# Calculus: One and Several Variables

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
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>

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 34

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online