Usando a regra da cadeia para z=f(x, y) e w=f(x, y, z), calcule: partial zpartial t, partial zpartial s e partial wpartial t, partial wpartial s e partial wpartial r: z=raiz1+x^2+y^2, x=s e^t, y=s e^-t

Passo 1

O enunciado pede para usar a regra da cadeia em duas etapas: primeiro para z=f(x,y) com z=√(1+x²+y²) e x=s eᵗ, y=s e^(–t), calculando ∂z/∂t e ∂z/∂s; depois para w=f(x,y,z), obtendo ∂w/∂t, ∂w/∂s e ∂w/∂r.

Passo 2

Calculamos primeiro as derivadas paramétricas de x e y:
x=s e^t partial xpartial t=s e^t,;partial xpartial s=e^t,;partial xpartial r=0
y=s e^-t partial ypartial t=-s e^-t,;partial ypartial s=e^-t,;partial ypartial r=0

Passo 3

A função z=√(1+x²+y²) tem derivadas parciais:
fx=xraiz1+x^2+y^2=xz, fy=yz

Passo 4

Aplicando a regra da cadeia para z em relação a t e s:
partial zpartial t=fx,xt+fy,yt=x,(s e^t)+y,(-s e^-t)z =s^2(e^2t-e^-2t)raiz1+s^2(e^2t+e^-2t),
partial zpartial s=fx,xs+fy,ys=x,e^t+y,e^-tz =s(e^2t+e^-2t)raiz1+s^2(e^2t+e^-2t)

Passo 5

Para w=f(x,y,z) aplicamos a regra da cadeia geral:
partial wpartial u=wx,xu+wy,yu+wz,zu, uint,s,r

Passo 6

Substituindo as derivadas parciais e z_u obtidas:
partial wpartial t=wx,s e^t - wy,s e^-t + wz,s^2(e^2t-e^-2t)raiz1+s^2(e^2t+e^-2t),
partial wpartial s=wx,e^t + wy,e^-t + wz,s(e^2t+e^-2t)raiz1+s^2(e^2t+e^-2t),
partial wpartial r=wxcdot0 + wycdot0 + wzcdot0 = 0

Resposta

partial zpartial t=s^2l(e^2t-e^-2tr)raiz1+s^2l(e^2t+e^-2tr), partial zpartial s=sl(e^2t+e^-2tr)raiz1+s^2l(e^2t+e^-2tr)
partial wpartial t=wx,s e^t - wy,s e^-t + wz,s^2(e^2t-e^-2t)raiz1+s^2(e^2t+e^-2t), partial wpartial s=wx,e^t + wy,e^-t + wz,s(e^2t+e^-2t)raiz1+s^2(e^2t+e^-2t), partial wpartial r=0

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