261E1-F2011

261E1-F2011 - EXAM 1 FALL 2011 MATH 26100 Name Student ID...

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Unformatted text preview: EXAM 1 FALL 2011 MATH 26100 Name Student ID Recitation Instructor Recitation Section and Time INSTRUCTIONS: 1. This exam contains 11 problems each worth 9 points (one point free). 2. Please supply a_ll information requested above on the scantron. 3. Work only in the space provided, or on the backside of the pages. You must Show your work. 4 . Mark your answers clearly on the scantron. Also circle your choice for each problem in this booklet. 5. No books, notes, or calculators, please. Mark TEST 01 on your scantron! DE E <1: BEE/4 om; i I i g g ; L 1. Let C be the curve given by fit) = (4%, t, 5 —- 252) for t > 0. At what point does the tangent line to C at (4, 1, 4) intersect the my plane? 2. The arclen’gth of the curve FOE) : 2t§+ £254 (111W? for 2 g t g 4 is: 17 I B. 4+In2 C. 16+ln2 15 T E. 12+1n2 A. D. 3. A particle moves in space With acceleration EL’(t) : at]? and initial velocity and position given by 17(0) 3 5, HO) :2 3+ k. Where is the particle at time t = 2? A. (1,1,62) B. (0,1,8) C. (0,1,6—1) D. (1,1,e2—2) E. (0,1,62—2) 4. Suppose that z is defined implicitly as a function of a: and y by the equation eyz + Sin(7ryz) — myz = 0. What is the value of g; at (e,1, 1)? A. J 8 B. l e C, #1 71' D. 3 7r E. 1 e fivr 5. The surface area of a rectangular box is given by the function 5(50, y, z) = 22:31 + Zyz + 2xz Where 13,31, 2 are its sides. These are measured as so : 10 cm, y = 20 cm, x 30 cm ; with possible errors in measurements as much as 0.1 cm. Use differentials to estimate i the maximum error in the calculated surface area. . 12 cm2 . 24 cm2 A B C. 36 cm2 D E . 48 cm2 E r i ¥ i i 9' 3 6. Given if z (1, —1,2) and 5: (2,1,0), find 75 such that the vector ('2’: (5,73 — 1,2) is perpendicular to (i X b. l A. 75:1 B. H 0. 732—1 1 D. t=——2 E. 15:0 7. The intersection of the hyperbolic paraboloid 3:2 — yz — z — 1 = O with the yz—plane consists of a hyperbola and a parabola a hyperbole, an ellipse two lines $50573?” a parabola 8. Let fix, y) = V902 + y. The equation for the tangent plane to z = f(x, y) at (2,1) is 9. The critical points of f (x, y) = 351:3 + 3y3 + 1133y3 are: A. 2x/5z—4m—yzl B. 2x/gz—4x—y = 10 ‘ C. 2z—2m—y=l D. 22—2m—y=10 E. 2\/gz—2a:—y=9 A. (0,0),(1,—1) B. (0,0) C. (1,1) D. (0, 0),(—31/3, —31/3) E. (—31/3,—31/3),(1,1) 10. The directional derivative of the function f(a:, y) = 4:1:y + emy at the point (0,1) and in the direction of 17 = (3, —4) is: (5, 0) <3) *4) —15 magma?» 11. If then 5:): is: (3t 2/.— l 152 + '1}! “(3,73 = 75+ 52, 'U(3)t) =1n(t) —(1+%~> ' (t+32)2+1nt —<2<t + s?) + 11-3?) (t+32)2+lnt ~(2(t+ 32) + %) ' ((t+32)2+lnt)2 —(u+1) ' (um—v)? —(2u+1) tag-H) ...
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This note was uploaded on 12/09/2011 for the course MA 261 taught by Professor Stefanov during the Fall '08 term at Purdue.

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261E1-F2011 - EXAM 1 FALL 2011 MATH 26100 Name Student ID...

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