14.2Ex46-57

# 14.2Ex46-57 - 478 C H A P T E R 14 C A L C U L U S O F VE C...

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478 CHAPTER 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) In Exercises 46–53, fnd the general solution r ( t ) oF the diFFerential equation and the solution with the given initial condition. 46. d r dt = h 1 2 t , 4 t i , r ( 0 ) = h 3 , 1 i SOLUTION We frst fnd the general solution by integrating d r : r ( t ) = Z h 1 2 t , 4 t i = ¿ Z ( 1 2 t ) , Z 4 tdt À = - t t 2 , 2 t 2 ® + c (1) Since r ( 0 ) = h 3 , 1 i ,wehave: r ( 0 ) = - 0 0 2 , 2 · 0 2 ® + c = h 3 , 1 i ⇒ c = h 3 , 1 i Substituting in (1) gives the solution: r ( t ) = - t t 2 , 2 t 2 ® + h 3 , 1 i = - t 2 + t + 3 , 2 t 2 + 1 ® 47. r 0 ( t ) = i j , r ( 0 ) = 2 i + 3 k The general solution is obtained by integrating r 0 ( t ) : r ( t ) = Z ( i j ) = ±Z 1 i ±Z 1 j = t i t j + c (1) Hence, r ( 0 ) = 0 i 0 j + c = c The solution with the initial condition r ( 0 ) = 2 i + 3 k must satisFy: r ( 0 ) = c = 2 i + 3 k Substituting in (1) yields the solution: r ( t ) = t i t j + 2 i + 3 k = ( t + 2 ) i t j + 3 k 48. r 0 ( t ) = t 2 i + 5 t j + k , r ( 0 ) = j + 2 k We frst fnd the general solution by integrating r 0 ( t ) : r ( t ) = Z ³ t 2 i + 5 t j + k ² = ±Z t 2 i + ±Z 5 j + ±Z 1 k = ± 1 3 t 3 i + ± 5 2 t 2 j + t k + c (1) The solution which satisfes the initial condition must satisFy: r ( 0 ) = ± 1 3 · 0 3 i + ± 5 2 · 0 2 j + 0 · k + c = j + 2 k That is, c = j + 2 k Substituting in (1) gives the Following solution: r ( t ) = ± 1 3 t 3 i + ± 5 2 t 2 j + t k + j + 2 k = ± 1 3 t 3 i + Ã 5 t 2 2 + 1 ! j + ( t + 2 ) k 49. r 0 ( t ) = h sin 3 t , sin 3 t , t i , r ( 0 ) = h 0 , 1 , 8 i We frst integrate the vector r 0 ( t ) to fnd the general solution: r ( t ) = Z h sin 3 t , sin 3 t , t i = ¿ Z sin 3 , Z sin 3 , Z À = ¿ 1 3 cos 3 t , 1 3 cos 3 t , 1 2 t 2 À + c (1) Substituting the initial condition we obtain: r ( 0 ) = ¿ 1 3 cos 0 , 1 3 cos 0 , 1 2 · 0 2 À + c = h 0 , 1 , 8 i = ¿ 1 3 , 1 3 , 0 À + c = h 0 , 1 , 8 i

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SECTION 14.2 Calculus of Vector-Valued Functions (ET Section 13.2) 479 Hence, c = h 0 , 1 , 8 i − ¿ 1 3 , 1 3 , 0 À = ¿ 1 3 , 4 3 , 8 À Substituting in (1) we obtain the solution: r ( t ) = ¿ 1 3 cos 3 t , 1 3 cos 3 t , 1 2 t 2 À + ¿ 1 3 , 4 3 , 8 À = ¿ 1 3 ( 1 cos 3 t ) , 1 3 ( 4 cos 3 t ) , 8 + 1 2 t 2 À 50.
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14.2Ex46-57 - 478 C H A P T E R 14 C A L C U L U S O F VE C...

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