SM221 Final Exam – May 7, 2012 – Answer Key Part I 1. c 6. c 11. d 16. a 2. a 7. b 12. a 17. d 3. b 8. c 13. b 18. d 4. e 9. c 14. b 19. b 5. b 10. a 15. a 20. b SM221 Final Exam – May 7, 2012 – Answer Key Part II 21. See Theorem 15, Section 14.6, p. 939 and Theorem 3, Section 16.5, p. 1092 of text. Part III 22. π (π₯, π¦) = π₯ 3 − 3π₯π¦ + π¦ 3. ππ₯ = 3π₯ 2 − 3π¦, ππ¦ = −3π₯ + 3π¦ 2 Setting ππ₯ = 0 and ππ¦ = 0 gives two critical points, (0,0) and (1,1). 2 The discriminant is π· = ππ₯π₯ ππ¦π¦ − ππ₯π¦ = (6π₯)(6π¦) − (−3)2 . At (0,0), we have π· = −9 < 0 so (0,0) is a saddle point of π. At (1,1), we have π· = 27 > 0 and ππ₯π₯ = 6 > 0 so π has a local minimum at (1,1). 23. οΏ½ 2π 0 2 οΏ½ οΏ½ 0 5 1+π 2 π ππ§ ππ ππ = 8π 24. a. π (π₯, π¦) = 3π₯π¦ + 2π¦ 2 . b. π (0,2) − π(1,0) = 8 − 0 = 8 c. π₯ = 1 − π‘, π¦ = 2π‘, 0 ≤ π‘ ≤ 1, ππ₯ = −ππ‘, ππ¦ = 2ππ‘ οΏ½ π β ππ = οΏ½ 3π¦ ππ₯ + (3π₯ + 4π¦)ππ¦ πΆ 1 πΆ 1 = ∫0 3(2π‘)(−1) + οΏ½3(1 − π‘) + 4(2π‘)οΏ½2 ππ‘ = ∫0 6 + 4π‘ ππ‘ = 8 25. π(π₯, π¦, π§) = 30 1+π₯ 2 +π¦ 2 +π§ 2 , ππ₯ = −60π₯ (1+π₯ 2 +π¦ 2 +π§2 )2 , with similar formulas for ππ¦ and ππ§ . The rate of change of temperature at the point (1,1,1) in the direction of 〈2, −2,1〉 is −60 1 5 1 〈1,1,1〉 β 〈2, −2,1〉 = − ∇π(1,1,1) β 〈2, −2,1〉 = 16 3 4 3 i.e., the temperature is decreasing at 1.25o C per meter. 26. a. b. ∫πΆ 2π¦ ππ₯ − 2π₯ ππ¦ 2π 5 =β¬interior of πΆ −4 ππ΄ = ∫0 ∫0 −4π ππ ππ = −4(Area of circle of radius 5)= -100π. c. No, π is not conservative because the integral of π around the closed curve πΆ is not zero. 27. The mass of the wire in grams is π π π = ∫0 0.1√1 + π‘ 2 οΏ½(−2 sin(π‘))2 + (2 cos(π‘))2 + 12 ππ‘ = ∫0 0.1√1 + π‘ 2 √5 ππ‘ ≅ 1.366. 28. The mass of the solid in grams is 2π π 3 π = ∫0 ∫02 ∫0 2π π2 sin(π) ππ ππ ππ = 81π. 29. Let π be the given surface. A parametrization of π is π(π₯, π¦) = 〈π₯, π¦, 3 − π₯ − π¦〉, for (π₯, π¦) in the region π· given by 0 ≤ π₯ ≤ 3 and 0 ≤ π¦ ≤ 3 − π₯. ππ₯ = 〈1,0, −1〉, ππ¦ = 〈0,1, −1〉, ππ₯ × ππ¦ = 〈1,1,1〉 (which is upward). 3 3−π₯ β¬π π β ππΊ = ∫0 ∫0 3 3−π₯ 〈π¦, 2π₯, 3 − π₯ − π¦〉 β 〈1,1,1〉 ππ¦ ππ₯ = ∫0 ∫0 3 + π₯ ππ¦ ππ₯ = 18. 30. Let π· be the region enclosed by π. By the Divergence Theorem, the charge π is 2 2 2 π0 βπ· div π¬ ππ = π0 ∫0 ∫0 ∫0 2 + 1 + 3 ππ§ ππ¦ ππ₯ = 48 π0 .