Homework 9 - This print—out should have 24 questions...

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Unformatted text preview: This print—out should have 24 questions. Multiple—choice questions may continue on 1. I = 3 the next column or page w find all Choices 2 before answering. 1 2. f = m 001 10.0 points 2 3 . . 3 v, Evaluate the definite integral 3. I 3 ~12: ‘ 3 M2 ‘ I : fldx. __ 1 new 1 4. I ~— —-——— 2 . yr t»ij g sz't“? to NIH % m 002 10.0 points Evaluate the integral 2 arr/4 3 138.112 3,; ____ 4 3‘ I z 3 a.) .21! xxx-gt W I : WWW—m— dac. , 0 W COS x 3 Tr r; 41’; __ ‘1 ‘ 7 MA. I ‘l a 4 I W "“ (I? D‘ 1. I = —2 ’of 17:22»: “ “ l W " “tn ,_-»/2*‘—-~\ ‘3’ “r?” w E /'/ If "H A.‘ J. 2 3 35‘ élmzwsté V L] "(MK C5 //g. I z L9 K m .‘ _- n : WK (3 133.,“ “2/ _,.—""“ If,“ a _,,,,. "' i i/ “f; ‘. ” — w K > , Co 1: *1 5N 1—] j l 0 f, w/«r h); ’2}. _,_,_ 3. I = —1 5’ Lf anx/B’. I m (ax/i 9' [a f H . ‘Q _ 0’) % w if: {305 10.0 points 5. I m 0 1 __.._._ I Determine the indefinite integral 003 10.0 pomts ‘ I _ 2 + m Evaluate the integral — 2 __ 1. dz” ‘ ’3 In fl/Slutanzwd' 1 a: m 0 1»I=2(ta,n' (~\~\W)+J. \2/ miller (dkm645) w HWUQ - hamrick A (57395) 2 »/-'"“"5'.§” I m ‘ ‘ ‘ 6- I 3”" 2 tan—1 + V 4 ""“ 372 + 0 Determine the indefinite integral "1 006 10.0 points I ____ 2mm x/SE n, ,,,,,, MW _. W 3 1. tan—1 m 21n(1 + :6) + 0 d1: . Evaluate the integral XLY'IW‘) L1 2. I m 4$taflm1$+2V1+CIfi+C [a 1 I m 4111 I‘V-f’f-w“ «3:2; MWELW [3235’ 3.1m4x/Etan“i\/§+2x/1+$+C 5 I?— ,-!+“" ‘ 2 I = 4111 ~— w M- r L» (3) ‘: O E“ g 4. I = 2M5Etan“1\/E~ln(1+m)+0 , :1 My raj) ML“ 5 (law E -— i éwmd.” M ,h w w. 3 I \m?! j 5. I = 2$tan 131m111(1+37) +0 6’: _1_1 1 '; Hy m. {.A mm) f “1 /‘ Q” “U ' _ 6.I=2mtan 33+ l—i—m-I—C’ _\ 71 \ 1W5) ' x (w W " .‘ ‘ "f s 009 10.0 oints .35 :m p 3 , ' M . . . I! “W” etermme 1f the Improper integral 6- I z "’3'" I «a ‘ "3 1 ilk-“1“ I I =-" / d’U 3:" «‘3 «y “00%;” [WX 007 10.0 points —m 6 ~ I? U k (-5-;- «AW? \. Evaluate the integral is conver em; or diver (ant, and if convéiiééh—fijii 59"} g g in 8 find its value. .75 I = x/reagwklda: "9 1m. j. i; EH3 M1“ LN»: wzfifi‘ 1 I m 3 i v M m / "M v* " Rel? ' )- __ 2-“th 2': agdu-f t ‘ ‘ 1 I '" 1+1” was)“ {gfi-ww [MEI—f1” 2 I = 6 «2;.wa M H 1 3:: db ma- ?” 7" ax Hw- .‘ V var“ .. 2. I m I ——— ~ in 3 u”"‘ m: "35:: m __ 3 p 2" {1‘ "'WM 1 ' ‘ Em 2 /" .D "‘ 3. I —-- “ fly /3 I. a ‘ I x‘ m 3. 6 DJ.” 2 c”. M (‘54): f-WW‘WW” ‘3 1 I h-Qf at. \ L E d”WWMummy; -- r I f, .\ I __ r m :1, L m a T. M; (/w‘ . , ‘ w ' 3. I —— 2 + Y I: 1 flaw u him-m 4° I 15 divergga r A . A. " n “Mm mm» 1.: Aron-O + mum} x —,‘ v 243%} MT,\ #7 gr“ \V\ “U :L u):¢"§tf:7fi*"‘ “we-:53” 5 . . | 7 . ‘F J; “1‘... «I... :1, gumilx'MrV‘ ,,_ "" um miller (dkm645) w ewes) »- hamrick — (57395) 3 is convergent or divergent, and if convergent, 5. I m __ find its value. 2 wwihmwfi. 1 Z {A 45 ’ M m- 2 a (V '7 010 )91) points 1' I ”"”“ 27‘" {, live) ( . . /. . ‘2-- 4, u. Dete erf’ffie 1mproper1ntegral 2. ’I is divergent 2..“ 6 CO __ ‘ "" 9 "2,73,! .2 3 1mg rmwawrfiml converges, and if it does, compute its value. , LIZ; w 1 4. I m — 1. I does not converge \ M47? 2. I = 5, I m g— 3 I: 6.?“% 013 10.0 points Determine if the improper integral \Mf 1“; 1 31 2 011 10.0 points it. 0 ’ Determine if the improper integral is convergent or divergent, and if convergent, .1. find its value. a DO 1 ’ I. (.1 ‘ ‘t 3w... CR" -.~ I z W dz: W /5 x -_ * f8 a:(1n 3.93)2 {re-9G5}; . o, 33.2 ’ converges, and if it does, compute its value. (f; f 5:4, Zuqudltm 2. I is divergent J __ r! "1.1%.!- O 7’ I; 1“ EU 3 I m 31 2 1” '3’” “Vi” ' if m (n + ) twiifiriwee “M m. i v“ .- ; QM _ t m £4}: “3. g. I .\ .3 w new be?“ / _ ' , 4. I 3 20112—1) Vim/2» I w 0% [A , ,u. Lt ‘ J, M. _, v .1“ m P, flu.» I g 1‘ (a if,“ u finely) $14,567“ Erwin E I _. “Ev ' 5. : 3(ln2w2); i din (DNCI-VVZ» '— @‘B “m V“ “ 6.1 w 2(1112-4-1) 1 ,_N_,r...______-_.. .. 5. I 2 m r ,_,.— ~~"' 36 ‘\ 7- I x 60:12 _ : ‘ fl 1 “Ema-MMM‘W"_,,,,.J=~*"“M 111 38 014 10.0 points 012 10.0 points u _ ‘ . , . ‘ _ The reglon R 18 bounded by the m—ams and Determlne 1f the IIIlpI‘OpGI‘ Integral the graphs of 1 - ~1 2' sm :2: 2 I m z _ din : W = . 0 WW 9 ma m 1 ; - S (V‘: I PK _ am“ ‘YLV‘ 2H5? = l V .4: miiier (dkm645) — HWOQ M--- hamrick A (57395) 4 A part of “R is ShOWIl as the shaded region in Determine Whether the partial derivativee fm, fy of f are positive, negative or zero at the point P on the graph of f shown in \‘\ I ¢ w ’i we age!!! #2" Compute the voiume of the solid of revolution obtained by rotating 7?, around the w»axis. ‘i-W" if; wr K1.» 1. volume m 11:“ Q<igflm N \3/ E _® H 5-5; :1 Jere ,2 33$ e: l x ’7‘ e H :3 L 3. volume = 2th '13,“:qu60 up?“ :1“ 5. voiume = 3w “\- ‘\_,/ M F l l c: 6. volume infinite 015 10.0 points Determine fm + fy when 7. fa; > 0, fy > O f(:c,jg 24m2+$y—2y2+3m+2y. 8. f$<0, f >0 9‘ 017 10.0 points Determine f3; when \-M__~W:W;W_M'M I ' f (m, y) m a: eoe(m + y) + sin(:r; + y) . 1. f3; = 2cos($+y)+$sin(m+y) 4 f$+fy = 7m+5y+1 2. fm = 231n(m+y)+93cos(cc+y) 5 fag—inf?) =2 7m+5y+5 3. fx 2 2mcos(a3+y) 6 fm+fy m 7x-——3y+1 4- ft m 2511101: + y) — meoskc «+~ y) 016 10.0 points 5. f3: ""23: sin(:a + y) miller (dkmfiéfi) HW09 e hamrick «» (57395) 5 6.fi;m—w$Mm+m a‘flfiw)m(x—&m+flymM) 7' z m COSW + 021 (part 2 of 4) 10.0 points m 2 w ' @x COS“: “if” 3mm“: + 10 (ii) Determine fy for the function f. 018 ' 10.0 oints Kn.) p 1. fy(m,y) = (:1:+2)(33+2y+2) Find the slope in the zit—direction at the point (2,42, 0) on the graph off when 2. ifywyy) = (x m 8X3; W 2.9, m 14) flnywmmfi+m%- VMW~ Ergflt§ 3. 'y(;v,y) 2 (a: +2) (53 + By —- 14) a 3; ‘L—wa} ~[2-f X tlbfii‘gy “'2‘? .019 10.0 points "1"" 4" L" 4mnww)fl(y+@@$+ym@ Find the vaiue of f3 and fy'at (1, ml) when *5“ 2 2 5- rim) = (yw8)(2rc+y~14) fling] :2 i—Zm +3y. 022 (part 3 of 4) 10.0 points 1.fin =1, 5‘ :21 ‘ _ (1?“1) (11—1) (iii) Determine fm + .ny for the function f. z w m9, ' =1 . M f‘ (1,—m1) j” (1,—1) 1- fyyxmay) = 113+9 —* 6 e . =1, m—41 , , , lm4> Eben [email protected])$2@+y+flfl n . mg, ‘ ==Wu f” (1,4) fy (1,—1} 3. (fm+ fWXImy) : 33+y+ 10 5. f f] z -8 44444 xarw yo4> ;“a(nw+tnew)=2e+ym6)“\ omumm1fi®1mmmms ‘““w -w we _ 5-mm+nnen)¥2e+y—m) A function f is defined by Hm, y) m (a: + 2m; ‘- 8m: + y W 6)” 023 (Part 4 of 4) 10.0 points ('1) Determine fa: for the function f. (iv) Determine fivy for the function f, wwwizfifirm‘f:-mhw\- L.hmw)=([email protected]+y+& _ .gfflWWw)mzfl+%_¢a Q , '2‘! j 5:. LE-v' i, S‘x-j -f«- 5% 3v w‘, 1%}, “4.1% . ‘ mgflfl/fl/ ii (y—&@m+yw® 2-hMey)=$+flymfl '3 X {j J“:31- "‘\‘"3 " “5‘4"” 842:" : ‘3‘?” '“' I U) 3‘5 i "‘7': ‘1‘”:3‘ fmngy) z 233 "1” y m" 1-6 zx+1 "‘1 J “j 4. jammy) fl 2x+2y~16 miller (dkm645) ~~ HWOQ ~~ hamrick — (57395) 6 5. f$y(zc,y) = 2$+2y+16 024 10.0 points Determine the second partial of 1” when 2 2 “PM? ‘ 4m 4L ‘ mm m — + i. v» y 63, 833 y x 1 z 33'" W“ EVE-é- r X) ._ 3 W = “1:2” "" {in 349 hmwm; « v 1(3) é}: 4. fmy = 8:1: + y va‘W-I ! 5 .— + y W m 11/2 3x2 ...
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This note was uploaded on 03/24/2010 for the course M 57395 taught by Professor Hamrick during the Fall '09 term at University of Texas.

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Homework 9 - This print—out should have 24 questions...

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