Chapt_09_1 - 9 Vector Calculus EXERCISES 9.1 Vector...

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9 9 Vector Calculus EXERCISES 9.1 Vector Functions 1. 2. 3. 4. 5. 6. 7. 8. 9. Note: the scale is distorted in this graph. For t = 0, the graph starts at (1 , 0 , 1). The upper loop shown intersects the xz -plane at about (286751 , 0 , 286751). 438
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9.1 Vector Functions 10. 11. x = t , y = t , z = t 2 + t 2 =2 t 2 ; r ( t )= t i + t j +2 t 2 k 12. x = t , y t , z = ± t 2 +4 t 2 +1= ± 5 t 2 1; r ( t t i t j ± 5 t 2 1 k 13. x = 3cos t , z =9 9cos 2 t = 9sin 2 t , y = 3sin t ; r ( t ) = 3cos t i + 3sin t j + 9sin 2 t k 14. x = sin t , z =1, y = cos t ; r ( t ) = sin t i + cos t j + k 439
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9.1 Vector Functions 15. r ( t )= sin2 t t i +( t 2) 5 j + ln t 1 /t k . Using L’Hˆ opital’s Rule, lim t 0 + r ( t · 2cos2 t 1 i t 2) 5 j + 1 /t 1 /t 2 k ¸ =2 i 32 j . 16. (a) lim t α [ 4 r 1 ( t )+3 r 2 ( t )] = 4( i 2 j + k ) + 3(2 i +5 j +7 k )=2 i +23 j +17 k (b) lim t α r 1 ( t ) · r 2 ( t )=( i 2 j + k ) · (2 i j k 1 17. r 0 ( t 1 t i 1 t 2 j ; r 0 ( t 1 t 2 i + 2 t 3 j 18. r 0 ( t h− t sin t, 1 sin t i ; r 0 ( t h− t cos t sin t, cos t i 19. r 0 ( t h 2 te 2 t + e 2 t , 3 t 2 , 8 t 1 i ; r 0 ( t h 4 te 2 t +4 e 2 t , 6 t, 8 i 20. r 0 ( t t i +3 t 2 j + 1 1+ t 2 k ; r 0 ( t i +6 t j 2 t (1 + t 2 ) 2 k 21. r 0 ( t 2sin t i + 6cos t j r 0 ( π/ 6) = i 3 j 22. r 0 ( t )=3 t 2 i +2 t j r 0 ( 1) = 3 i 2 j 23. r 0 ( t j 8 t (1 + t 2 ) 2 k r 0 (1) = j 2 k 24. r 0 ( t 3sin t i + 3cos t j k r 0 ( 4) = 3 2 2 i + 3 2 2 j k 25. r ( t t i + 1 2 t 2 j + 1 3 t 3 k ; r (2) = 2 i j + 8 3 k ; r 0 ( t i + t j + t 2 k ; r 0 (2) = i j k Using the point (2 , 2 , 8 / 3) and the direction vector r 0 (2), we have x =2+ t , y =2+2 t , z =8 / 3+4 t . 26. r ( t t 3 t ) i + 6 t t +1 j +(2 t +1) 2 k ; r (1) = 3 j +9 k ; r 0 ( t )=(3 t 2 1) i + 6 ( t 2 j +(8 t +4) k ; r 0 (1) = 2 i + 3 2 j +12 k . Using the point (0 , 3 , 9) and the direction vector r 0 (1), we have x t , y =3+ 3 2 t , z =9+12 t . 27. d dt [ r ( t ) × r 0 ( t )] = r ( t ) × r 0 ( t )+ r 0 ( t ) × r 0 ( t r ( t ) × r 0 ( t ) 28. d dt [ r ( t ) · ( t r ( t ))] = r ( t ) · d dt ( t r ( t )) + r 0 ( t ) · ( t r ( t )) = r ( t ) · ( t r 0 ( t r ( t )) + r 0 ( t ) · ( t r ( t )) = r ( t ) · ( t r 0 ( t )) + r ( t ) · r ( t r 0 ( t ) · ( t r ( t )) = 2 t ( r ( t ) · r 0 ( t )) + r ( t ) · r ( t ) 440
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9.1 Vector Functions 29. d dt [ r ( t ) · ( r 0 ( t ) × r 0 ( t ))] = r ( t ) · d dt ( r 0 ( t ) × r 0 ( t )) + r 0 ( t ) · ( r 0 ( t ) × r 0 ( t )) = r ( t ) · ( r 0 ( t ) × r 0 ( t )+ r 0 ( t ) × r 0 ( t )) + r 0 ( t ) · ( r 0 ( t ) × r 0 ( t )) = r ( t ) · ( r 0 ( t ) × r 0 ( t )) 30. d dt [ r 1 ( t ) × ( r 2 ( t ) × r 3 ( t ))] = r 1 ( t ) × d dt ( r 2 ( t ) × r 3 ( t )) + r 0 1 ( t ) × ( r 2 ( t ) × r 3 ( t )) = r 1 ( t ) × ( r 2 ( t ) × r 0 3 ( t r 0 2 ( t ) × r 3 ( t )) + r 0 1 ( t ) × ( r 2 ( t ) × r 3 ( t )) = r 1 ( t ) × ( r 2 ( t ) × r 0 3 ( t )) + r 1 ( t ) × ( r 0 2 ( t ) × r 3 ( t )) + r 1 ( t ) × ( r 2 ( t ) × r 3 ( t )) 31. d dt h r 1 (2 t r 2 ³ 1 t ´i =2 r 0 1 (2 t ) 1 t 2 r 0 2 ³ 1 t ´ 32. d dt [ t 3 r ( t 2 )] = t 3 (2 t ) r 0 ( t 2 )+3 t 2 r ( t 2 )=2 t 4 r 0 ( t 2 t 2 r ( t 2 ) 33. Z 2 1 r ( t ) dt = · Z 2 1 tdt ¸ i + · Z 2 1 3 t 2 dt ¸ j + · Z 2 1 4 t 3 dt ¸ k = 1 2 t 2 ¯ ¯ ¯ 2 1 i + t 3 ¯ ¯ ¯ 2 1 j + t 4 ¯ ¯ ¯ 2 1 k = 3 2 i +9 j +15 k 34. Z 4 0 r ( t ) dt = · Z 4 0 2 t +1 dt ¸ i + · Z 4 0 ¸ j + · Z 4 0 sin πtdt ¸ k = 1 3 (2 t +1) 3 / 2 ¯ ¯ ¯ ¯ 4 0 i 2 3 t 3 / 2 ¯ ¯ ¯ ¯ 4 0 j 1 π cos πt ¯ ¯ ¯ ¯ 4 0 k = 26 3 i 16 3 j 35. Z r ( t ) dt = · Z te t dt ¸ i + · Z e 2 t dt ¸ j + · Z te t 2 dt ¸ k =[ te t e t + c 1 ] i + h 1 2 e 2 t + c 2 i j + h 1 2 e t 2 + c 3 i k = e t ( t 1) i + 1 2 e 2 t j + 1 2 e t 2 k + c , where c = c 1 i + c 2 j + c 3 k .
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This note was uploaded on 02/17/2010 for the course MATHEMATIC MAS201 taught by Professor Xingqin during the Spring '10 term at Korea Advanced Institute of Science and Technology.

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Chapt_09_1 - 9 Vector Calculus EXERCISES 9.1 Vector...

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