Geometria de superficies

# Geometria de superficies - Geometria de Superficies y...

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Unformatted text preview: Geometria de Superficies y Variedades 26 de marzo de 2011 Ejercicio 1. Dada la curva de revoluci´ on dada por la siguiente paramatrizaci´ on: x = ψ ( u ) z = ϕ ( u ) Hallar la primera forma cuadr´ atica I. Desarrollo: La primera forma cuadrarica I viene dada por la siguiente f´ ormula: I = g ij dr i dr j Puesto que r = r ( u,v ) = ( ψ ( u ) cos v,ψ ( u ) sin v,ϕ ( u )), obtenemos que : I = g 11 du 2 + 2 g 12 dudv + g 22 dv 2 (1) donde: g 11 = r 1 .r 1 (2) g 12 = r 1 .r 2 (3) g 22 = r 2 .r 2 (4) Entonces para encontrar I basta calcular ( 2 ),( 3 ),( 4 ), para esto hallemos r 1 y r 2 r 1 = ( ˙ ψ ( u )) cos v, ˙ ψ ( u ) sin v, ˙ ϕ ( u )) r 2 = (- ˙ vψ ( u ) sin v, ˙ vψ ( u ) cos v, 0) Luego g 11 = r 1 .r 1 = [ ˙ ψ ( u ))] 2 cos 2 v + [ ˙ ψ ( u )] 2 sin 2 v + [ ˙ ϕ ( u )] 2 = [ ˙ ψ ( u ))] 2 + [ ˙ ϕ ( u )] 2 (5) g 12 = r 1 .r 2 =- ˙ vψ ( u ) ˙ ψ ( u ) sin v cos v + ˙ vψ ( u ) ˙ ψ ( u ) sin v cos v + 0 = 0 (6) g 22 = r 2 .r 2 = ˙ v 2 [ ψ ( u )] 2 sin 2 v + ˙ v 2 [ ψ ( u )] 2 cos 2 v = ˙ v 2 [ ψ ( u )] 2 (7) Finalmente reemplazamos ( 6 ),( 7 ),( 7 ) en ( 1 ) Respuesta: I = [ ˙ ψ ( u ))] 2 + [ ˙ ϕ ( u )] 2 du 2 + ˙ v 2 [ ψ ( u )] 2 dv 2...
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## This note was uploaded on 03/27/2011 for the course MATHEMATIC 504 taught by Professor Carlostrujillo during the Winter '09 term at Buena Vista.

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Geometria de superficies - Geometria de Superficies y...

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