ'ioult  you may wish to complete the remainder
of the exam before sinking lots of time into it. Suppose that. P is the plane described by the equation
2 = 4:1: — 3y+ 2. If A is the point (1, 1,2) and B is the point (4, 5, 2), then A. A and B lie on P B. A lies on P but B doesn’t. C. B. lies on P but A doesn’t.
& both A and B do not lie on P, and A and B lie on opposite sides of P.
both A and B do not lie on P, and A and B lie on the same side of P. 'P has ezbumﬁmt‘ixBgE‘fZTO. ILemce,
(“xi—D L, normal do 'P and (0,031} (A130,. Wt ﬁlm.“ comma“ “hm. “OMS‘OCHEM “cam VA‘: A (0.0.2.3 = (4,4,0) Vat": BLO:°:’\ : (‘q3'9,o». mmm RV; m W73 win dawn/Hie Lyn[1,“
NH A Mud! 8 Cu. Via ’hu picks“, and whm or M
H cw. CM Sam [opposite scales.
\— A a a
SIM FLA/A W H‘VG m Lam ”“8th k )
E V) 'huz MMV‘. Problem 2. (10 = 5 + 5 points)
(a) ( 5 points) The function f(:c, y) = y3 + 3:1:2  63y — 153; + 61: has two critical points. These are
A (2 —3) and (—2, 1).
B. (—2,—3) and (~21).
C. (— —4, ~3) and (0,1).
D. (.— 2, 3) and (2, 1).
®2 (2 .3) and (~2,—1). V? : < 6% .b\j+b, 3‘1"“ox~l§> V¥=8 =5 (ox—(03+B=o mt 331Lurs=o =9 “‘3" M44 \f‘zx 5’=o =3 'ngw mA L51 '2‘1“3 TO % KayI Mel €333 “3"“ =0
X: Z .333 (b) (5 points) The function f(:t:,y) = yam  a: + l/y has critical points 0 = (1,1/2) and D =
(—1, —1/2). We have A. . f obtains a. local minimum at C and a local maximum at D. B. j obtains a local maximum at C and a local maximum at D. C. C is a saddle point for f and f obtains a local maximum at D. ® both 0 and D are saddle points for f. _
E. not enough information to determine the nature of the critical points 0 and D. m?— Mal in (mm ‘Pﬁ‘r‘m—gﬁfgx Q 5 31" $3 = 241?" '3
?xy = F,“ = 258
4:»! 3 ‘3. k “MO ‘ua LLSS m 0 &f‘
W C {1), but tmckuch. 414d"
LoHA. wu. Sadlb. 901%  Problem 3. (10 = 5 + 5 points)
(a) (5 points) At the point (OJ/4), the rate of change of the function f(:z:,y) = e” cos(y) is
maximized in the direction
® (11', #4)
x/irrE + 16
B. (1,0) Vi: ‘1 < v55" Camp , x 6’61 (105(5) — 5"? 364013» E. none of the above. 9‘? (Off/q) 1" < E“: g, " v—E 7 V1? (OWN) A
WHOSTLM ' ' (b) (5 points) At the point (1,0), the rate of change of the function f(z:,y) = ’9 cos(y) in the
direction of the vector (3, 4) is A. 7/5 13 (3  8iI“~(1))/5 C. —4 Problem 4. ( 10 = 5 +5 points)
(a) (5 points) We can rewrite the iterated integral 2 fizmwem _1 :3
as
AﬁJﬂﬁmmnw
B. f_1 I”? sin(y) do: dy + L f_‘/— l" sin(y) da: :13;
© f: I”? sin(y) dz £13; + I; f_1ﬁsin(y) dxdy + In If; sin(y) dz: 033;
D. f_1 [”1” sin( y)d:cdy+ fol f:1ﬁsin(y)d:cdy+ f0 f1; 5in(y) dxdy
E. none of the above (b) (5 points) Consider the triple integrals «f2 IIR 4
E = f f / sin(go)p2dpd6dr,a
0 0 3 «f2 4 5
F=/ fjrdzdrdﬂ.
O 3 0
Wehave A. E is the volume of the region in the ﬁrst octent bounded between the spheres of radii 9 and
16 while F is the volume of the region in the ﬁrst oetant below the plane 7. = 5 and bounded
between the cylinders of radii 9 end 16. B. F is the volume of the region in the ﬁrst octant bounded between the spheres of radii 9 and
16 while E is the volume of the region in the ﬁrst octent below the plane z = 5 and bounded
between the cylinders of radii 9 and 16. © E is the volume of the region in the ﬁrst octant bounded between the spheres of radii 3 and
4 while F is the volume of the region in the ﬁrst octant below the plane 2: = 5 and bounded
between the cylinders of radii 3 and 4. D. F is the volume of the region in the ﬁrst octsnt bounded between the spheres of radii 3 and
4 while E is the volume of the region in the ﬁrst octant below the plane 2: —= 5 and bounded
between the cylinders of radii 3 and 4. E. none of the above. and Problem 5. (10 points) Suppose F (?,t2.z2,2:::g,4rz2 + 7rcos(7ry), 23.31922) and C is the curve parame—
terized by r(t) =< 2m” +t/2,t(t2 t/2+ 1) > for g t g 1/2
We have ch  dr is equal to A. 15/16
. —15/16
17/16
D. 17/16
E. none of the above. IA No46 we F V? mm “gay x1121+smmj\ “Cm '53 ‘hM. Mum“ Macon QM “it ‘ﬂkﬁ‘l‘ds
gm: = Hal/m .C mm»
C. = {3 ( U, was}  mom 1‘ “E’ﬂl'v/LY * ﬂaw/2.3 o C. 10 Problem 6. (20 = 0 + 10 + 10 points) Suppose S is the surface of the cylinder described by
S = {(x,y,z) I0 5 z :5 4 and 3:2 +y2 =16} with orientation away from the origin.
(a) (0 points) Please sketch 5'. (b) (10 points) Please ﬁnd a parameterization of 3. Make sure you give the domain, equations, and
check that associated normal vector has the correct orientation. 0695211 —3
V‘(e,v\ 2 <4 Costs), 45m (93, v‘)
aeveq “‘19 = <  45mm , 4005(6)”),0) %{:<OJ03 '3 ‘° 4  <4 ~
”(2, x (‘v  caste); HISM(9\’ 0) COT We? 6“!th 7 Veg 11 (c) (10 points) For the vector ﬁeld F = (xe‘, ye”, e1“), please compute ffSF  :15. g ﬁg? 3 111' ‘4
lécadexfm
f S f 44mm» e",qsrutexe", e “ V> .
o 0 (4%(93ﬂahl93, 0» ads '11;
=f {J (No (051(9) 9" + [(9 stCOXGV) JuJQ
O D __ m 4
= no S) f e'aV =iC821T3LeH— h l 12 Problem 7. (I5 = 5 + 10 points) Suppose S is the orientable surface obtained by taking the union
of the cylinder described by  31={(x.y,z)IOst4andm2+y9=4}
and the (upper) hemisphere of radius two centered at (0, 0, 4)
’93 = {(x,y,z) 32+y2+ (Z — 4)2 = 4 and z 2 4}. The orientation of S = (S; U 5'3) is away from the origin.
(a) (5 points) Please sketch the surface S. (b) (10 points) Please compute f [S (cur1(F)) dS where
F = (yam2 + cos(za:), e"If — my”, 6””).
Note: Your calculations can be continued onto the back of this page. ﬁ‘uwn W5 Moreenx, S‘s]; Mtﬁ)‘é? in “‘12.; Same.
as 85‘ (WEBAs“ mm s" L, in out (yr mm...
SI a" On ﬂu K—«a film. 13 (continuation of problem 7(b)) I
vmamhuig, S" 3 EH (1 a) : (V‘COSCG) '1”an (O5,0> oéeezr
air4.51. = (costs), saute), o 5 H < “(Sixley’mtml 05 a» m A A oﬁcwﬁhoﬁ OK 7? Yes 2. 7W
$(V‘F)'GQSI‘§£ {m1 ‘M:’X2"*jz> <°JOHF> «AvA9 14 Problem 8. (15 points) Let E be the unit cube in the ﬁrst octant ($23., E has vertices at (1, 1, 1),
(0,0,0), (1, 0, 0), (1, 1, 0), (0,1, 0), (0, 1, 1), (0,0, 1), and (1, 0, 1)). Let 8 denote the surface obtained
by considering the surface of E without its top (37.8., the union of the ﬁve remaining sides of E). S
is given the outward orientation. Please compute ff F  dS . s where F = (a: cos(1ry), —ye‘,yz). Note: Your calculations can be continued onto the back of this page. 1.3+ S' dim/wt: 1w swim a; 9 MA on
'1" 41min m “a?“ a" E. w [mu 6‘ 1533451153. Edf’  if Fiat"?
.s' ,5" T “FIRM “ha; 0&le WWW} (continuation of problem 8) D 34 dcreet c telAM?
"f c
but How) “<me 0 av“ .3 1"“ ~ {1.0.05 (Era : {0, D: I> a I YT; = (D. MD) carter\ ova‘ﬂ‘hm I Y“. END OF EXAM Before leaving the examination room:
(1) Make sure you have transferred your answers for problems 0 — 5 to the chart on page one. (2) Make sure you have placed your name and section number on page one.
(3) Make sure you turn in your exam to the proctor at the front of the room. 16 ...
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