spprelim2 - ® ‘1 Let Vi be the volume of the region...

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Unformatted text preview: ® ‘1»; Let Vi be the volume of the region coordinate planes. Then 3 3“$ 3—5—3] 27 Vl = f f / dzdydzn = --—. o n o 6 Let M; be the volume of the region in the first octant enclosed by the plane m*+y+z = planes. Then 10 10"“3 10*3‘11 1000 D 0 D in the first octant enclosed by the plane a: +3; + z = 3 and the 10 and the coordinate . 2 9 3 The volume of the region between the planes in the first octant is then V2 = V1 1800 7 i . =.._.._.._._..~__ G) Let f(x, y, z) = x2 + y2 +22 be the square ofthe distance fmmflu: origin. We wantto minimize f(x, y,z) subject tothe«constraintsg1(x,y,z}=2y+42——5=Oamdgz(x,y,z)=4x2 +4y2—z2=0. Thus V‘f=2xi+2yj+22k, Vgl =2j+4k,and vg2=8xi+8yj——szsotha1 vf=AVgl+11Vg2 => 2xi+2yj+22k =A(2j+4k)+p(8xi+8yj—sz) => 2x=8x11,2y=2A+8yfl,and22='4z\—22p : x=00rp=%. CASE 1: x=0 2:} 4(0)2+4v2-22=D => 22 i2y' 2} 2y+4(2y}~5=0 => y: %,or2y+4(—2y)—~5=0 => y = — g yielding the points (0,;a 1) and (a, — 52, g . CASE2: p=§ => y=A+y => A=0 => 22:4(o)-—2z(4i) => z=0 => 2y+4(o)=_5 z; y=gmd (O)2 = 43:2 +4 (.3)2 => no solution. Thenr(o,§,1) =§andf(o,—§,§) =25(§‘5 +5) = '32; => thepoint (o, g, 1) isclosestto meofigin. .m\ (if: .‘i‘X-J-i-ZKV . ’ @ 31(X,3g,:)=x2+32+%2~211,50 Vfifibfli— QgJ—rZaK gzmyl‘z)=x+‘3+%’5° v22: 1+3“!- K 508V? 151.? 585475104 arzx‘Aaj/L ,xc XgAf/u 12.1% 1?sz x2+ldz+tz =21; X+g+%r0. I" W 50611250715 are {2,2,'“‘U}(—,?,»Q/4/JJ (1:35; (7'31? [4- 313 ’1) ~_,__.. ; ) 9'2; 2. ~_—-—-—— «2 #361,) I 7 ‘ ' , r - ..- firm-wry, 75/4/°?z’/)=/o2, f/figfiégé, 1} /3, mad ,2, 2. {Mpera’tlure is {.2 mo! #12 W'm‘mum {mfg/raflre 15 43. @ lid Pifigfl). ’52 14w Var/48X #9 {fie 59X i’M M6 for?! odor/ml. VoKumgr fcxggahgxga cansz’rainzf 3(4g,—2)=12+g1+%2—l=0 V’f" 3g?” 1‘ 63X??? 3’ng v3: 2x” ng+22z 3061/2 Hug 555190”: 57x62 =' 22M 232 7 25M «ME = 2—2/\ Xz+tal+22 =1 77ng if; {— J4.- 5' -2#=0,i‘7" A); a, (9. M'ZZZfég {he (fin?! eyaficm 55X, %6 sewna/ 5;; 09/, W61 7%? Mrs? 56;; 2. 73m @514 flame/Jim; 6557532 11344 ‘ ”95,75 are #12 Same (avg/2); 5'0 fine 12 f—flcmq/ gig/,6 We? mg Meg, QflzzzxgzroZ/laz. #51qu 5:” 5% Xzzgarié. Simce X70, 63/70, 270[iu7%g [icy—5 voila/142‘) , M? mafia!“ 1%“?! X”#:£' Xz+yz+2asi {Mb/£565 X:y:2 g'L. (3‘ m Maximum W6umgg‘5 CF. (17%“? :gF. 5 £(L‘flgl {+3E2")r~/B. Hamel-Ha mafimum . f(x,y,z)= ficpsxsinw—i-z) atPo (0,0,{3 =; “9,0,3 —_- 1; f" =—\/2—$inxsin(y+z), fy = ficosxcosQ—i-fljl: ficosxcosti—z) z) 13,351) = 1 «fix—03+{yw93+ (z- E) =y+z~ §+ 1; f“ =—\/§C08xsin(y+z), fyy = '“VEWSXSinU'l'Zqu =- Zcosxsin(y+z), fxy =‘\/§SinXDOS(y+Z),fu= ~ 2sinxcosfy+z),fyz=-— 'lcosxsiniyd—z). Theabsoiutevalueof each of these second yarfial derivatives is bounded above by V5 => M: v5; thus 5130‘, y, 2)} s (é) («5) (0.01 + 0.01 +0.91)? 2 0.0006: ® ”1'; {3.) Put .113: y) = (1] h232‘83my2‘ and find the linear approximation at (x: y} = (1’ 1). ‘RIB have: ‘ 1.01 1.03 33:13 7-" m1, 13 = 412, so 13(sz = fia—am—fl—fiyd). Thus / Mm) dm dy = .99 .99 .0091 111?. ' (”D {b} We are given that 1311,1393, and if”! are less than L So the error in the finear apprmdfiaation is bounded by E = 1/2(.{31 + .0132 = .0002. But, we need to find the error in the integration. Since we know that, {Lhay} ~ f{:v,y)i S E on our region, we have that IL!!! 1,01 1.01 1.01 Lizzy) mtg- my) dandy] = I .99 .99 .99 .99 1.01 1.01 f L(a:1y)~f(z,y) dz dy .99 .99 “1.01 1.01 S / dedy .99 .99 = £003.02)? = 8- 10-8. 9 (a) We ham: fags, y) = kzez—ysinlc+6——5k. and fy(a:,y) = 131111;. Therefore 12(1), 0) = k2—5k+5, and 13(1), 0) = a. The point (0,0) is critical for f ifand only if k2 — 51:: +5 = 0. That is is = 2,3. (b) If (0,0) were a. local extreme point then by (a) is: = 2, 3. We have fm(m,y) = 11:21:”, fw(m,y) = 0,f$y(3,y) = —sink, and 3340,13) = k2,fyy({),0) = 0,fxg(6,fi) = —sink. Hence fmmfi) ’ fyymfi) — {fmy(fi,0))2 = —(sink)2 < B when Is: = 2, 3. Therefore (0, 0) cannot be an enreme point for f. a} it =332+3ya fy = 33-1-3112, firm: 63;: fry =31 fyy =69: fmfyy "fgg =362yfi9. Solving f3. = f3, = 0 we get that the critical points are (0,0) and (—1, ~13- The point: (0,8) is a. saddle point. The function has a local madman. at, {~1, ~13. . ‘ ' . , 1. , . ._ . , b) Onm=9,05y_<_lwehavef{9,y3=2y+1,wbichhasanabsolutemmdmum3aty=landanabsolute minimumlatyrafl. 0ny=0,0$z$1wehavethatflmfi)=m+1,whichhasanabsolutema:dmum2at x=1andanabscluteminimumlatm=0. Ony=1-:z;, 05:31wehaveg(x)=f(z,l-a:)=4a:2-—5:x:+3. Since g’iz} =8z-5 =8 forz: g, weconclude thatghasanabsolute minimum 1-1—2- at arr—1% and anabsolute madmum3atz=fl Solvingfwr. 1-4y=flandfy =~4z+2=0weget $318330th (1%”: , whichliwinthe interior of the triangle, so it. is a. criticai point. Then f(%, i) = g Therefore, the absolute mmdmum of f is 3 at (O, l) and the absolute minimum is 1 at {i}, D). ‘ We have: f(m.y)=zz+y'*’—'-my—m fn{z,y)=2 1:41;?!) = 2‘1 - 9 ~ 1’ ' 139$”: 93 '= 2‘ EA”: 3]) = 2‘9 “ 3" ’ fmg(:z:,y) = *1 Om: fimction is a. polynemial so the gradient exzsts everywhere, so our only critical points occur when the gradient is 0. This happens when: \ So our (Daily critical point is I, . Now we must check if it a. local min, lecal max, or a saddle point. We WHO E ’3 have fmfww 39:4-1=3>0 so our cn‘tical point is not a saddle point. Since fm(2]3, 1/3) = 2 we have that our critical point is a. local minimum by the 2nd derivative t . To find the critical points of f we have to solve the system f: = fg : 9 (none than there are no points at which the derivatives do not exist}. This system is 44:3 + y = 0 and 2: + 43,13 = {3. The first equation gives 3/ 2 —433 and substituting this into the second equation we get. 3: -— #39 = 0. Thus either 2 E. O or as = 1/44 :2 if? and so I = 1—1/2. Using the first equation to find 3; from; we find that the criticai points are (0,0), (ix—~23) and {~7i,-, % . To classify them we use the second derivative test. The discriminant is D : ft: fyy~f3y = (1223} (12112) — 1 = 1:444:33;2 —— 1. Thus D(O,Q) = —-l < 0 and so (0,0) is a. saddle point. Further, D(%,———§-) := D(——%,-§-) = 3 > G and f“ %,———§-) = f::(~%.-§ 2: 3 > 0 and so (:31 -—._%} and (‘%z%) are local minimum points. ® a) f, 2 3t + 332 and ft = 33 —— 3132. Solving 31 + 352 = 0, 3.5 — 3152 = O we obtain the critical points (0,0) and (1, —1). Furthermore, f5, = 65, f5, = 3, f” = —6t, so fssftt — ff,- 2 —36st — 9. The second derivative test shows that (0,0) is a saddle point and that f has a local minima at (1, —1). 1)) Let A = (0,0), B = (1,0), 0 = (1,1). 011 AB we have that t = 0 so f = 33, and it has maxima at (1,0). On BC we have that s = 1 so f = 315+1—t3, and it has maxima at (1,1). On AC we have that s = 1: so f = 352, and it has maxima at (1, 1). Part a) shows that there exist no critical points in the interior of the triangle. Hence the absolute maxima is 3 and it is achieved at (1, 1). Interior Critical Points: fm(:c, y); — 230211, fy(:z:, y) = 21:2 Therefore a: = 0 and ~1 <2 y < 1 We have f(0,y) = 0 Boundary Points: 11:2 + y2 = 1 implies m2 = 1—y 2. Let g(y)= (1 — 2)y with g(3,r )= _ f 1 i __1_ 1 31,12. The extreme points of f on the boundary are (212%, \7—3)a nd (fig? fl) We have f (iL/f-E, %) = M and f <i%7_%> = —%§. The greatest value of f inthe given region is a? and the smallest value is —2—‘9/_—. @ Note that V f (512,1!) 2 4(y — $3)i + 2(29: — y)j. If the professor is neither ascending nor descending, then he must be walking along a level curve. Since level curves are perpendicular to the gradient, we just need to solve the equation Vf(t,12) ~v(t) = 0, where v is the professor’svelocity vector. Since v(t) =, i+2tj, we have 4052-153) +2(2t-—t2)2t = 0. Which simplifies to 121;2 —8t3 2: 0, solving for it gives us solutions at t = 0 and t = 3/2. So, the solution points are (22,3!) = (0,0) and (my) 2 (3/2, 9/4). @ fm— - 2:10 —- — ‘0, f,-- - 2y 4—- - 0; (1,2) is the only critical point. On the boundaries: 112:0, 0<2y<31 f(0, gy)=y2 —-4yandf(0, y) =2y— 4:0, soy=2and(a:, y): (0,2). 1' We also keep the endpoints (0,0) and (0,3) y = 3,0 < a: g 3: f(.’L‘,3) = 3:2 —2:1:-3 and f($,3)’=2:c—2 = 0, a: = 1, (any) = (1,3). We also keep the endpoint (3,3). y=$,0<:1:<3:f(:z:,:r)=2$2—6mandf(:r,a:)’=4ac—6=0,:1:——~2—,(xy)=(%,-23-). The candidates are (1,2), (0,3), (0,0), (1,3), (3,3) and (2, 5). The corresponding values of f are —5, —3, O, -4, 0 and —— .The absolute maximum is 0 attained at (0,0) and (3,3) and the absolute minimum is ~5 attained at (1,2). ® fm = —-3+2:z:+a:2, fy = 3-2y-y2. Solving ~3+2$+£E2 =0, 3~2y~y2 :0, we get the points (—3,——3), {—3, 1), (1,—3), (1,1). The point (1,-3) is not in the domain. The points (~3,——3) and (1,1) are on the boundary. The point (~3,1) is a critical point. The second derivative test shows that f has a local maxima at (—-3, 1). The function f measures the area between the graph of 3 —- 2t — t2 and the s—axes, so it has an absolute maxima. Since 3 —- 2t - t2 = O has solutions ~3, 1 it follows that f has an absolute maxima at (—3, 1). Alternatively, f has a local maxima at (—3, 1) and f vanishes onwth’e’boundary of'the domain, so f has an absolute 1 . t3 32 mamma at (~3, 1). The maxima is / (3 — 21‘ — t2)dt— __ (3t —— t2- )3 = 3.. u _3 _3 . (a) The region D of integration is bounded below by the parabola y— — a“ and above by the circle , z'q— ‘ 7231,! = 2 (b) We evaluate the integral: x/Z—zz 1 1 U [I f: ydydz = 31(2 - 3‘) - 241113 = -1-4 [22: ~ 33/3—3: ROIH 22 ~2—1/3—1/5=E 1 fl v’i 2—11 (c) Reversing the order of' integration, we get / / ydzdy + / / ydmdy. ' ' 0 we 1 4m: mi (b) T he region is bounded above by :—r .— y?“ and below by z = y W 1,2 ‘1 1:: (Cl / / scygdydm‘ 0 . .733 El . 1 g // cosinydsc dy. D arcsiny 1112 ey ‘ f / (:5 +11) dccdy—i— / Lt +y) dmdy+/ez /:(x2+y)dxdy. O (l 1112 ® Reversing the order of integration we get 4 4‘74 2:627! 4 e29 1:2 H“ 11,] <~—> a o 4 “ 31 o 4 * y 2 <2.) The equivalent integral in polar coordinates is .71 1 1 .1 . , I f2 / e-rzrdrdh—l/z / e—TZ d(-r2)d19= if e": o o 2 o o 2 o @(a )The region is a triangle bounded by the y—axis the line 34* —— 3, and the line y = 9:. The integral is 3 34 35 (b/s/Oyfl; +113 dmdy= Bag-+114 dy=l—2—+—. )The fregion is bounded by the y-axis the line 7; = \/—, and the curve :1: = \/— The integral is l (b/f 77/0112 y71' cos(:c )da: dy— — fofi y7r sin(y2) dy = 71'. 1 (c) The region is bounded by the w—axis the line {17: 2, and the 2curve y~ — :32 1 4 1 d=— 294:- 8—1. 0 1; 2/08 3/ 4(6. 3 _ l % -1_ -Z _ —1 1119- 2/0 (2 l)d19—4(1 e ). 1 0 Beginning by using Fu— 2 biniHstheorem the 1ntegral 13/ /: rye—“’5 d9: dy 2/02 [:2 ye‘m5 dy dcc =/ (1/2):c4 e ”5 d2 7—- 1 8—32 I O 10 10 ' 1 1/5 @ a) Reversing the order of integration we get / f f (3:, y) do: dy. We evaluate the integral ,0 , . 12/2, , ,,,,1, t/ol/ii/qjifiefl else at = fifiw >< )yfdy ‘ Alhfiflfififl‘ 11%; = ./ol(y y My ‘" (y: _ £31: 3 2 5 0 10' [+19 'P+1 w/E'Z—P . b>/ / fix 1) d1 11:] / 11w) am an 2+2? P ( 11-19? The value of the integral 15— 30(the same as in 6a)) , since we obtain the picture of the region in 6b) by shifting the picture of the region in 6a)1p steps upward. a The r limits of integration are i} to 5311(38), and the i9 limits of integration are 0 to 7r/3. So we nave the area. is givén by the integral: x13 shiaa) 1:13 - 2 13 _. f / may = $5291 :19 = 5 f 1 “”5“” d9 = 1W3) = mm a o o - 3 o 2 4 The curve 7' = sin(29) is a fourleaved rose. The carve r = (203(8) is the circle of radius 1 and center (é, O) (in cartesian mordinates). 'I'he gaph shows that there are three i Intersection points. Soive r = gnaw) and r = sin(28) simmmusly. We get 2sin(93 (335(0) = cosw). So either sosfi?) = 0 or 3111(6) =3 1. Hence, the intersection points are {139) = (13,325) = (i)= =32!) and (1', H) = ($95) and (7319) = {—Aéfi, 5T") = (1;, .JE" , The mass is the area, so M = 5(427 — 7r) = 3“ 7;. The y coordinate is 3‘} = 34%, Where ME = f f y (114. Because of the symmetry about they—”axis, we may write Ma; as: 1 J31? 2 v22}: Mm: // ydydx+2// ydydx. G W 1 o it follows that Mac = is? and so 17 = 3 gr ' @ Step 1 Find the area of the region. The formula. for the area cf 3. parabola is Area. -—- :- - base - height. 013: parabola has a base If 2 and a height :1 oil so its area is 3:. Step 2 Find the centmid. Since our region is symmetric about the ycaxis we must have that the centroid lie: on the y—axis. So i = O. Nowweneedtofindfi. 1 1—5: 1(1_$2)2 1 2 1 .3 4 2.1 _ 8 Ms*/_1/fl ydyd$= 4 ,2 dz—fo{1—m3)dz—- a 1—2x-+zdz-1_§T5_15 s --Mz_§_.§._?. 03""M”15 4‘5 30 the centroid is {i5} = (0, :). ...../ ...
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