1. Seja f uma função diferenciável de três variáveis e sejam w=f(x, y, z), x=rho sen phi cos theta, y=rho sen phi sen theta e z=rho cos phi. Expresse partial w / partial rho, partial w / partial phi e partial w / partial theta em termos de partial w / partial x, partial w / partial y e partial w / partial z.

Passo 1

O enunciado pede para usar a regra da cadeia na função w=f(x,y,z), onde x,y,z estão dados em coordenadas esféricas:
x=rhosinphicostheta, y=rhosinphisintheta, z=rhocosphi
Precisamos expressar partial w/partialrho, partial w/partialphi e partial w/partialtheta em termos de partial w/partial x, partial w/partial y e partial w/partial z aplicando a regra da cadeia.

Passo 2

Para rho: calculemos primeiro
partial xpartialrho=sinphicostheta, partial ypartialrho=sinphisintheta, partial zpartialrho=cosphi
Então, pela regra da cadeia,
partial wpartialrho=partial wpartial xpartial xpartialrho+partial wpartial ypartial ypartialrho+partial wpartial zpartial zpartialrho

Passo 3

Para phi: temos
partial xpartialphi=rhocosphicostheta, partial ypartialphi=rhocosphisintheta, partial zpartialphi=-rhosinphi
Logo,
partial wpartialphi=partial wpartial xpartial xpartialphi+partial wpartial ypartial ypartialphi+partial wpartial zpartial zpartialphi

Passo 4

Para theta: obtemos
partial xpartialtheta=-rhosinphisintheta, partial ypartialtheta=rhosinphicostheta, partial zpartialtheta=0
Assim,
partial wpartialtheta=partial wpartial xpartial xpartialtheta+partial wpartial ypartial ypartialtheta+partial wpartial zpartial zpartialtheta

Resposta

partial wpartialrho=partial wpartial xsinphicostheta+partial wpartial ysinphisintheta+partial wpartial zcosphi,
partial wpartialphi=partial wpartial x,rhocosphicostheta+partial wpartial y,rhocosphisintheta-partial wpartial z,rhosinphi,
partial wpartialtheta=-partial wpartial x,rhosinphisintheta+partial wpartial y,rhosinphicostheta

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