Find the distance from the point P(0,0,1) to the surface z=x^2+2y^2

Oscar Burton 2022-10-21 Answered
Find the distance from the point P(0,0,1) to the surface z=x^2+2y^2
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Answers (1)

Jovanni Salinas
Answered 2022-10-22 Author has 18 answers
Find the distance from the point P(0,0,1) to the surface z = x 2 + 2 y 2
Let d ( x , y , z ) = ( x 0 ) 2 + ( y 0 ) 2 + ( z 1 ) 2 be the distance from point P.
We need to find the shortest distance wr.t constant
z = f ( x , y ) = x 2 + 2 y 2 i . e .   z x 2 2 y 2 = 0
d is always positive , so , for simplicity we can work on d 2 = x 2 + y 2 + ( z 1 ) 2 .
Subject to z x 2 2 y 2 = 0
Lagrangian function
L ( x , u , z , λ ) = d ( x , y , z ) + λ g ( x , y , z ) = x 2 + y 2 + ( z 1 ) 2 + λ ( z x 2 2 y 2 )
where λ is lagrange multiplier
By the lagrangian method we need to solve
L x = 0 ,     L y = 0 ,     L z = 0 ,     L λ = 0 i . e     2 x 2 λ x = 0       2 y 4 λ y = 0 2 ( z 1 ) + λ = 0       z x 2 2 y 2 = 0 2 x ( 1 λ ) = 0         2 y ( 1 2 λ ) = 0
when
x = y = z = 0 d ( x , y , z ) = 1 x = ± 1 2 , y = 0 , z = 1 2 d ( x , y , z ) = 1 2 + 0 + ( 1 2 1 ) 2 = 1 2 + 1 4 = 3 2 x = 0 , y 2 = 3 8 , z = 3 4 d ( x , y , z ) = 0 + 3 8 + ( 3 4 1 ) 2 = 3 8 + 1 16 = 7 4
shortest distance is from (0,0,0)
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