Se f(x, y)=3 x^{2}+2 x y-y^{2}, ache:
(a) \Delta f(1,4), o incremento de f em (1,4);
(b) \Delta f(1,4) quando \Delta x=0,03 e \Delta y=-0,02;
(c) d f(1,4, \Delta x, \Delta y), a diferencial total de f em (1,4);
(d) d f(1,4,0,03,-0,02).

Passo 1

O enunciado dá a função
f(x,y)=3x^{2}+2xy-y^{2}
e pede:
(a) o incremento \Delta f(1,4), isto é
\Delta f(1,4)=f(1+\Delta x,\,4+\Delta y)-f(1,4),
(b) esse mesmo \Delta f(1,4) quando \Delta x=0,03 e \Delta y=-0,02,
(c) a diferencial total
df(1,4,\Delta x,\Delta y)=f_{x}(1,4)\,\Delta x+f_{y}(1,4)\,\Delta y,
(d) e por fim df(1,4,0,03,-0,02).

Passo 2

Calcular primeiro f(1,4):
f(1,4)=3\cdot1^{2}+2\cdot1\cdot4-4^{2}=3+8-16=-5.
Agora
f(1+\Delta x,4+\Delta y) =3(1+\Delta x)^{2}+2(1+\Delta x)(4+\Delta y)-(4+\Delta y)^{2}.
Expandindo e agrupando em potências de \Delta x e \Delta y
f(1+\Delta x,4+\Delta y)=-5+14\Delta x+3\Delta x^{2}-6\Delta y+2\Delta x\,\Delta y-\Delta y^{2}.
Logo, cancelando o -5 de f(1,4), temos
\Delta f(1,4)=14\Delta x+3\Delta x^{2}-6\Delta y+2\Delta x\,\Delta y-\Delta y^{2}.

Passo 3

Para \Delta x=0,03 e \Delta y=-0,02 substituímos na expressão de \Delta f(1,4):
\Delta f(1,4)=14\cdot0,03+3\cdot(0,03)^{2}-6\cdot(-0,02)+2\cdot0,03\cdot(-0,02)-(-0,02)^{2}
=0,42+0,0027+0,12-0,0012-0,0004=0,5411.

Passo 4

Para a diferencial total, primeiro achamos as derivadas parciais:
f_{x}(x,y)=6x+2y, \quad f_{y}(x,y)=2x-2y.
Em (1,4) isso dá
f_{x}(1,4)=6\cdot1+2\cdot4=14, \quad f_{y}(1,4)=2\cdot1-2\cdot4=-6.
Portanto
df(1,4,\Delta x,\Delta y)=14\,\Delta x-6\,\Delta y.

Passo 5

Substituindo \Delta x=0,03 e \Delta y=-0,02 em
df(1,4,\Delta x,\Delta y)=14\Delta x-6\Delta y
temos
df(1,4,0,03,-0,02)=14\cdot0,03-6\cdot(-0,02)=0,42+0,12=0,54.

Resposta

(a) \Delta f(1,4)=14\Delta x+3\Delta x^{2}-6\Delta y+2\Delta x\,\Delta y-\Delta y^{2}.
(b) \Delta f(1,4)=0,5411.
(c) df(1,4,\Delta x,\Delta y)=14\Delta x-6\Delta y.
(d) df(1,4,0,03,-0,02)=0,54.