Fall 2005 Calc 3- Exam 1

Fall 2005 Calc 3- Exam 1 - MA 261 Exam 1 September 22, 2005...

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Unformatted text preview: MA 261 Exam 1 September 22, 2005 Name: Student ID Number: Lecturer: Recitation Instructor: Instructions: 1. This exam contains 12 problems worth 8 points. 2. Please supply all information requested above and on the mark—sense sheet. 3. Work only in the space provided, or on the backside of the pages. Mark your answers clearly on the scantron. Also circle your choice for each problem in this booklet. 4. No books, notes, or calculator, please. MA 261 EXAM 1 September 22, 2005 1. If M = (0,3,1) and N = (2, 3, —1) then a unit vector in the direction of MN is: A. (2,0, —2) B. (2,0,0) C. (—2,0,2) 2 2 D. —, 0,—— (\/§ \/§> 2 2 E. ———, 0, — ( \/§ «55) 2. The angle between the vectors 3?+ and I; is: 7r A. — 6 7r B. — 3 7r C. Z 7r D. — 2 E. 7r 3. Which of the following statements is true for all three—dimensional vectors (1', 1;, and E, if 0 is the angle between (i and b? (1)5x525x5 A. All are true B. (i), (ii), and (iii) only C. (i), (ii), and (iv) only D. (i), (iii) and (iv) only E. (ii) and (iv) only 4‘ The equation p = 2 cos HsinqS in spherical coordinates defines a sphere of radius R = .m .U .0 w .> PP Nn-lklI—ll—‘NJID—t 5. Find the equation of the plane that contains the points (1, 2, 1), (2, —1,0) and (3, 3, 1). A. B. $+2y+7z=12 —-$—2y+92=4 C. m—2y+7z:4 D. E. —:I:+2y+9,z:12 $+2y+z=6 6. Find a vector function F(t), 0 S t g 27r, that represents the curve of intersection of the cylinder 4:132 + 43/2 = 1 with the plane :1: + y + z = 2. A. F(t) = 2cos15§+ 2sintf B. 91.6.0 + (2 — 2cost — 2sint)E + (2 + 2cos15+ 2sint)E écos 1517+ %sin 155+ (2 — — s1n t)l€ 1 +( + ) +< — cos 1517+ % sin 15? 2 2 —¢ 2 — cost — sint)k 7. Find parametric equations for the tangent line to the curve F(t) = (t2 + 3t + 2, at cost, ln(t + 1)) at the point (2, 1,0). A. $=2+3t y=1+t z=t 13- 3, :t t—.t7 2—— x + y e(c0s s1n)z t+1 C. $=3+2t y=1+t 2:1 D. 113:3t y:2t z=1+t E. 93:2—t y=1+t 2:3—3t x/é 8. Find the arc—length of the curve defined by F(t) = (t, ——2— t2, t3), —1 S t S 1. A. 5 B. 4 C. 3 D. 2 E. 6 9. A particle starts at the origin with initial velocity i+ ;— Its acceleration is 6(t) z ti+i+ tic. Find its position at t = 1. 1s 1s 1» A.—' —' — 62+2J+3 k 7* 1~ 5a B.—' —'——k 6Z+2J 6 C. f+f+i£ 7a 3a 5a D.— —'——k 6Z+2J 6 E. {+2343 10. The level surface of f(:1:, y, z) = 3:2 + y2 — 22 corresponding to f(:1:, y, z) E 1 intersects the my-plane in a A. circle B. parabola C. ellipse D. hyperbola E. line 11. UL: lim ($,y,z)-+(0,010) $+2y—3z ‘hfl+y1+fi’ then A. L = B. = —-2 C. = —3 D. L = 0 E. the limit does not exist 12. If f (:5, y) = 111(302 + 2.142), then the partial derivative fzy equals 4mg (:52 + 2y2)2 4W2-y5 W ~8$y ($2 + 2y2)2 ($2 + 2y2)2 —2x (:52 + 2y2)2 ...
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This test prep was uploaded on 04/07/2008 for the course MA 261 taught by Professor Stefanov during the Spring '08 term at Purdue University-West Lafayette.

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Fall 2005 Calc 3- Exam 1 - MA 261 Exam 1 September 22, 2005...

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