Use a Regra da Cadeia para determinar as derivadas parciais indicadas.
(g) w=x y+y z+x z, x=r cos theta, y=r sen theta, z=r theta, partial w / partial r, partial w / partial theta, quando r=2 e theta=pi / 2.

Passo 1

O enunciado define
w = x y + y z + x z
com as mudanças de variável
x = rcostheta, y = rsintheta, z = r,theta,
e pede as derivadas parciais partial w/partial r e partial w/partial theta no ponto (r,theta)=(2,pi/2). Vamos usar a Regra da Cadeia em variáveis múltiplas.

Passo 2

Primeiro, calculamos as derivadas parciais de w em função de (x,y,z). Como
w = x y + y z + x z,
temos
partial wpartial x = y + z, partial wpartial y = x + z, partial wpartial z = x + y

Passo 3

Agora obtemos as derivadas das variáveis x,y,z em função de r e theta.
De x = rcostheta, y = rsintheta e z = r,theta segue
partial xpartial r = costheta, partial ypartial r = sintheta, partial zpartial r = theta,
partial xpartial theta = -rsintheta, partial ypartial theta = rcostheta, partial zpartial theta = r

Passo 4

Pela Regra da Cadeia em derivadas parciais,
partial wpartial r = partial wpartial x,partial xpartial r + partial wpartial y,partial ypartial r + partial wpartial z,partial zpartial r,
partial wpartial theta = partial wpartial x,partial xpartial theta + partial wpartial y,partial ypartial theta + partial wpartial z,partial zpartial theta

Passo 5

No ponto r=2, theta=pi/2 temos
x = 2cospi2=0, y = 2sinpi2=2, z = 2cdotpi2=pi
Logo
partial wpartial r = (y+z)costheta + (x+z)sintheta + (x+y)theta = (2+pi)cdot0 + (pi)cdot1 + (2)cdotpi2 = pi + pi = 2pi,
partial wpartial theta = (y+z)(-rsintheta) + (x+z)(rcostheta) + (x+y),r = (2+pi)(-2cdot1) + (pi)cdot0 + (2)cdot2 = -2(2+pi) + 4 = -4 -2pi +4 = -2pi

Resposta

partial wpartial rg|(2,pi/2) = 2pi, partial wpartial thetag|(2,pi/2) = -2pi.

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