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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+4v222=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‘XJiZKV . ’
@ 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;
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ﬁrmwry, 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.
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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.
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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. HamelHa maﬁmum . f(x,y,z)= ﬁcpsxsinw—iz) atPo (0,0,{3 =; “9,0,3 —_ 1; f" =—\/2—$inxsin(y+z), fy = ﬁcosxcosQ—iﬂjl: ﬁcosxcosti—z) z) 13,351) = 1 «ﬁx—03+{yw93+ (z E)
=y+z~ §+ 1; f“ =—\/§C08xsin(y+z), fyy = '“VEWSXSinU'l'Zqu = Zcosxsin(y+z),
fxy =‘\/§SinXDOS(y+Z),fu= ~ 2sinxcosfy+z),fyz=— 'lcosxsiniyd—z). Theabsoiutevalueof
each of these second yarﬁal derivatives is bounded above by V5 => M: v5; thus 5130‘, y, 2)} s (é) («5) (0.01 + 0.01 +0.91)? 2 0.0006: ® ”1'; {3.) Put .113: y) = (1] h232‘83my2‘ and ﬁnd the linear approximation at (x: y} = (1’ 1). ‘RIB have:
‘ 1.01 1.03
33:13 7" m1, 13 = 412, so 13(sz = ﬁa—am—ﬂ—ﬁyd). Thus / Mm) dm dy =
.99 .99 .0091
111?. ' (”D {b} We are given that 1311,1393, and if”! are less than L So the error in the ﬁnear apprmdﬁaation
is bounded by E = 1/2(.{31 + .0132 = .0002. But, we need to ﬁnd the error in the integration.
Since we know that, {Lhay} ~ f{:v,y)i S E on our region, we have that IL!!! 1,01 1.01 1.01 Lizzy) mtg my) dandy] = I
.99 .99 .99 .99 1.01 1.01
f L(a:1y)~f(z,y) dz dy
.99 .99 “1.01 1.01
S / dedy .99 .99
= £003.02)? = 8 108. 9 (a) We ham: fags, y) = kzez—ysinlc+6——5k. and fy(a:,y) = 131111;. Therefore 12(1), 0) = k2—5k+5, and 13(1), 0) = a. The point (0,0) is critical for f ifand only if k2 — 51:: +5 = 0. That is is = 2,3.
(b) If (0,0) were a. local extreme point then by (a) is: = 2, 3. We have fm(m,y) = 11:21:”, fw(m,y) =
0,f$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 = 3313112, 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,la:)=4a:2—5:x:+3. Since g’iz} =8z5 =8 forz: g, weconclude thatghasanabsolute minimum 11—2 at arr—1% and anabsolute madmum3atz=ﬂ Solvingfwr. 14y=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 39:41=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/fE, %) = 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 — $3)i + 2(29: — y)j. If the professor is neither ascending nor descending,
then he must be walking along a level curve. Since level curves are perpendicular to the gradient, we
just need to solve the equation Vf(t,12) ~v(t) = 0, where v is the professor’svelocity vector. Since
v(t) =, i+2tj, we have 4052153) +2(2t—t2)2t = 0. Which simpliﬁes to 121;2 —8t3 2: 0, solving for it gives
us solutions at t = 0 and t = 3/2. So, the solution points are (22,3!) = (0,0) and (my) 2 (3/2, 9/4). @ fm—  2:10 — — ‘0, f,  2y 4—  0; (1,2) is the only critical point. On the boundaries: 112:0, 0<2y<31 f(0, gy)=y2 —4yandf(0, y) =2y— 4:0, soy=2and(a:, y): (0,2).
1' We also keep the endpoints (0,0) and (0,3) y = 3,0 < a: g 3: f(.’L‘,3) = 3:2 —2:1:3 and f($,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$2—6mandf(:r,a:)’=4ac—6=0,:1:——~2—,(xy)=(%,23). The candidates are (1,2), (0,3), (0,0), (1,3), (3,3) and (2, 5). The corresponding values
of f are —5, —3, O, 4, 0 and —— .The absolute maximum is 0 attained at (0,0) and (3,3)
and the absolute minimum is ~5 attained at (1,2). ® fm = —3+2:z:+a:2, fy = 32yy2. Solving ~3+2$+£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 = 14 [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 / erzrdrdh—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 yaxis 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 ( 1119? 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+zdz1_§T5_15 s Mz_§_.§._?. 03""M”15 4‘5 30 the centroid is {i5} = (0, :). ...../ ...
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 Fall '06
 PANTANO
 Multivariable Calculus

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