Encontre a derivada parcial indicada usando a regra da cadeia:
u=x^2+y^2+z^2 ; x=r sen phi cos theta ; y=r sen phi sen theta ; z=r cos phi ; partial upartial r ; partial upartial theta ; partial upartial phi

Passo 1

O enunciado nos dá
u = x^2+y^2+z^2,
com as coordenadas esféricas
x = r,senphicostheta, y = r,senphisentheta, z = r,cosphi
Queremos calcular as derivadas parciais partial upartial r, partial upartial theta e partial upartial phi usando a regra da cadeia para funções de várias variáveis:
partial upartial r = partial upartial xpartial xpartial r + partial upartial ypartial ypartial r + partial upartial zpartial zpartial r,
e análogas para theta e phi.

Passo 2

Calcular partial upartial r.
Primeiro,
partial upartial x=2x, partial upartial y=2y, partial upartial z=2z
Em seguida,
partial xpartial r=senphicostheta, partial ypartial r=senphisentheta, partial zpartial r=cosphi
Logo
partial upartial r =2x(senphicostheta)+2y(senphisentheta)+2z(cosphi)
Substituindo x=rsenphicostheta, y=rsenphisentheta, z=rcosphi, obtemos
partial upartial r =2l[rsin^2phi(cos^2theta+sen^2theta)+rcos^2phir] =2r(sin^2phi+cos^2phi) =2r

Passo 3

Calcular partial upartial theta.
Usando a regra da cadeia:
partial xpartial theta=-rsenphisentheta, partial ypartial theta=rsenphicostheta, partial zpartial theta=0
Então
partial upartial theta =2x(-rsenphisentheta)+2y(rsenphicostheta)+2zcdot0
Substituindo x e y nota-se que os termos se cancelam, dando
partial upartial theta=0

Passo 4

Calcular partial upartial phi.
Novamente pela cadeia:
partial xpartial phi=rcosphicostheta, partial ypartial phi=rcosphisentheta, partial zpartial phi=-rsenphi
Logo
partial upartial phi =2x(rcosphicostheta)+2y(rcosphisentheta)+2z(-rsenphi)
Substituindo x,y,z e simplificando, todos os termos se cancelam, resultando em
partial upartial phi=0

Resposta

partial upartial r=2r, partial upartial theta=0, partial upartial phi=0

Exercícios relacionados