challenge_problem_3.20100309.4b9712050dcd61.21432083

challenge_problem_3.20100309.4b9712050dcd61.21432083 - ,...

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Challenge problem Harry Dankowicz Mechanical Science and Engineering University of Illinois at Urbana-Champaign danko@illinois.edu 1. Consider the functional F ( x;y;z ) = Z t 2 t 1 p _ x 2 + _ y 2 + _ z 2 c ( t ) dt under the constraint n 1 x + n 2 y + n 3 z ¡ t = 0 for some constant scalars n 1 , n 2 , and n 3 . Show that an extremum is obtained for x = x ¤ , y = y ¤ , and z = z ¤ provided that (_ x ¤ ; _ y ¤ ; _ z ¤ ) ¢ ( n 1 ;n 2 ;n 3 ) = 1 and (_ x ¤ ; _ y ¤ ; _ z ¤ ) £ ( n 1 ;n 2 ;n 3 ) = ( k 1 ;k 2 ;k 3 ) c ( t ) ¡ _ x ¤ 2 + _ y ¤ 2 + _ z ¤ 2 ¢ for some constant scalars k 1 , k 2 , and k 3 . Interpret the meaning of the functional. 2. Consider the functional F ( x;y;z ) = Z t 2 t 1 p _ x 2 + _ y 2 + _ z 2 dt under the constraint that x 2 + y 2 + z 2 ¡ R 2 = 0 : Show that an extremum is obtained for x = x ¤ , y = y ¤
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Unformatted text preview: , and z = z provided that k 1 x + k 2 y + k 3 z = 0 for some constant scalars k 1 , k 2 , and k 3 . Interpret the meaning of the functional. 3. Derive the equations of motion for a particle in three-dimensional space, for which the constraint _ xz + _ y = 0 holds. Compare these to the formulation that results from nding an extremum to the action integral in the presence of this constraint. Reference material: Study sheet { Calculus of Variations. 1...
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