14.2%20Ex%2053-57

14.2%20Ex%2053-57 - 480 C H A P T E R 14 C A L C U L U S O...

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480 CHAPTER 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) r ( 0 ) = h 0 , 0 , 0 i + c 1 · 0 + c 2 = h 2 , 1 , 1 i c 2 = h 2 , 1 , 1 i Combining with (2) yields the solution: r ( t ) = ¿ 0 , 0 , 1 2 t 2 À + h 3 , 1 , 1 i t + h 2 , 1 , 1 i = ¿ 2 + 3 t , 1 + t , 1 + t + 1 2 t 2 À 53. r 00 ( t ) = - e t , sin t , cos t ® , r ( 0 ) = h 1 , 0 , 1 i , r 0 ( 0 ) = h 0 , 2 , 2 i SOLUTION We perform integration componentwise on r 00 ( t ) to obtain: r 0 ( t ) = Z - e t , sin t , cos t ® dt = - e t , cos t , sin t ® + c 1 (1) We now integrate r 0 ( t ) to obtain the general solution: r ( t ) = Z (- e t , cos t , sin t ® + c 1 ) = - e t , sin t , cos t ® + c 1 t + c 2 (2) Now, we substitute the initial conditions r ( 0 ) = h 1 , 0 , 1 i and r 0 ( 0 ) = h 0 , 2 , 2 i into (1) and (2) and solve for the vectors c 1 and c 2 . We obtain: r 0 ( 0 ) = h 1 , 1 , 0 i + c 1 = h 0 , 2 , 2 i c 1 = h− 1 , 3 , 2 i r ( 0 ) = h 1 , 0 , 1 i + c 2 = h 1 , 0 , 1 i c 2 = h 0 , 0 , 2 i Finally we combine the above to obtain the solution: r ( t ) = - e t , sin t , cos t ® + h− 1 , 3 , 2 i t + h 0 , 0 , 2 i = - e t t , sin t + 3 t , cos t + 2 t + 2 ® 54. Show that w ( t ) = h sin ( 3 t + 4 ), sin ( 3 t 2 ), cos 3 t i satis±es the differential equation w 00 ( t ) =− 9 w ( t ) .
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This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

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14.2%20Ex%2053-57 - 480 C H A P T E R 14 C A L C U L U S O...

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