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Review Exam 3

# Review Exam 3 - Vincent(jmv m Review ExamS Gilbert...

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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 = 6-73: 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°(1-cos£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 = ln-y. 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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