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Unformatted text preview: Vincent (jmv???) m Review ExamS  Gilbert — (57495) 1 This print—out should have 22 questions.
Muitiple—choice questions may continue on the next coiumn'or page w find all choices before answering. 001 0.0 points A rectangular box with no top isto be constructed having a
volume of % cubic inches. What width :22, in
inches, will minimize the amount of materiai
to he used in its constrnction. (Assume no
material is wasted.) 1. width 4 inches 2. width 5 inches 3. width 2 3inches
4. width mm 2 inches 5. width : 1 inches correct Explanation:
When the box has width $, length 3; and
height z, the materiai A in the box is given by (l) A($>y1 Z) = my + 2$z + Zyz since the box has no top. On the other hand,
the box has volume % cubic inches, so (i) wyz = %
Eliminating z from (T) and (i) we thus see
that 1 l A = m w.
(93,24) int/+3} +33 The minimum value of A occurs at a critical
point. Now, after differentiation, 1 1
‘332, Am
SC 92 The criticalpoint of A thus occurs at the
soiution of the equations 1
: —, 3f", : m,
y 332 y2 i.€., when a: m y m i. To check this gives a
local minimum of A(:c, y), observe that 2
A = A m e2
m (1,1) 393 52:1 > 0’
C w'A‘ ~31] —2>0
W (1.1)? y3 ya? ’
and
B 2 A :
W as 1’ my so AC m .32 > 0 at the critical point. Conse
quently, A :: A(cc, y) has a iocal minimum at
(l, 1), showing that the least amount of ma
terials will be used in the construction of the boxw
width = 1 inches . 0.0 points Use Lagrange multipliers to ﬁnd the maxi—
mum value of the function ﬂay) m 6932 — 41:2
subject to the constraint 6w2+4y2=8. CMH 3.4 Consequently, Vincent (jrnv???) M Review Exam?) — Gilbert — (57495) 3 volume = 185cu.units . 005 0.0 points Find the value of the double integral I = /L(S$—5y)dccdy when A is the region {(wiylzySmSﬁ, 0§ygl}. fﬁ (8m v—w 5y)da: : [4mg  5mg]
2; But then 3/2 m 4y—5y + if. 1
In] (eym5y3/2+y2)dm
{l 2 [2y2 — 2315/2 + e3 l 8
1. I z w
1.5
13
» 2. I m ——
80
11
3. I m {5
‘ 19 '
4. I m ——
30
l
5. I m — correct
3
Explanation:
The integral can be written as the repeated
integral
1 W7
I 2/ / (8m—5y)dm nip.
0 3:
NOW ﬂ .1" 1
0. Consequently, I
I:—.
3 006 The temperature at a point (ac, y) in the
plane is Tm, 1;) measured in degrees Celsius.
If a bug crawls so that its position in the piene
after t minutes is given by 0.0 points 2
y m cieterrnine how fast is the temperature rising
on the bug’s path after 3 minutes when (1:: 1+t, Tx(2,5) = 12, Ty(2,5) = 6.
1. rate 2 7° C/min correct
2. rate 3 SOC/min
3. rate = 5” C/min
4. rate 2 4° C/min
5. rate m 6° C/rnin
Explanation: By the Chain Rule for partial differential"
tion, the rate of change of temperetuure T on
the bug’s path is given by all" m dT(;r(t), y(t)) 3:? dt
ee+ne
6:2: alt 6y dt 1 2 = TewT.
2./i+t 5” 3 1" But when t m 3. the bug is at the point (2, 5),
While  1 l
21/1+t't=3m 1' Consequently, after 3 minutes the tempera—
ture on the bug’s path is changing at a rate m age—2x6 = 7°C/min. Vincent (jmv???) '~ Review Exam3 ~ Gilbert ~ (57495) 5 4. local max value ~55 at (3, 2) 5. local min value —55 at (3, M2) Explanation:
First we have to locate and classify the
critical points of f. Now after diﬁerentiation, "f3, =3m2+3y—33, f, me—Q, Thus the criticai point (229,3;9) of f is the
solution of the equations 3332+3y m 33, ac m 3. 216., (mmyo) m (3,2). But after éiﬁerentiaw
tion once again, ' xx = 673: yy : 0: = 3
At (3, 2), therefore,
A m m = ' ::
While
0 = fyylme) = O; in particular,
ACWBQ = we < 9. Thus by the second derivative test, f has a saddle—point at (3, 2) . 010 0.0 points
Find the guintercept of the straight line
'9! 3 ma: + b
that best ﬁts the data points
(—2: 0), (0: 4): (2, 9)
in the sense of ieast squares error. 13 correct
3 W13
6 1. y—intercept H 2. y—intercept 3. y—intercept = 13
—1
4. y—intercept : mug—33
. 26
5. y—intercept w» ~3— Explanation:
To ﬁnd the Regression line for the given data set we have to minimize the expression E(m, b) x: (—2m+b)g+(bw4)2+(2m+b—9)2. It wisz occur at the critical point of E, 316., at
the solution of the equations 8E
5% _ m4(m~2m+ b) +4(2m+b~9) = 0, 32% = 2(w2m+b)+2(bm4)+2(2m+b~9) m 0. Solving these for m and b, we obtain 9 13 Now, the y—intercept of the line y 2 mm —i— 5
occurs at ya u b, so the Regression iine has . 13
y—zntercept = :3— keywords: 011 0.0 points Evaiuate the iterated integrai I =f14{f:(§+g)dy}a. 15 W41—
II :12
15 211——
2I 5n2 Vincent (jmv???) “— Revievv ExamB m Gilbert W {57495) 7 formed by the lines a: = 2, y m :0 and y = 0.
Integrating ﬁrst with respect to a: means in—
tegrating along the darkened segment of the line 3; m a, 0 S a g 2, lying inside this tri—_ angle, and then letting a range from 0 to 2,
The oniy difﬁculty is that the integral with re—
spect to :5 cannot be determined, so we have
to reverse the order of integration. Reversing the order of integration means
integrating with respect 3; along the darkened
segment 'of the line :r = b, 0 g b g 2, lying
inside this triangle, and then letting 6 range
from 0 to 2. This gives a repeated integral 2 LI: 2
I = f 26”: (lg) dm.
0 0 But after integration the inner integral be— comes 2 $ 2
[2yem L) m Zmeﬁ, so
2 2 1
I =f Zacexzda: = [~26x2] . 0 2 0 Consequently, I w (e4~l). 014 0.0 points Which, if any, of the following are correct?
A, If Jeff meme,
m2+y2$i then
1 271'
I:// rf{rcost9,rsin6)d6dr.
0 i} B. If
In x/inm2—~92d:rdy,
$9+y251 then I n.— 27r/3. ' 1. both of them correct 2. A only 3. neither of them 4. B oniy Explanation:
A. TRUE: when changing to polar coordi
nates,
dzrdy m rdﬂdr. B. TRUE: the integral is the volume of the
upper hemisphere of the sphere of radius 1
centered at the origin, hence has value 27r/3. 015 10.0 points The soiid shown in lies inside the sphere
:32 + y?“ 4» 22 x 16
and outside the cylinder
m2 ~34 yz a: 4. Find the volume of the part of this solid lying
above the icy—plane. 1. volume = 8%
2. volume m Bait/E
lfix/é it correct lib/g7? 5. volume m 8x/é7r 3. volume 4. voinme Vincent (jaw???) — Review ExamB ~ Gilbert w (57495) 9 Thus
3 2 W t
I = m / r°(1cos£16)d9dr
4.0 0
3 2 5 3 1f 2
z m d = m ._ ’ ,
till/0 r T 4W[6T l0
Consequently, . 017 10.0 points Reverse the order of integration in the inte— gral
4 lny
I=f (f f<x,y)dm)dy,
I 0 but make no attempt to evaluate either into
gral. 1. I a f($,y)dy)dzr
2. I = /:(/e: f(m,y)dy)drt
3. I = fly/16$ f(a:,y)dy)d:c 1114 4
4. I 2 f f(22, y)dy)ds: correct
. 0 CI 5. I = Aln4(£ey f(m,y)dy)d$
6. I x f04(/:x f(x1y)dy)dst Explanation:
The region of integration is the set of all
gnoints {($,y):0_<_m£lny,1§y§4} in the plane bounded by the y—axis and the
graphs of since y m em when .z = lny. This is the shaded
region in (not drawn to scale). Integration is taken first
with respect to y for ﬁxed m along the soiid
horizontal line. To change the order of integration, now ﬁx
:1: and let y vary along the solid vertical line in (not drawn to scale). Integration in y is along
the line from (33, 1:13;) to (w, 4) for fixed as,
and then from a: m 0 to cc m1n4. Consequently, after changing the order of
integration, Vincent (jrnv’???) w Review Exam?) w Gilbert  (57495) ‘ ll Consequently, 2.
keywords: integral, triple integral, re
peated integral, iinear fnnetion,polynomial
integrand, exponential integrand, monomial
integrand, evaluation of triple integral 020 (part 1 of 3) 10.0 points 3
The solid E in the ﬁrst octant of 3—spaee
Shown in ‘
4. is bounded by the parabolic cylinder 3; = 3:2 and the planes
23:31, mmz, 3x20. Which of the following shaded regions is the 5.
projection of E onto the say—mplane? ‘ correct Explanation: Since E lies in the first oetant, its projection
onto the my~p1ane lies in the ﬁrst quadrant of
the mymolane. On the other hand, the trace ...
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 Fall '09
 Gilbert

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