Questão resolvida - Encontre dz_dx,dz_dy em (0,0,0) se xzye^xz z cos (y) 0 - Cálculo II - UNIBTA

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Encontre em se . dz / dx, dz / dy 0, 0, 0( ) x + z + ye + z cos y = 03 2 xz ( )
 
Resolução:
 
Vamos realizar as derivadas implícitas e, em seguida, substiutir o ponto ;0, 0, 0( )
 
x + z + ye + z cos y = 0
dz 
dx
3 2 xz ( )
 
3x + 2z + 0 ⋅ e + e z + x ⋅ y + cos y + 0 z = 0 2
dz 
dx
xz xz dz 
dx
dz 
dx
( ) ( )
 
3x + 2z + 0 + zye + xy + cos y + 0 = 02
dz 
dx
xz dz 
dx
dz 
dx
( )
 
2z + xye + xy + cos y = - 3x - zye
dz 
dx
xz dz 
dx
dz 
dx
( )
dz 
dx
2 xz
 
2z + xye + xy + cos y = - 3x - zye
dz 
dx
xz ( ) 2 xz
 
=
dz 
dx
-3x - zye
2z + xye + xy + cos y
2 xz
xz ( )
Em , temos que é dado por;0, 0, 0( )
dz 
dx
 
0, 0, 0 = = = = =
dz 
dx
( )
-3 0 - 0 ⋅ 0e
2 ⋅ 0 + 0 ⋅ 0e + 0 ⋅ 0 + cos 0
( )2 0⋅0
0⋅0 ( )
-3 ⋅ 0 + 0 ⋅ e
0 + 0 ⋅ e + 0 + 1
0
0
0 + 0
0 + 0 ⋅ 1 + 1
0
0 + 0 + 1
0
1
 
0, 0, 0 = 0
dz 
dx
( )
 
Agora x + z + ye + z cos y = 0→
dz 
dy
3 2 xz ( )
 
 
 
(Resposta - 1)
0 + 2z + 1 ⋅ e + ye ⋅ x + cos y + -sen y z = 0
dz 
dy
xz xz dz 
dy
dz 
dy
( ) ( ( ))
 
2z + e + yxe + cos y - zsen y = 0
dz 
dy
xz xz dz 
dy
( )
dz 
dy
( )
 
2z + yxe + cos y = - e + zsen y
dz 
dy
xz dz 
dy
( )
dz 
dy
xz ( )
 
2z + yxe + cos y = - e + zsen y
dz 
dy
xz ( ) xz ( )
 
=
dz 
dy
-e + zsen y
2z + yxe + cos y
xz ( )
xz ( )
Em , temos que é dado por;0, 0, 0( )
dz 
dy
 
0, 0, 0 = = = =
dz 
dy
( )
-e + 0 ⋅ sen 0
2 ⋅ 0 + 0 ⋅ 0 ⋅ e + cos 0
0⋅0 ( )
0⋅0 ( )
-e + 0 ⋅ 0
0 + 0 ⋅ e + 1
0
0
-1 + 0
0 ⋅ 1 + 1
-1
1
 
0, 0, 0 = - 1
dz 
dy
( )
 
 
(Resposta - 2)