Given the relations: <mtable columnalign="right left right left right left right left right lef

ga2t1a2dan1oj

ga2t1a2dan1oj

Answered question

2022-05-08

Given the relations:
F 1 ( x , y , u , v ) = u e y + ( y + 1 ) e v x + x y + 1 = 0 F 2 ( x , y , u , v ) = x + y + x u + e y + v 1 = 0
find the linear approximation of ( u , v ) in terms of ( x , y ) near ( x , y , u , v ) = ( 0 , 0 , 2 , 0 ).

Answer & Explanation

arbotsck8sg

arbotsck8sg

Beginner2022-05-09Added 22 answers

First of all you have to adapt your notation to the multivariable case. Since your (implicit) function ( u , v ) = f ( x , y ) is f : R 2 R 2 the linear approximation is
f ( x , y ) f ( a , b ) + J f ( a , b ) ( x a , y b )
where J f is the jacobian matrix, i.e. the matrix of partial derivatives:
J f = ( u x u y v x v y )
Now, the point is that partial derivatives can be obtained from the implicit function theorem without explicitly writing ( u , v ) as a function of ( x , y ). So, differentiating both equations with respect to x gives:
F 1 x = e y u x + v ( y + 1 ) e v x + x ( y + 1 ) e v x v x + y = 0 F 2 x = 1 + u + x u x + e y + v v x = 0
and you can obtain both partial derivatives (observe that now the system is linear). Repeating the process with y you obtain the other two. Finally, substituting for the point with the given values of ( u , v , x , y ) gives the answer.

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