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math_273_(spring_2009)_practice_test_2_solutions

# math_273_(spring_2009)_practice_test_2_solutions - MATH 273...

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Unformatted text preview: MATH 273 SPRING 2009 PRACTICE TEST 2 90 L uTio/vf NAME: 1. (14 points). Find the local maxima and minima and saddle points of the function f (x, y) = m2 + 3/2 — 6x + 2y + 5. A: u—e a> =s,g=—-1 .3 m 0,4 cm, Answer: local min at (3, —1). 2. (12 points). Use Lagrange multipliers to ﬁnd the maximal and minimal values of f (as, y) = x2 + y subject to the constraint g(a:,y) = :32 + y2 = 1, v,l:421(ld> I ‘77:.42437) 2x .-, >le ”(mm 1 2 A27) PUSH‘JQ‘J’IPS 4), X=0 => 31:1 =>7=11 Two yam/+8! {0,1), (0,4) 1) )‘zi => l:23=>‘1=-?l_ \$1h¢==ax=i§- Tm fonds: (§’{)J ('eg Answer: min at (0,—1), f(0,—1)= —1, max at (iiéj), f (ilzéj) = 2. £(qﬂ=1 1(( 00-1) '3 ‘1- (3.411: 1 3. (12 points). Find the integral of f(z,y) = x2 + y2 over the region D = {(\$,:g)|1‘2 + y2 S 4}. Hint: pass to the polar coordinates. 2. 1K 2 SW) 1A: gjrc‘ww = M f W” ’D o o o z :1]. %25.117: 27'1‘6‘27. Y 0 Answer: 87r. 4.(12 points). Evaluate / /D<x — mam, D={(\$:y):OSySLy2SmSy}- where 5.(12 points). Find the Jacobian of the transformation m=u3—v3, y=u3+v3. 4047) ”L "W; W: 5 (MZVL \$[hltlj 3‘41 3V}. . Answer: 18u2v2. 6. (12 points). Evaluate the line integral where the curve C is given by /gds, c\$ \$2154 y=t3 withégtgl. 1%} = % I = Véc'/+uiw’/w‘d+= = V679)?» (31‘71044: =V/H5f9p1y‘: t V/éf-rall/ [SLAB J: ffﬂ [bf-71 (“W—4+ [9%be ’0 6 C 4 ’2, Answer: ﬁ(253/2 — 133/2). Le! “3 [”43”, 1‘“: 3th+ 3 lé'3 7. (14 points). Use the transformation x: 211 ,y: 512 to evaluate the integral f [R 1/4m2 + 25312 dA where R 15 the region bounded by the ellipse 42:2 + 25y2— — 16. H 1153‘: 9-8“) + 251—5?) vi- /( [a \$1») m 19"” af ’Wﬁ‘ﬂ ’2' ("I V/ 4902/4100?! '3 g t [/(‘W/ MW“) "5} *4 my s2 "' = 3- Answer: 157T Q(¢IV) V : .— 0 3' 5' ”“4 rs i( unﬁt/(k- 7? HMW‘IQAXU £2” ﬂ 5 [MW “vi/tug, 4 IT .3} J _ 32 ‘7} 1 31 2/? 57/? y I (T wiQdﬁ v E‘ '2” E— /a : -S-; '3 5/;- 8. (12 points). Find the work done by the force ﬁeld F(m,y) = msin yi + yj in moving an object along the parabola y = \$2 from (—1,1) to (27 4). t”): <6, +2>/ t€[—1,2] \$7“): (1, 2+) 120%”.qu «6 sm“, 1"» (1, if) .= = + m4 t2+ 4H 2.. ”h J (m +‘+z+’)4* =6: cos '1 Answer: —%(cos4 — cos 1) + % Ixﬁui-iw ml fﬁml‘zdf : +2.1» 2 ‘ if y .. éfakuéﬂ :- M; t2. I 01%; LW 3' CD‘S'fL‘f’C , 2. ...
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