Calculus o f Variation s 9 Now we can apply the fundamental lemm a to conclude that th e Euler - Lagrange equatio n i n this cas e consist s o f the two equation s ( , X ' V = 0 an d ( , V ' V = 0. The easies t wa y t o handl e thes e equation s i s t o us e th e paramete r s as th e arc-lengt h paramete r fo r th e curve , s o tha t x'2 + y'2 = 1 . Then th e Euler-Lagrang e equatio n become s x" = 0 an d y" = 0. That is , x and y are linear functions o f s, so that th e curve is a straight line. C. Isoperimetri c problem . Ther e i s a hos t o f problem s o f thi s nature, bu t w e giv e just on e illustration . Th e exampl e w e are goin g to handle can be stated thi s way: amon g all closed curves in R2 whic h have a give n ar c lengt h L , fin d on e whic h enclose s a maximu m are a A. This proble m ha s a n interestin g twist , i n tha t a maximu m i s sought unde r a constrainin g equation . I n fact , thi s i s reminiscen t of th e Lagrang e multiplie r techniqu e o f finite-dimensional calculus . Rather tha n approac h th e proble m tha t way , however , w e ca n ri g things t o fi t ou r pattern . Namely , th e quantit y A -= - L2 i s invarian t under a change o f scale, an d therefor e tha t i s the functio n w e seek t o extremize. Suppose w e parametriz e ou r close d curve s i n M 2 a s x = x(s), y = y(s), 0 5 5 o - The n fro m vecto r calculu s w e have 1 f s ° A — - / {xy f — yx)ds (assumin g positiv e orientation) , 2 Jo L = / y/x' 2 + y ,2 ds. Jo Then w e want t o fin d th e variatio n i n A I2'

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