Suponha que f seja uma função diferenciável de x, y e z e u=f(x, y, z). Então, se x=r sen phi cos theta, y=r sen phi sen theta e z=r cos phi, expresse partial upartial r, partial upartial phi e partial upartial theta em termos de partial upartial x, partial upartial y e partial upartial z.

Passo 1

Temos a função u = f(x,y,z) diferenciável e as transformações esféricas x = r sinphi costheta, y = r sinphi sintheta e z = r cosphi. Vamos usar a regra da cadeia para expressar partial upartial r, partial upartial phi e partial upartial theta em termos de partial upartial x, partial upartial y e partial upartial z.

Passo 2

Aplicamos a regra da cadeia para funções compostas. Para partial upartial r:
partial upartial r = partial upartial xpartial xpartial r + partial upartial ypartial ypartial r + partial upartial zpartial zpartial r
Para partial upartial phi:
partial upartial phi = partial upartial xpartial xpartial phi + partial upartial ypartial ypartial phi + partial upartial zpartial zpartial phi
Para partial upartial theta:
partial upartial theta = partial upartial xpartial xpartial theta + partial upartial ypartial ypartial theta + partial upartial zpartial zpartial theta

Passo 3

Agora vamos calcular as derivadas parciais de x, y e z em relação a r, phi e theta. Para x:
partial xpartial r = sinphicostheta, partial xpartial phi = rcosphicostheta, partial xpartial theta = -rsinphisintheta
Para y:
partial ypartial r = sinphisintheta, partial ypartial phi = rcosphisintheta, partial ypartial theta = rsinphicostheta
Para z:
partial zpartial r = cosphi, partial zpartial phi = -rsinphi, partial zpartial theta = 0

Passo 4

Substituindo as derivadas das coordenadas nas fórmulas da regra da cadeia, chegamos a:
partial upartial r = partial upartial xsinphicostheta + partial upartial ysinphisintheta + partial upartial zcosphi
partial upartial phi = rl(partial upartial xcosphicostheta + partial upartial ycosphisintheta - partial upartial zsinphir)
partial upartial theta = rsinphil(-partial upartial xsintheta + partial upartial ycosthetar)

Resposta

partial upartial r = partial upartial xsinphicostheta + partial upartial ysinphisintheta + partial upartial zcosphi
partial upartial phi = rl(partial upartial xcosphicostheta + partial upartial ycosphisintheta - partial upartial zsinphir)
partial upartial theta = rsinphil(-partial upartial xsintheta + partial upartial ycosthetar)

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