Business Calc Homework w answers_Part_83

Business Calc Homework w answers_Part_83 - Section 10.3 22....

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13. 1 E sec t tan t dt 2 i 1 1 E tan t dt 2 j 5 (sec t 1 C 1 ) i 1 (ln ) sec t ) 1 C 2 ) j 5 (sec t ) i 1 (ln ) sec t ) ) j 1 C 14. 1 E } 1 t } dt 2 i 1 1 E } 5 2 1 t } dt 2 j 5 (ln ) t ) 1 C 1 ) i 1 ( 2 ln ) 5 2 t ) 1 C 2 ) j 5 (ln ) t ) ) i 2 (ln ) 5 2 t ) ) j 1 C 15. r ( t ) 5 ( t 1 1) 3/2 i 2 e 2 t j 1 C , and r (0) 5 i 2 j 1 C 5 0 , so C 52 ( i 2 j ) i 1 j r ( t ) 5 (( t 1 1) 3/2 2 1) i 2 ( e 2 t 2 1) j 16. r ( t ) 5 1 } t 4 4 } 1 2 t 2 2 i 1 1 } t 2 2 } 2 j 1 C , and r (0) 5 C 5 i 1 j ,so r ( t ) 5 1 } t 4 4 } 1 2 t 2 1 1 2 i 1 1 } t 2 2 } 1 1 2 j . 17. } d d r t } 5 ( 2 32 t ) j 1 C 1 and r ( t ) 5 ( 2 16 t 2 ) j 1 C 1 t 1 C 2 . r (0) 5 C 2 5 100 i , and } d d r t } ) t 5 0 5 C 1 5 8 i 1 8 j . So r ( t ) 5 ( 2 16 t 2 ) j 1 (8 i 1 8 j ) t 1 100 i 5 (8 t 1 100) i 1 ( 2 16 t 2 1 8 t ) j . 18. } d d r t }52 t i 2 t j 1 C 1 , and r ( t ) 5 1 2} t 2 2 } 2 i 1 1 2} t 2 2 } 2 j 1 C 1 t 1 C 2 r (0) 5 C 2 5 10 i 1 10 j , and } d d r t } ) t 5 0 5 C 1 5 0 , so r ( t ) 5 1 2} t 2 2 } 2 i 1 1 2} t 2 2 } 2 j 1 (10 i 1 10 j ) 5 1 2} t 2 2 } 1 10 2 i 1 1 2} t 2 2 } 1 10 2 j 19. v ( t ) 5 (1 2 cos t ) i 1 (sin t ) j and a ( t ) 5 (sin t ) i 1 (cos t ) j . Solve v ? a 5 0: (sin t 2 sin t cos t ) 1 (sin t cos t ) 5 0 implies sin t 5 0, which is true for t 5 0, p ,or2 p . 20. v ( t ) 5 (cos t ) i 1 j , and a ( t ) 5 ( 2 sin t ) i . Solve v ? a 5 0: 2 sin t cos t 5 0, which is true for t 5 } k 2 p } , k any nonnegative integer. 21. v ( t ) 5 ( 2 3 sin t ) i 1 (4 cos t ) j , and a ( t ) 5 ( 2 3 cos t ) i 1 ( 2 4 sin t ) j . Solve v ? a 5 0: (9 sin t cos t ) 2 (16 sin t cos t ) 5 0, is true when sin t 5 0 or cos t 5 0, i.e., for t 5 } k 2 p } , k any nonnegative integer. 22. v ( t ) 5 ( 2 5 sin t ) i 1 (5 cos t ) j , and a ( t ) 5 ( 2 5 cos t ) i 1 ( 2 5 sin t ) j . Solve v ? a 5 0: (25 sin t cos t ) 1 ( 2 25sin t cos t ) 5 0, which is true for all values of t . 23. v ( t ) 5 ( 2 2 sin t ) i 1 (cos t ) j , and a ( t ) 5 ( 2 2 cos t ) i 1 ( 2 sin t ) j . So v 1 } p 4 } 2 5 ( 2 ˇ 2 w ) i 1 1 } ˇ 1 2 w } 2 j , and a 1 } p 4 } 2 5 ( 2 ˇ 2 w ) i 1 1 2} ˇ 1 2 w } 2 j . Then ) v ) 5 ) a ) 5 ! } 5 2 } § , v ? a 5 } 3 2 } , and u 5 cos 2 1 1 } v ) v ? )) a a ) } 2 5 cos 2 1 1 } 3 5 } 2 < 53.130 8 . 24. v ( t ) 5 3 i 1 (2 t ) j , and a ( t ) 5 2 j . So v (0) 5 3 i , and a (0) 5 2 j . These are perpendicular, i.e., the angle between them measures 90 8 . 25. (a) Both components are continuous at t 5 3, so the limit is 3 i 1 1 } 3 2 3 2 1 2 3( 9 3) } 2 j 5 3 i . (b) Continuous so long as t 2 1 3 t ± 0, i.e., t ± 0, 2 3 (c) Discontinuous when t 2 1 3 t 5 0, i.e., t 5 0 or 2 3 26. (a) Use L’Hôpital’s Rule for the i -component: lim t 0 1 } sin t 2 t } 2 i 1 lim t 0 (ln ( t 1 1)) j 5 lim t 0 1 } 2co 1 s2 t } 2 i 1 lim t 0 (ln ( t 1 1)) j 5 2 i 1 0 j 5 2 i . (b) Continuous so long as t ± 0 and t 1 1 . 0, i.e., ( 2 1, 0) < (0, ). (c) Discontinuous when t 5 0 or t 1 1 # 0, i.e., ( 2‘ 2 1] < {0}. 27. v ( t ) 5 (sin t ) i 1 (1 2 cos t ) j , i.e., } d d x t } 5 sin t , and } d d y t } 5 1 2 cos t Distance 5 E 2 p /3 0 ˇ (s w in w t w ) 2 w 1 w ( w 1 w 2 w c w o w s w t ) w 2 w dt 5 E 2 p /3 0 ˇ 2 w 2 w 2 w c w o w s w t w dt 5 E 2 p /3 0 2 sin 1 } 2 t } 2 dt 5 3 2 4 cos 1 } 2 t } 24 5 2 2 p /3 0 Section 10.3 411
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28. (a) r (0) 5 1 } 1 4 } e 0 2 0 2 i 1 ( e 0 ) j 5 } 1 4 } i 1 j , r (2) 5 1 } 1 4 } e 8 2 2 2 i 1 ( e 4 ) j Initial 5 1 } 1 4 } ,1 2 , terminal 5 1 } 1 4 } e 8 2 2, e 4 2 (b) v ( t ) 5 ( e 4 t 2 1) i 1 (2 e 2 t ) j ; } d d x t } 5 e 4 t 2 1, and } d d y t } 5 2 e 2 t .
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Business Calc Homework w answers_Part_83 - Section 10.3 22....

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