( 1,5 pt) Dada a função de duas variáveis f(x, y)=x cos (x^2 y), obtenha partial fpartial x, partial fpartial y, partial^2 fpartial x^2, partial^2 fpartial y^2 e partial^2 fpartial x partial y no ponto (2,1) com passos hx=hy=0,01. Utilize Diferenças Finitas Centradas em todos os casos.

Passo 1

O enunciado dá a função de duas variáveis
f(x,y) = x cosl(x^2yr)
e pede as derivadas parciais de primeira e segunda ordem em (2,1) usando diferenças finitas centradas com hx=hy=0,01. Vamos usar as fórmulas:
partial fpartial xapprox f(x+h,,y)-f(x-h,,y)2h, partial fpartial yapprox f(x,,y+h)-f(x,,y-h)2h,
partial^2 fpartial x^2approx f(x+h,y)-2f(x,y)+f(x-h,y)h^2, partial^2 fpartial y^2approx f(x,y+h)-2f(x,y)+f(x,y-h)h^2,
partial^2 fpartial xpartial yapprox f(x+h,y+h)-f(x+h,y-h)-f(x-h,y+h)+f(x-h,y-h)4h^2

Passo 2

Calculamos f nos pontos necessários com h=0.01.
f(2,1)=2cos(4) approx -1.3072872
f(2.01,1)=2.01cos(2.01^2cdot1)approx -1.25166, f(1.99,1)approx -1.36079
f(2,1.01)=2cos(4cdot1.01)approx -1.24570, f(2,0.99)approx -1.36679
f(2.01,1.01)approx -1.18773,; f(2.01,0.99)approx -1.31425,; f(1.99,1.01)approx -1.30020,; f(1.99,0.99)approx -1.41575

Passo 3

Aplicando as diferenças centradas para as derivadas de primeira ordem:
partial fpartial x(2,1) approx -1.25166-(-1.36079)2cdot0.01 = 0.109130.02approx 5.4565,
partial fpartial y(2,1) approx -1.24570-(-1.36679)2cdot0.01 = 0.121090.02approx 6.0548

Passo 4

Para as segundas derivadas puras:
partial^2 fpartial x^2(2,1) approx -1.25166 -2(-1.3072872)+(-1.36079)(0.01)^2 = 0.00212440.0001approx 21.244,
partial^2 fpartial y^2(2,1) approx -1.24570 -2(-1.3072872)+(-1.36679)(0.01)^2 = 0.00208840.0001approx 20.884

Passo 5

Para a derivada mista:
partial^2 fpartial xpartial y(2,1) approx -1.18773 -(-1.31425)-(-1.30020)+(-1.41575)4,(0.01)^2 = 0.010970.0004approx 27.425

Resposta

partial fpartial x(2,1)approx 5.4565
partial fpartial y(2,1)approx 6.0548
partial^2 fpartial x^2(2,1)approx 21.244
partial^2 fpartial y^2(2,1)approx 20.884
partial^2 fpartial xpartial y(2,1)approx 27.425

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