spring02prelim2sol - Spring 2002, Prelim 2 Solutions 1a....

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Unformatted text preview: Spring 2002, Prelim 2 Solutions 1a. Since a ( t ) = d v ( t ) dt , v ( t ) = Z a ( t ) dt = •Z µ 1- 2 t 2 ¶ dt ‚ i + •Z (cos πt ) dt ‚ j + •Z (sin πt ) dt ‚ k , = µ t + 2 t ¶ i + µ 1 π sin πt ¶ j- µ 1 π cos πt ¶ k + C 1 . The constant vector, C 1 , is computed from the initial velocity vector: v (1) = µ 1 + 2 1 ¶ i + µ 1 π sin π ¶ j + µ 1 π cos π ¶ k + C 1 , = 3 i + 1 π k + C 1 = i + 1 π k , ⇒ C 1 =- 2 i , ∴ v ( t ) = µ t + 2 t- 2 ¶ i + µ 1 π sin πt ¶ j- µ 1 π cos πt ¶ k . Since v ( t ) = d r ( t ) dt , r ( t ) = Z v ( t ) = •Z µ t + 2 t- 2 ¶ dt ‚ i + •Z µ 1 π sin πt ¶ dt ‚ j- •Z µ 1 π cos πt ¶ dt ‚ k , = µ 1 2 t 2 + 2 ln t- 2 t ¶ i- µ 1 π 2 cos πt ¶ j- µ 1 π 2 sin πt ¶ k + C 2 . The constant vector, C 2 , is computed from the initial position vector: r (1) = µ 1 2 + 2 ln 1- 2 ¶ i- µ 1 π 2 cos π ¶ j- µ 1 π 2 sin π ¶ k + C 2 , =- 3 2 i + 1 π 2 j + C 2 = 4 i + 2 j + k , ⇒ C 2 = 11 2 i + µ 2- 1 π 2 ¶ j...
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This note was uploaded on 02/12/2008 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell University (Engineering School).

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spring02prelim2sol - Spring 2002, Prelim 2 Solutions 1a....

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