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Unformatted text preview: MATH 215 — Winter 2007 Final Examination — Key Problem Indicate your answer @.®
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@ 10 ——— Problem II]. (t? = {3' + ti points) {a} {6' points} 'Which of the following statements might not be useful for this exam. A. single] + cos2[:rj = 1 and cosl[2;t} = cosils} — singlisc}
1 — cos[2:c}
2 E. Sinﬂm} = 2 Hints) cosh} and ail12m] = [1 c032{g;] = ﬂ D. The formulae for spherical coordinates are :r = pcos{9}sin{,uj y = ssi11{ﬁ'} sin{:,a} z = ems{ts}
E. Less than lll'Ft of US. citizens are vegetarians. {b} (i? points) Which of the following statements might be usaful for this exam.
A. cos{sf3} = 1;“2 and sinhrﬁ] = JEI2
E. cosh?!) = J2I2 and sinfirfs} = xii2,32.
C. Both the wife and the mother of Alexander lGraham Bell were deaf.
D. cos{1rrfﬁ] = JEN and sin[wrfE‘+}=1f2.
E. A one minute kiss burns about 31:] calories. Problem 1. [HI = 5 + 5 points} fa] {5 points} Suppose 11' = {1,—1,” and s = {—1,li, 2}. If ii and t?" are perpendiculan then b can
be A. —2
E. —1 0.1:]
[Jul
E2 {h} [5 points} Suppose 11' = {1, a, 1, 1} and ﬁ' = {—1,b.2}. If the parallelogram spanned by 11' and a?
has area m. then b can be A. —2
E. —1
C. G
D. 1
E. 2 Problem 2. {5 points] Consider the curve C parametrized by r{t} = {ooe3{t}.2ooe2{t}.sin3{t}} for
[I E t 5 nfﬁ. The length of U is A. in}?
B. 3 C. 311'
D. sfe
E. None of the above. Problem 3. [5 points} A thin lamina of density p = If H’s? + 1:2 grams per square meter occupies
the [bounded] region in the plane hounded by the curves H = irfdl, = Tiff11 r = If? and r = 53.
[We assume that "aft! 5 It? 5 Tnfsi in the latter two equations.) The mass of the lamina (in grams]
13 A. 5172 13. Sag C. tiargvﬁ D. 4{ln{T4rf4] — ln{rrf4}]l E. 12s2 Problem 4. (It? points} {a} [5 points} The direction of steepest assent for the function ﬁr, y, z} = r3 + sin(yz} at the point
(1.0.1) is A. {BN’E 12mm
n. (are. a. 1f2}
0 own, —1NT£'I+.G} D. {1,11, o}
E. {ah/TI. Wﬁ. 13m} [h] {5 points] Let C be any Curve from the point (D, 1,11" f2] to the point (35¢, 1}. The integral / 3:32 its: + z eoa[yz} sly + yeoslfys] dz
CT is equal to A. —2ﬁ
E. —13
C. I]
D. 13
E. 26 Problem 5. (It? points} A sketch ef the vector ﬁeld F = {F2 Q) is prrwided helew. hi"llich of the
feilewlng statements are true? A. If D is the. street ennhnf‘reti h}! the cleaned curve 02, then f I 1,}in — ﬂ.) HTA :3: U.
E. fchdr}ﬂ. C. 56;}? F . elr e e. D. diva} at the point A is appmxim'cttely zem. E. F is conservative. 4D 4!!! 43L“  I' 41" 4L5 {"1 4L! 0.2 J.' DID III I 112 '13 [IA IJ.5 IIIJE L‘t? uz. Qu ! [I Problem 5. {115' points} If the Surface 3 is parametersed by r{u,u]l = {a — 11,1: + 11,112 — 112:], then
A. ru{2,1} = {1,1,4}.
B. r,,{2,1} is tangent to the Surface 3 at {2, 1}.
C.r,,{2,1}={1,—1,2}.
D. r,,[2,1) is tangent to the surface 5' at the point (1,3,3).
E. The point [2,4, T) lies on the tangent plane to the surface S at the point {1,3,3}. Problem 7. [1‘8 = 3 + 5 + It? points} Suppose S is the parabolic cap out from the parabaloid
z = 15—32 hp? by the cone 2 = 2px? +y2. That is, S is that part of the parabaloid z = lifu—a‘ii—y2
above the cone 3 = 21ft” + ya. We orient 5' away from the origin. {a} (3 points) Please sketch 3 '1 {b} {5 points) Please ﬁnd a parameterization of 5. Make sure you give the domain for the variables
in your parameterization. iiir151= 4*? ”“91, rsu'n (a): Lgrrl) l—f. Ema}
as [man] [c] {It} points) Suppose that S is a thin lamina of constant density 6f 11' {in hwy11,2]. 1illﬁ'hat ia the
mass of 5' [in kg]? s At itemM91 av”) 
a ’ 1:} 1‘} xﬁ, : (Lur‘cmwu, artam}, . > “or ﬁt 1 4r5mm} J Wannay _ by J 1
1:.) lTr “Earl. 1' T1 [4r1+1\1 3 2.11
L
‘wtagg ; g gelling '= S g FUGFHDTZ olE'Jtr
‘55 :! Ll!
:  —. (32331—1 a] Problem 8. {15 points) Suppose U is the boundary of the triangle with vertieea (E, D, [I], {D1 12. El},
and {l}, [L 3] with counterclockwise orientation when viewed hon: above. Suppose Fm; y, z] = {2.122 + aretan(e=], me + {y + 1]”, h1{1 + 32} + my + 13,52}.
Compute ch . dr. (Hint: Stokea’ theorem may be uaeful.) "ltllﬁl‘jilL 1"? ‘j‘llFln'H'ﬁ I
if lit—'1}; E—aﬁ ‘F{X.E‘)= {XI I'lIx‘lil a“? D 9'2 53; DEFé {57"11: 41.","1'5; t wwe
Fla. '1: (OJ‘Lll‘) Liv? Hlﬁah'm’ '3. Wiiﬂd'ﬂ'hﬁvx eg a?“ '" .1
rh:’<ll‘110‘} runeé 1+ 3 E
L??? elf‘ 1 Egg (ﬁxfjnéﬁ ': SS <351y“3}2'>'<Z,B+)clﬁéi‘ t3 .=~ "" =19? Problem 9. [1?“ = 2‘ + 15 points] Let E he the rectangular solid cut from the ﬁrst octant by the
planeez=2,$=1andy=1. {Thatie,E={{:t1y,z}ﬂ:TIE l,ﬂ£y£1,ﬂ£z£2}.} [a] {2 points] Sketch E. [h] {15 points} Let 3’ be the lzlouncleu'}r of E and let 3 he the surface obtained by removing the top
face {i.e., the rectangle in the plane a = 2} from 3". We euppoSe that 5' has “outward” orientation.
Find the ﬂux acroae S of the 1Hector ﬁeld F{:t, y, z} = {arctanfy} —..'I:2. coe{$) sin[c} —y3, z[2:c+ﬁy2]).
(Hint: the divergence theorem my be useful.) ' ‘5‘; E 43'": 5‘5“ to?" — 55 rigT
5“ A?” T I i 1
. f E. 43*“: S25 AME‘HU' r ES 5 (ax311 haywire) Jaéxda, 1!.
ll
'31.; ...
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