Se f(x, y)=3 x^{2}+2 x y-y^{2}, ache:
(a) \Delta f(2,-1), o incremento de f em (2,-1);
(b) \Delta f(2,-1) quando \Delta x=-0,01 e \Delta y=0,02;
(c) d f(2,-1, \Delta x, \Delta y), a diferencial total de f em (2,-1);
(d) d f(2,-1,-0,01,0,02).

Passo 1

O enunciado define a função
f(x,y)=3x^2+2xy-y^2
e pede:
(a) o incremento real \Delta f(2,-1) em termos de \Delta x e \Delta y;
(b) o valor numérico de \Delta f(2,-1) quando \Delta x=-0,01 e \Delta y=0,02;
(c) a diferencial total d f(2,-1,\Delta x,\Delta y);
(d) o valor numérico dessa diferencial em \Delta x=-0,01 e \Delta y=0,02.

Passo 2

Vamos calcular primeiro
f(2,-1)=3\cdot 2^2+2\cdot 2\cdot(-1)-(-1)^2 =3\cdot4-4-1 =12-4-1 =7.
Em seguida consideramos
f(2+\Delta x,\,-1+\Delta y) =3(2+\Delta x)^2 +2(2+\Delta x)(-1+\Delta y) -(-1+\Delta y)^2.

Passo 3

Expandindo cada termo:
3(2+\Delta x)^2=12+12\Delta x+3\Delta x^2,
2(2+\Delta x)(-1+\Delta y) =2\bigl(-2+2\Delta y-\Delta x+\Delta x\,\Delta y\bigr) =-4+4\Delta y-2\Delta x+2\Delta x\,\Delta y,
-(-1+\Delta y)^2 =-(1-2\Delta y+\Delta y^2) =-1+2\Delta y-\Delta y^2.
Somando tudo e subtraindo f(2,-1)=7
\Delta f(2,-1) =f(2+\Delta x,-1+\Delta y)-7 =10\Delta x+6\Delta y+3\Delta x^2+2\Delta x\,\Delta y-\Delta y^2.

Passo 4

Agora avaliamos em \Delta x=-0,01 e \Delta y=0,02:
\Delta f(2,-1) =10(-0,01)+6(0,02)+3(0,0001)+2(-0,01\cdot0,02)-(0,02)^2
=-0,10+0,12+0,0003-0,0004-0,0004 =0,0195.

Passo 5

Para a diferencial total calculamos as derivadas parciais
f_x(x,y)=6x+2y,\qquad f_y(x,y)=2x-2y
em (2,-1):
f_x(2,-1)=6\cdot2+2\cdot(-1)=12-2=10, \quad f_y(2,-1)=2\cdot2-2\cdot(-1)=4+2=6.
Logo
d f(2,-1,\Delta x,\Delta y) =10\,\Delta x+6\,\Delta y.

Passo 6

Substituindo \Delta x=-0,01 e \Delta y=0,02 na diferencial:
d f(2,-1,-0,01,0,02) =10(-0,01)+6(0,02) =-0,10+0,12 =0,02.

Resposta

(a) \Delta f(2,-1)=10\Delta x+6\Delta y+3\Delta x^2+2\Delta x\,\Delta y-\Delta y^2
(b) \Delta f(2,-1)\bigl|_{\Delta x=-0,01,\;\Delta y=0,02}=0,0195
(c) d f(2,-1,\Delta x,\Delta y)=10\Delta x+6\Delta y
(d) d f(2,-1,-0,01,0,02)=0,02

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