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REVIEW 2 - Review Quiz#2(20 points Math 200 Name Key Group...

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Unformatted text preview: Review Quiz #2 (20 points) Math 200 Name: Key 04/07/10 Group: 1. (a) Find the gradient of f (x, y)=5xy2—4x3y at the point P(l,2). VH’W) : <¥X)¥?> 2 <$yknllxayj lOXy ~Lf33> W012) = (20—20, 20.») : <4, 16> (b) Find the directional derivative of f at P(1,2) in the direction of v =(—1,3) and explain its meaning. ——I "' - "‘ A -' : {£114.33 0“ W’97) “ View] , u. “L 7%? m —-. __ _ _ g _: 52 ““013 ~ V¥(I,2)-<de-g)fio> —(—Lr,:e>-<-,%ro)g%5> - fit‘fifa i=5, UTE 15 1406 rcd-e oi: aka-n39 ai- -F {in like du‘rec‘ix‘an. d'F 7 '5 (*1; 33 at (3036* [30,1]. (c) Find the maximum rate of change of f at P(l,2) and give the direction in which it occurs. maximum. wad-e of change : u VFW-1)“ '1 H (“HEW = m WEI-“w? d‘ficfié" oi- maxing“ mi'e a“? change e-P -F at {30,23 is Vitus) = ~43 +153 m «3 +43 2. Suppose 2: f (x, y),where x=g(s,t), y=h(s,t), g(1,2)=3, gS(l,2)=—l, g,(l,2)=4, h(1,2)=6, hs(1,2)=—5, h,(1,2)=10, fx(3,6)=7, fy(3,6)=8. Find%whens=landt=2. ”Lemon mi i=2, x=g(1,2)=3 owe! y=l~(l,2):é a- 3e 31 91 3 — a?" "a; 3t + 3) Sit ‘ +"("I’)3t("Jt) MC?("I7“‘t(‘Jt) )3; 9t 3:; = ix (3/6)gt(l,2)+¥,(3,6)Lt(112) : 7(+)+8(I0)=103 3. Find the critical points of f (x, y) = 4+x3 + y3 ~—3xy and classify them as local maximums, minimums, or saddle points. "c” 2 3X13; = O =>7=Xi can“ pm 1 'P)’ 3 37"“ 3x —'-'- 0 30(1) -3): =0 :1) Xz—x:° :) ”(xiii-to are {(6,5)} - . x=o orx=l QM ’1‘ ivy—6""? £7726‘11;Xy =53 7 7:0 m-yzi DOW) 7- “i 1 . 0(00) - m” {m ”MM-H)“ "35x74 I 6‘9 <0 . 719QO J (0101;; a Saddle fond" DIM): '36-“: >0 Qnd fixflilidé >0 .Timcm) item 1'; q local minimum at J (‘1‘ . 4. Use Lagrange multipliers to find the right circular cylinder of smallest surface area that has a volume of it; include the top and the bottom circles. L Q t '5 represent Sacrifice. a. pea) r rfidl‘UJ) (a) Give the function to be minimized. in M65“ and U Uolu e M . 3(7)“ : Qfi‘r‘k r- .3an (b) Give the constraint. U (wk) = THEL : 17' (c) Give the system of three equations and three unknowns that needs to be solved. Us=avv < IrrL +44.er l‘fl‘r ) ‘-'- 2(21rr4c) Wrz> 2111'\‘f"-!-1I‘vr‘: A'l’n'rk h+1ralrln lTrr I Aan r; 2 -'-’ AY‘ Wr‘ln : Tr r‘k = l (d) Solve the system of equations to find the correct dimensions of the cylinder. h+2f 3 Ark , I" 3 Z 2 3 2;” => +1r 1k 3 t, 3‘, $Ub$+\'+0+|.n} 11 2 3y- {n+0 —H\e Conj‘l-ra'm‘l- gird-fer». r115 :1 ~ 3 3 _ yield; )1» =1 => r '"i 3 ‘3 item) “if; M 11:20? 5. Calculate the value of the multiple integral ”x32 dA , where D = {(x, y) | 0 S y S 1, y2 S x S 2y} . D l I 6. Reverse the order of integration for Jo I 2 ysin(x2) dx dy. Do not evaluate. y 7 x371 ‘ ’ . "r; 0" Elwlrlfxf’, 057“} 54’“ ‘I‘SMx‘dey 3 7. Use polar coordinates to evaluate I: I0 (1322 + x3) dy dx. J5?" 9 Ens-m - w 5720, on“? 13096)] @5913m05r53g 3 211' 5 ’5 s -5 COS 9A9 rfidr 1‘. Sine/3 ' é‘r o 3 (Oaf-LI).§L3 o y?- _ 13:3 "- 5 8. Evaluate ”L 22 d V , where Wis bounded by the planes y = O, z = 0, and x+ y =1 and the cylinder y2 + 22 =1 in the first octant. ”fhdu‘jfli 2,9921” =CIDIU— flow» =f£y 371;” b ‘ x a 042._ W : {[(1‘*l‘é(\v)]dx: [1(l-X)LIJ§+ (1-29:10 m gn-S‘l' Odors-T11" : 0 —-I:_J~2 +11: :- 3]: “TL: 2 '52 9 Rewrite the lutegral Il I12 ID “V f(x y,z) dz dydx as aniterated integral in the orderdxdy dz. 7:)‘1- Lu: ?(X,y/2)l-fo5l)xt_{),f.fl 65351-7} I 7 ‘ {7 H0“?- 4} AJ Qmflflédydx - ojf 16g>$géjdxdyda J1? IJl—xZ—wz 6 k (x2 + y2 + z2 )2 dz dy dx to spherical coordinates. Do not evaluate. Lu 1: part of Spkere U '5 gflgytajloéxfiljofyiUl-x‘Jdi‘35m} [NHL C03,.) 0) and. Pi] _ ‘ _ rn 4cm Lam: ' EOACPJIOEJ’EB “595 V5 05 “P ‘ Vt} 7- J SW m 1. I j (x+y +ildad,dx - fijjff sf=n¢00lquflae jfj/j’SWPde-WG 10. Convert the integral I; I01 2 11. Find the volume of the solid bounded by the cylinder x2 + y2 = 4 and the planes 2 = 0 and y + z = 3. BTW 911‘ 1 3 Psi-n6 2:3“; x+7 0+ 6 M] fJfra'zarae" O O D Q'fi'fl r‘(’3-rsia6)dr—de : jm 3— 1L '3 - e as o [1?" 3r5m1r 2 =5w(é~§sine)ae e“ [69+3cos9] 12. Find the mass of the lamina that occupies the region D bounded by the parabola x =1— y2 and the coordinate axes in the first quadrant with density function p(x, y)— - y. xzt-jt ma” m ‘gfb‘ljjdnn — Iii!) )dXd-‘f I l p # 13. Set the up following; do not evaluate. (a) A double integral that gives the volume of the tetrahedron bounded by the plane 2x+ y + z = 2 and the three coordinate planes. 13): V: g5( H11X7dydx (b) A triple integral in cylindrical co-ordinates that gives the volume above the paraboloid z = x2 + y2 2 and below the half-cone z = ‘fo +y2 . ‘2 : X2171 .. W 1.. a? | r r =r .. ‘L V‘J Jf rdaarae r "“3 O 0 rz' \‘(t‘~ll=o 7 r: 6 "rt-l 7i and below the plane 2 = 2. . airy/Lt lSkCP ftoSQ=l " 95 if Mame faiwma ...
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