Exam 3 - Version 070 m Exam03 »- Gilbert ~ (57495) 1 This...

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Unformatted text preview: Version 070 m Exam03 »- Gilbert ~ (57495) 1 This print—out should have 12 questions. Multiplechoice questions may continue on the next column or page — find all choices before answering. 001 10.0 points Locate and classify all the local extrema of I‘f(a:,y) m 333~y3—-322y+2. 1. local min at (m1, 1), saddle point at (8, 0) 2. local min at (0, 0'), saddle point at (-1, I) 3. local max at (0, 0), saddle point at (~1', 1) 4. local max at (—1, 1), saddle point at (9, 0) correct 5. local min at ("1, 1), local max at (0, 0) Explanation: Since f has derivatives everywhere, the crit— ical points occur at the solutions of Vflflia y) : fssi + .ny “we” 0- But fm m 0 when . 6f 5; m 3:52m3y = 0, i.€., y = 3:2, while fy : 0 when gui- = w3y2m3a: = 0, 16,33 2 —y2 Substituting the first into the second yields 2: m —(a:2)2 x "334, which can be written as sc(1+:l:3) = 0, m, a: = 0, m1. ' Thus, the critical points are (0, 0), (—1, l). and to classify these critical points we use the Second Derivative test. Now Jam: m 633: fyy m “61’: m “3- But then, at (0, 0), A m 'fmm, 0) e 0, s m 11¢,th ~3, ' while 0 == fyy(0,0) = 0. Consequently, D : $10—82 e —9 < 0. and so there is a saddle point at (0, 0) .' On the other hand, at (~1, 1), A:f$$(wli 1):“"63 B=fxylmla DEWB: while C‘ : fyy(—~1,1) = “6. Thus D = ACWB2 = 27 > c. and so, since A, C < 0, there is a local maximum at (—1, I) . keywords: 002 10.0 points The paraboloid 1 z = fine? +312) is shown in Version'O'F'O ~ ExemOB ~ Giibert “— (57495) 3 When v = 4i—3j therefore, the graph of f has at P in the direction of v. keywords: 7 004 10.0 points Find the J ecobian of the transformation T: (fit?) me (3:,y) when m=efcoséh y H m ~4€2T cos 29 W462?” ll 4. m 482? cos 28 ——3 cos 29 Explanation: For general functions the Jacobisn of the transformation 36'"? sin 6 . 3 cos 26 correct T: (u, U) "+ 0): ylui 0)) is defined by Qt 8(u, ’0) 6y But when a: : creosfl, y = 3eWTsin6. then by the Product Rule, 823 T 8:1: ,, , a? m e c036, 5mg — we 81119, while as _ _1« i as _ MT 5—; — ~36 51:19, 5(9- _— 35 0089. In this case, am, y) _ ercosfl we? sin6 3(7": a) — ——3e”7" sinfl 3e” costl m 313Te""7"(cos2 6 w sin2 3) . Consequently, aim, y) 6&1 9) —— 3cos26 because ere"? r: 1, c0826 = cosgélwsin26. keywords: 005 10.0 points - A rectangular box with no top is to be constructeci having a. volume of 4 cubic inches. What length 3;, in inches, wiil minimize the amount of material to be used in its construction. (Assume no material is wasted.) 1. length :2 2inches correct Version 070 —- Exam03 m Gilbert m (57495) 5 we see that I can be written as a repeated integral 1 m A4(f03(/02($wy)dz)dy)dm. Now 3 2 fowmyldzfi [(xwlzloeflwwy), while Thus 4 I :2 f9 (633~—9)d.’13 — l r—l 0.3 55-3 M i CD a :_.1 Consequently, I348—36m12. keywords: integral, triple integral, repeated integral, linear function,integrai over a speci— fied volume, limits of integration, polynomial integrand, exponential integrand, evaluation of tripie integral 007 10.0 points Use Lagrange Multipliers to determine the minimum value of fins, y) m 21:2 + 392 +1 subject to the constraint May) = $2+yg~1 e 0- 1. min value = 1 2. min value m 2correct 3. min vaiue = 4 ii ca 4. min value ii (:0 5. min value Explanation: The extreme values of f subject to the con— straint g m 0 occur at the solutions of (WWI, y) e A(V9)(mi y), 9'02, 3;) e 0- NOW ‘ (WW; :1) =1 (411?, 2:1), (WW: 3;) = (2%, 21!)- Thus 43: m ZASC, 2y 2 ZAy, and so 2:5(Am2) =e, 2y()\~ 1) =0. From the first equation, it follows that either mm0mX=2 (i) If A = 2, then y z 0 by the second equation, and so the constraint equation gives g($,0)=m2ml : 0, m, $=i1. i (ii) However, if m m 0, then by the con— straint equation, 963,31) = y2-1 ii 0, i.e.,y==i1. Consequently, the critical points are (i1,0), (flit). Since f(:l:1, O) m 3, f(0, i1) = 2, we thus see that f has minimum value 2 2 subject to the constraint 9(3):, y) x 0. keywords: 008 10.0 points Version 070 — Exemfl?) — Gilbert M (57495) 7 in the plane bounded by the xeaxis end the graphs of (not drawn to scale). Integration is taken first with respect to'y‘for fixed :1: along the dashed vertical line. To change the order of integration, first fix 3; in the interval [0, 4] and let x vary along the solid horizontal line. Then a: varies from 92 to 16. Hence, after changing the order of integration, keywords: double integral, reverse order inte— gration, perebole, 810 10.0 points In the contour map below identify the points P, Q, and R as local minima, local maxima, or neither. A. local minimum at R, B. local maximum at Q, C. local minimflm at P. 1. all of them 2. A and C only 3. l3 and C only 4. IA and B only 5. C only 6. B only 7. A only correct 8. none of them Explanation: A. TRUE: the contours near R are closed curves enclosing R and the contours decrease in value as we appeoach R. So the surface has a local minimum at R. B. FALSE: the point Q lies on the 0— contour and this contour divides the region near Q into two regions. In one region the contours have values increasing to 0, while in the other the contours have values decreasing to 0. So the surface does not have a local minimum at Q. C. FALSE: the contours near P are closed curves enclosing P and the contours increase Version {)70 w Exan103 ~ Gilbert w (57495) by converting to polar coordinates. 1. I flew w 1) correct 2. I = 27r(e4——i) 3. I m Airflow“ i) 4. I : 27r(616- 1) 5. I = «(ell—l) 6. I : 47r(e4—1) Explanation: In Cartesian coordinates the region of inte— gration is {'(aégy) : OSysV16ww2,Osm‘s4}, which is the shaded region in The graphic shows that in polar coordinates the region of integration is {(r,8):0£r§4,0$05n/2}. Thus in polar coordinates 4 7r/2 2 I 2 46?" ngdi" 0 0 4 16 2 = 27:] re?" d?" = n/ eudu, 0 :0 setting u m: r2. Consequently, I 2 flow m— 1) ...
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This note was uploaded on 03/24/2010 for the course M 57500 taught by Professor Gilbert during the Fall '09 term at University of Texas at Austin.

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Exam 3 - Version 070 m Exam03 »- Gilbert ~ (57495) 1 This...

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