Utilize a Regra da Cadeia para determinar as derivadas parciais indicadas.
Y=w arctan (u v), u=r+s, v=s+t, w=t+r;
partial Ypartial r, partial Ypartial s, partial Ypartial t quando r=1, s=0, t=1

Passo 1

O enunciado define
Y = w;arctan(u,v), u = r+s, v = s+t, w = t+r,
e pede partial Ypartial r, partial Ypartial s e partial Ypartial t no ponto (r,s,t)=(1,0,1). Vamos usar a regra do produto e a regra da cadeia, além da derivada de arctan.

Passo 2

Definimos
theta = arctan(u,v), Y = w,theta
As derivadas parciais de u,v,w são
ur=1,;us=1,;ut=0, vr=0,;vs=1,;vt=1, wr=1,;ws=0,;wt=1

Passo 3

Para partial Ypartial r aplicamos produto e cadeia:
partial Ypartial r = wr,theta + w,partial thetapartial r,
partial thetapartial r = 11+(u,v)^2l(ur,v + u,vrr) = v1+(u,v)^2
Logo,
partial Ypartial r = 1cdotarctan(u,v) + w,v1+(u,v)^2

Passo 4

De modo análogo,
partial thetapartial s = us,v + u,vs1+(u,v)^2 = v+u1+(u,v)^2, partial thetapartial t = ut,v + u,vt1+(u,v)^2 = u1+(u,v)^2
Logo
partial Ypartial s = ws,theta + w,v+u1+(u,v)^2 = w,v+u1+(u,v)^2,
partial Ypartial t = wt,theta + w,u1+(u,v)^2 = 1cdotarctan(u,v) + w,u1+(u,v)^2

Passo 5

No ponto (r,s,t)=(1,0,1) temos
u=1,;v=1,;w=2,;u,v=1,;arctan(1)=pi4,;11+(u,v)^2=12
Substituindo:
partial Ypartial r =pi4 + 2cdot12 =pi4+1, partial Ypartial s =2cdot1+12 =2, partial Ypartial t =pi4 + 2cdot12 =pi4+1

Resposta

partial Ypartial r=pi4+1, partial Ypartial s=2, partial Ypartial t=pi4+1

Exercícios relacionados