exam2w06sol - Name: Lab Section: MATH 215 — Winter 2006...

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Unformatted text preview: Name: Lab Section: MATH 215 — Winter 2006 Second Examination READ THIS This exam contains eight problems, worth a total of 100 points. The first four questions are multiple choice. Your answers for these four 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 four 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. b , 0—3 TOTAL —— ————_ Points Problem 1. (20 = 10 + 10 points) (a) (10 points) For the function f (x, y) = cos(e’2y), the gradient of f at the point (0, 37r/ 2) is equal to A. 1 ‘ ‘ ‘ K1 2 i ,1 z\ _1) (KWD : <—€Iv\ (e CK‘Z7‘)'SM(€ 736/ t ,S\;A&€XL‘/\ex1 7.) \> go Vi (OWL) = - mhfime" (313701) ‘5 (b) (10 points) Suppose g(x, y) = x3 — ya: and 17 = (3, 4). The derivative of g in the direction of 17 at the point (—1, —1) is equal to A. 13 5 a mo. wank V03 (“.43 ‘ X— C. 13 \\7‘\ D. 16 E. 9/5 Va (n53: <3X1—‘3, "x> _ A : (LLB Va) (4."): <3—(d3‘ (X) ) _———-——‘ 1' ’- {L—‘W 7 <“ 30 (73("1'W‘e—fl-fl : 7;: (LN) (35‘? l “fir Problem 2. (10 = points) The iterated integra /_Z / inwzwbdydfi is equal to A. 3 s/3-_y / / sin(y2x) dandy 0 -\/!7 B. 3 W7 / / sin(x2y)dmdy 0 -\/3:? C. 4 «a 8 «st—y / / sin(:c2y) d1: dy + / / sin(a:2y) dx dy 0 «A? 4 —\/s——y 4 «a s ¢8——y / / sin (y2x) da: dy + / / sin(y2x) dz: dy 0 —\/g 4 -¢8:§ E. None of the above. Problem 3. (10 = 5 + 5 points) Let F(:1:,y) = —sin(ye‘2)e22 (2mg, 1). (a) ( 5 points) The vector field F is @ conservative because it is the gradient of the function cos(ye‘2). . conservative because it is the gradient of the function cos(ey’2). C. conservative because it is the gradient of the function — cos(ye’2). D. conservative because it is the gradient of the function sin(ye’2). E. not conservative. gum (TED) (b) (5 points) Recall that F(a:, y) = — sin(ye’2)e’2(2$y, 1). If C’ is the curve beginning at Maud ending at (0,0) which travels along the upper-half of the circle of radius 7r / 2 centered at (1r / 2, 0) then the value of / F - dr C I 1 3km. E}: VF bakes]: 4‘?- (050361), F 7 Problem 4 (20 - = 5 + 5 + 8 + 2 ' I I pomts Th - (a) (5 Pomts) Fmd the critical point(S) of) f (xiighout thls pmblem’ “a”, y) = 2/3 + 32:2 — 3.2;2 +11 V‘?(X.uD-: < (0x, 3%1_.(aj> B‘L'bj’,‘ \jto)z (b) (5 points) Classify the critial point(s) of f (a: y) VOA, 9&9: Cdcm 5” 3; $3 ’ Q31 " 3 ‘1 $2,“ = (a gnu) = \b 5 $53 : gcjv‘ :0 (0,0) ~Dmp\= ——3‘O 40 g) sedan Quwdl' 0 . , 3L, 70 SW 1‘0 7° ( .Lx mot» :5 (05m v, -——-- + 4 \2 a (6&th MM .\ (c) (8 points) Use the method of Lagrange multipliers to find the extrema for f (as, y) restricted to the circle of radius one centered at the origin. E New): 5W9" Km?“ L“) —%—/ :) (ox ,, )2): (a 3V1,(0\jc (c) Vi: =X72) (d) (2 points) What are the extrema of f (x, y) on the disc of radius one centered at the origin? Sweat 43% (0)03 14 a flack“: poi}: ) r H (aghth «d (\on «MAJ AQYMM (c): m MM m 4 Located)“; Lo,—b\ (“10) \ MOQ m WC“ (4 Problem 5. (20 = 10 + 10 points) (a) (10 points) Let Cl be the arc of the parabola y = 4 — 2132 from (—1, 2) to (2, —4). Compute / my dx. 01 0 an: @5440) for {G ["113 ° we»: i Z ‘L 850de = f 4.— w—uz) 1 cu = [GM 4834* c. fl " 7. (lei zt" (41"224) (i_’z_) ’ l ’— ‘ ’ =-' L ‘,_I 2. .4 (b) (10 points) Let 02 be the cosine curve y = cos(m) for 0 S a: S 7L Compute fF-dr. G]. where F(a:,y) = (y, 6”). . em: 4+, com» 5w *6 CWT/2 ?_«(h z 0, ’antfl) '“72. S‘ F. AV" 2 S < ecoslfx> ' )'S\1’\(‘L\» Cfli “‘- a 1m, 0 a g Coslf) + CC “(V-mm)ch “If/L (05(6) 4 [ shin +5 Problem 6. (10 = 5 + 5 points) (a) (5 points) Use polar coordinates to evaluate /4 fmflfi + yx2)d$ dy ° ‘W If N, 494v- “j (~174’X1’) 0‘74 :f/(V‘SU’1(9))@‘L)V-Oé=ée ° 0 q ‘/ r = f / r" («one») on = fr‘*(—(—n—(—I\Bau 0 ° 0 q q 5— W = 2fr'v ,4” 14.2,.“ :2; = O (b) (5 points) Set up, but do not evaluate, the following integral in spherical coordinates 10 Problem 7. (10 points) An ice cream cone (filled with ice cream) may be modeled by consider'ng the bounded region above the graph of (p = 7r / 6 and below p = 16. You are in charge of designing he conical paper cup which will hold the ice cream cone. The only requirement the holder must sati fy is that the upper lip of the holder lie above the center of mass of the ice cream cone. Assum'ng that the ice cream (and cone) have uniform density, what is the minimum height of the holder? Lu ‘34 F. We Mal _ _ SSS/*3 0N 201V =1 / V \l END OF EXAM Before leaving the examination room: (1) Make sure you have transferred your answers for problems 0 — 3 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. 10 ...
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exam2w06sol - Name: Lab Section: MATH 215 — Winter 2006...

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