# s1109 - t 2 t 1 dt X i ∂L ∂q i-d dt ∂L ∂ ˙ q i δq...

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ADVANCED DYNAMICS — PHY 4241/5227 HOME AND CLASS WORK – SET 2 Solution for assignment 9. Deviation of the Euler-Lagrange equations from the least action principle in general coordinates q i , ˙ q i , i = 1 , . . . n : 0 = δ Z t 2 t 1 dt L ( q i , ˙ q i , t ) = Z t 2 t 1 dt { L ( q i + δq i , ˙ q i + δ ˙ q i , t ) - L ( q i , ˙ q i , t ) } = Z t 2 t 1 dt ( L ( q i , ˙ q i , t ) + X i ∂L ∂q i δq i + X i ∂L ˙ q i δ ˙ q i - L ( q i , ˙ q i , t ) ) = Z t 2 t 1 dt ( X i ∂L ∂q i δq i + X i ∂L ˙ q i d dt δq i ) = Z t 2 t 1 dt X i ( ∂L ∂q i - d dt ∂L ˙ q i ) δq i + " ∂L ˙ q i δq i # t 2 t 1 = Z
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Unformatted text preview: t 2 t 1 dt X i ( ∂L ∂q i-d dt ∂L ∂ ˙ q i ) δq i . The last equality holds because of δq i ( t 1 ) = δq i ( t 2 ) = 0. As the variations are inde-pendent, the Fnally obtained relation is equivalent to ∂L ∂q i-d dt ∂L ∂ ˙ q i = for i = 1 , . . . , n ....
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## This note was uploaded on 11/10/2011 for the course PHY 4241 taught by Professor Berg during the Spring '11 term at University of Florida.

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