(2,0) Seja u=f(x, y), onde x=e^2 s cos 2 t e y=e^2 s sin 2 t, verifique que:
partial^2 upartial s^2+partial^2 upartial t^2=4 e^4 s[partial^2 upartial x^2+partial^2 upartial y^2]

Passo 1

Beleza, vamos lá! A gente tem uma função partial^2 upartial s^2 + partial^2 upartial t^2 = 4 e^4s ( partial^2 upartial x^2 + partial^2 upartial y^2 ) , onde e . O objetivo é mostrar que:
Vamos precisar usar a regra da cadeia para derivadas parciais e calcular as derivadas de em relação a e .

Passo 2

Primeiro, vamos calcular as primeiras derivadas de partial upartial s = partial upartial x cdot partial xpartial s + partial upartial y cdot partial ypartial s em relação a partial xpartial s = 2 e^2s cos 2t e partial ypartial s = 2 e^2s sin 2t usando a regra da cadeia.
Para partial upartial s = 2 e^2s cos 2t cdot partial upartial x + 2 e^2s sin 2t cdot partial upartial y :
Calculando as derivadas de e em relação a :
Então,

Passo 3

Agora, vamos calcular a segunda derivada de partial^2 upartial s^2 = partialpartial s ( 2 e^2s cos 2t cdot partial upartial x + 2 e^2s sin 2t cdot partial upartial y ) em relação a partial^2 upartial s^2 = 4 e^2s cos 2t cdot partial upartial x + 4 e^2s sin 2t cdot partial upartial y + 4 e^4s cos^2 2t cdot partial^2 upartial x^2 + 8 e^4s cos 2t sin 2t cdot partial^2 upartial x partial y + 4 e^4s sin^2 2t cdot partial^2 upartial y^2 :
Aplicando a regra do produto:

Passo 4

Vamos fazer o mesmo para partial upartial t = partial upartial x cdot partial xpartial t + partial upartial y cdot partial ypartial t :
partial xpartial t = -2 e^2s sin 2t
Calculando as derivadas de partial ypartial t = 2 e^2s cos 2t e partial upartial t = -2 e^2s sin 2t cdot partial upartial x + 2 e^2s cos 2t cdot partial upartial y em relação a :
Então,

Passo 5

Agora, calculamos a segunda derivada de partial^2 upartial t^2 = partialpartial t ( -2 e^2s sin 2t cdot partial upartial x + 2 e^2s cos 2t cdot partial upartial y ) em relação a partial^2 upartial t^2 = -4 e^2s sin 2t cdot partial upartial x + 4 e^2s cos 2t cdot partial upartial y + 4 e^4s sin^2 2t cdot partial^2 upartial x^2 - 8 e^4s sin 2t cos 2t cdot partial^2 upartial x partial y + 4 e^4s cos^2 2t cdot partial^2 upartial y^2 :
Aplicando a regra do produto:

Passo 6

Agora, somamos as duas segundas derivadas:
partial^2 upartial s^2 + partial^2 upartial t^2 = [4 e^2s cos 2t partial upartial x + 4 e^2s sin 2t partial upartial y] + [-4 e^2s sin 2t partial upartial x + 4 e^2s cos 2t partial upartial y]
+ [4 e^4s cos^2 2t partial^2 upartial x^2 + 8 e^4s cos 2t sin 2t partial^2 upartial x partial y + 4 e^4s sin^2 2t partial^2 upartial y^2]
+ [4 e^4s sin^2 2t partial^2 upartial x^2 - 8 e^4s sin 2t cos 2t partial^2 upartial x partial y + 4 e^4s cos^2 2t partial^2 upartial y^2]
Simplificando os termos:
Os termos lineares em 4 e^2s cos 2t partial upartial x - 4 e^2s sin 2t partial upartial x + 4 e^2s sin 2t partial upartial y + 4 e^2s cos 2t partial upartial y = 0 e 4 e^4s (cos^2 2t + sin^2 2t) partial^2 upartial x^2 + 4 e^4s (sin^2 2t + cos^2 2t) partial^2 upartial y^2 = 4 e^4s ( partial^2 upartial x^2 + partial^2 upartial y^2 ) se cancelam:
E os termos de segunda ordem somam:

Resposta

Conseguimos verificar que:
partial^2 upartial s^2 + partial^2 upartial t^2 = 4 e^4s ( partial^2 upartial x^2 + partial^2 upartial y^2 )

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