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kof9595995
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I tried to solve the equation of catenary by variational method the other day. The integral we want to minimize is the potential energy:
[tex]U = \int_{{x_2}}^{{x_1}} {\rho gy\sqrt {1 + y{'^2}} } dx[/tex]
Then I got stuck at the constraint problem, and in this book,page55:http://books.google.com.sg/books?id...epage&q=catenary, variational method&f=false"
It used the fact that the total length of the rope is constant:
[tex]l = \int_{{x_2}}^{{x_1}} {\sqrt {1 + y{'^2}} } dx[/tex]
Then it just made use of lagrange multipliers,change the integral to
[tex]U + \lambda l = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{'^2}} } + \lambda \sqrt {1 + y{'^2}} )dx[/tex]
and minimized it.
I got confused here, when I learned lagrange multipliers in Lagrangian mechanics, if we want to minimize the action integral
[tex]I = \int_{{t_1}}^{{t_2}} L dt[/tex]
with a constraint g(x,y...)=constant, we change the integral to
[tex]I = \int_{{t_1}}^{{t_2}} {(L} + \lambda g)dt[/tex],
e.g.,the constraint's equation goes inside the integral you want to minimize, so why not in this catenary problem change the integral into
[tex]{U^*} = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{'^2}} } + \lambda l)dx = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{'^2}} } + \lambda \int_{{x_1}}^{{x_2}} {\sqrt {1 + y{'^2}} dx} )dx[/tex]
And if I do it in this way, I can't see how to solve it.
(edit: typo in the title: equation of catenary )
[tex]U = \int_{{x_2}}^{{x_1}} {\rho gy\sqrt {1 + y{'^2}} } dx[/tex]
Then I got stuck at the constraint problem, and in this book,page55:http://books.google.com.sg/books?id...epage&q=catenary, variational method&f=false"
It used the fact that the total length of the rope is constant:
[tex]l = \int_{{x_2}}^{{x_1}} {\sqrt {1 + y{'^2}} } dx[/tex]
Then it just made use of lagrange multipliers,change the integral to
[tex]U + \lambda l = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{'^2}} } + \lambda \sqrt {1 + y{'^2}} )dx[/tex]
and minimized it.
I got confused here, when I learned lagrange multipliers in Lagrangian mechanics, if we want to minimize the action integral
[tex]I = \int_{{t_1}}^{{t_2}} L dt[/tex]
with a constraint g(x,y...)=constant, we change the integral to
[tex]I = \int_{{t_1}}^{{t_2}} {(L} + \lambda g)dt[/tex],
e.g.,the constraint's equation goes inside the integral you want to minimize, so why not in this catenary problem change the integral into
[tex]{U^*} = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{'^2}} } + \lambda l)dx = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{'^2}} } + \lambda \int_{{x_1}}^{{x_2}} {\sqrt {1 + y{'^2}} dx} )dx[/tex]
And if I do it in this way, I can't see how to solve it.
(edit: typo in the title: equation of catenary )
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