Certamen 1 Reemplazo - Matemáticas III (2010) Stgo.pdf - Universidad T´ ecnica Federico Santa Mar´ıa Departamento de Matem´atica Coordinaci´ on

Certamen 1 Reemplazo - Matemáticas III (2010) Stgo.pdf -...

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Universidad T´ ecnica Federico Santa Mar´ ıa Departamento de Matem´ atica Coordinaci´on de Matem´ atica 3 (MAT023) 1 er Semestre de 2010 Certamen Acumulativo de Reemplazo - Pauta: Viernes 9 de Julio 1. En el elipsoide 4 x 2 + y 2 + 4 z 2 - 16 x - 6 y - 8 z + 25 = 0 hallar el punto m´ as cercano y el punto as lejano de la superficie 2 x + 2 y + 2 z = 0 . Soluci´ on: La mirada geom´ etrica. Considerar f ( x , y , z ) = 4 x 2 + y 2 + 4 z 2 - 16 x - 6 y - 8 z + 25 . Basta encontrar los puntos en el elipsoide tal que f ( x , y , z ) = λ (1 , 1 , 1) . f ( x , y , z ) = (8 x - 16 , 2 y - 6 , 8 z - 8) , y se tiene el sistema: 8 x - 16 = λ 2 y - 6 = λ 8 z - 8 = λ x = λ + 16 8 y = λ + 6 2 z = λ + 8 8 Reemplazando en la ecuaci´ on del elipsoide se tiene: 4 λ + 16 8 2 + λ + 6 2 2 + 4 λ + 8 8 2 - 16 λ + 16 8 - 6 λ + 6 2 - 8 λ + 16 8 + 25 = 0 Se ve feo, pero termina siendo: 6 16 λ 2 - 4 = 0 λ = ± 4 r 2 3 = ± 8 6 Luego: (a) El punto m´ as cercano es P 1 = 1 6 + 2 , 4 6 + 3 , 1 6 + 1 . (b) El punto m´ as lejano es P 2 = - 1 6 + 2 , - 4 6 + 3 , - 1 6 + 1 . En efecto: d ( P 1 , π ) = 6 + 6 3 y d ( P 2 , π ) = 6 - 6 3 .
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