Determine as derivadas parciais de:
z=2 x^2 y-4 x y^2-y^3, onde x=rho^2 theta sen alpha e y=rho theta cos 2 alpha

Passo 1

O enunciado nos dá
z = 2x^2y - 4xy^2 - y^3,
com
x = rho^2,theta,sinalpha, y = rho,theta,cos2alpha
Queremos calcular as derivadas parciais
partial z/partial rho,
partial z/partial theta e
partial z/partial alpha usando a regra da cadeia.

Passo 2

Calculamos as derivadas parciais de z em relação a x e y:
partial zpartial x = partialpartial xl(2x^2y - 4xy^2 - y^3r) = 4xy - 4y^2,
partial zpartial y = partialpartial yl(2x^2y - 4xy^2 - y^3r) = 2x^2 - 8xy - 3y^2

Passo 3

Agora as derivadas de x e y em relação a rho, theta e alpha:
partial xpartial rho = 2rho,theta,sinalpha, partial xpartial theta = rho^2,sinalpha, partial xpartial alpha = rho^2,theta,cosalpha,
partial ypartial rho = theta,cos2alpha, partial ypartial theta = rho,cos2alpha, partial ypartial alpha = -2,rho,theta,sin2alpha

Passo 4

Pela regra da cadeia, para cada variável uinrho,theta,alpha vale
partial zpartial u = partial zpartial x,partial xpartial u + partial zpartial y,partial ypartial u

Passo 5

Substituindo os resultados anteriores, obtemos
partial zpartial rho = (4xy - 4y^2),(2rho,theta,sinalpha) + (2x^2 - 8xy - 3y^2),(theta,cos2alpha),
partial zpartial theta = (4xy - 4y^2),(rho^2,sinalpha) + (2x^2 - 8xy - 3y^2),(rho,cos2alpha),
partial zpartial alpha = (4xy - 4y^2),(rho^2,theta,cosalpha) + (2x^2 - 8xy - 3y^2),(-2,rho,theta,sin2alpha),
em que
x=rho^2thetasinalpha
e
y=rhothetacos2alpha.

Resposta

partial zpartial rho = (4xy - 4y^2),(2rho,theta,sinalpha) + (2x^2 - 8xy - 3y^2),(theta,cos2alpha),
partial zpartial theta = (4xy - 4y^2),(rho^2,sinalpha) + (2x^2 - 8xy - 3y^2),(rho,cos2alpha),
partial zpartial alpha = (4xy - 4y^2),(rho^2,theta,cosalpha) -2,(2x^2 - 8xy - 3y^2),(rho,theta,sin2alpha),
com x=rho^2thetasinalpha e y=rhothetacos2alpha.

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