Calculus: One and Several Variables

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-14 JWDD027-Salas-v1 November 30, 2006 13:43 724 SECTION 14.1 CHAPTER 14 SECTION 14.1 1. f ± ( t )=2 i j +3 k2 . f ± ( t ) = sin t k 3. f ± ( t )= 1 2 1 t i 1 2 1+ t j + 1 (1 t ) 2 k4 . f ± ( t 1 t 2 ( i j ) 5. f ± ( t ) = cos t i sin t j + sec 2 t k6 . f ± ( t e t ± i +(1+ t ) j +(2 t + t 2 ) k ² 7. f ± ( t 1 1 t i sin t j +2 t k8 . f ± ( t e t ( i j ) 2 e 2 t ( j k ) 9. f ± ( t )=4 i +6 t 2 j t +2) k ; f ±± ( t )=12 t j k 10. f ± ( t ) = (sin t + t cos t ) i + (cos t t sin t ) k f ±± ( t ) = (2cos t t sin t ) i +( 2sin t t cos t ) k 11. f ± ( t 2sin2 t i + 2cos2 t j +4 t k ; f ±± ( t 4cos2 t i 4sin2 t j 12. f ± ( t 1 2 t 1 / 2 i + 3 2 t 1 / 2 j + 1 t k f ±± ( t 1 4 t 3 / 2 i + 3 4 t 1 / 2 j 1 t 2 k 13. (a) r ± ( t 2 te t 2 i e t j ; r ± (0) = j (b) r ± ( t ) = cot t i tan t j + (2cos t + 3sin t ) k ; r ± ( π/ 4) = i j + 5 2 k 14. (a) r ±± ( t ( 2 e t 4 te t + t 2 e t ) i 2 e t + te t ) j ; r ±± (0) = 2 i 2 j (b) r ±± ( t 1 t i + 2 2ln t t 2 j 1+2ln t 4 t 2 (ln t ) 3 / 2 k ; r ±± ( e e 1 i 3 4 e 2 k 15. ³ 2 1 ( i t j ) dt = ± t i + t 2 j ² 2 1 = i j 16. ³ π 0 (sin t i + cos t j + t k ) dt = ´ cos t i + sin t j + t 2 2 k µ π 0 =2 i + 1 2 π 2 k 17. ³ 1 0 ( e t i + e t k ) dt = ± e t i e t k ² 1 0 =( e 1) i + 1 1 e · k 18. ³ 1 0 ( te t i e 2 t j + e t k ) dt = ± ( te t e t ) i e 2 t j e t k ² 1 0 e 1 i e 2 j e 1 k 19. ³ 1 0 1 t 2 i + sec 2 t j · dt = ± tan 1 t i + tan t j ² 1 0 = π 4 i + tan(1) j
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-14 JWDD027-Salas-v1 November 30, 2006 13:43 SECTION 14.1 725 20. ± 3 1 ² 1 t i + ln t t j + e 2 t k ³ dt = ´ ln t i + 1 2 (ln t ) 2 j 1 2 e 2 t k µ 3 1 =ln3 i + 1 2 (ln3) 2 j + 1 2 ( e 2 e 6 ) k 21. lim t 0 f ( t )= ² lim t 0 sin t 2 t ³ i + lim t 0 e 2 t · j + ² lim t 0 t 2 e t ³ k = 1 2 i + j 22. Does not exist ( because of t | t | k ) 23. lim t 0 f ( t lim t 0 t 2 · i + ² lim t 0 1 cos t 3 t ³ j + ² lim t 0 t t +1 ³ k =0 i + 1 3 ² lim t 0 1 cos t t ³ j +0 k = 0 24. lim t 0 f ( t j + k 25. (a) ± 1 0 te t i + te t 2 j · dt = i + e 1 2 j (b) ± 8 3 ² t t i + t ( t +1) 2 j + t ( t 3 k ³ dt = ¸ 5+ln ( 4 9 i + ¸ 5 36 +ln ( 4 9 j + 295 2592 k 26. (a) 3 4 i + 1 4 j + k (b) 2 e 2 i +2 e 4 j k 27. 28. 29. 30. 31. 32. 33. 34. 35.
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-14 JWDD027-Salas-v1 November 30, 2006 13:43 726 SECTION 14.1 36. 37. 38. 2 π 4 π 6 π x 1 2 y 39. (a) f ( t ) = 3cos t i + 2sin t j (b) f ( t ) = 3cos t i 2sin t j 40. (a) f ( t ) = (1 + cos t ) i + sin t j (b) f ( t ) = (1 + cos t ) i sin t j 41. (a) f ( t )= t i + t 2 j (b) f ( t t i + t 2 j 42. (a) f ( t t i + t 3 j (b) f ( t t i t 3 j 43. f ( t )=(1+2 t ) i +(4+5 t ) j +( 2+8 t ) k , 0 t 1 44. f ( t )=(3+4 t ) i +2 j 5+14 t ) k , 0 t 1 45. f ± ( t 0 i + m j , ± b a f ( t ) dt = ² 1 2 t 2 i ³ b a + ²± b a f ( t ) dt ³ j = 1 2 ( b 2 a 2 ) i + A j , ± b a f ± ( t ) dt =[ t i + f ( t ) j ] b a =( b a ) i d c ) j 46. f ( t ´ 1 2 t 2 +1 µ i 1+ t 2 +1) j te t e t +4) k 47. f ± ( t i + t 2 j f ( t )=( t + C 1 ) i + ( 1 3 t 3 + C 2 ) j + C 3 k f (0) = j k = C 1 =0 ,C 2 =1 3 = 1 f ( t t i + ( 1 3 t 3 ) j k 48. f ( t e 2 t i e 2 t k = e 2 t ( i k ) 49. f ± ( t α f ( t f ( t e αt f (0) = e αt c 50. For each ±> 0 there exists δ>
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This note was uploaded on 04/30/2011 for the course MATH 1431 taught by Professor Any during the Spring '08 term at University of Houston.

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ch14[1] - P1: PBU/OVY JWDD027-14 P2: PBU/OVY...

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