Usando a regra da cadeia para z=f(x, y) e w=f(x, y, z), calcule: \frac{\partial z}{\partial t}, \frac{\partial z}{\partial s} e \frac{\partial w}{\partial t}, \frac{\partial w}{\partial s} e \frac{\partial w}{\partial r}.
z=x^{2}+y^{2}, x=\operatorname{cosh} (s) \cos (t), y=\operatorname{senh}(s) \operatorname{sen}(t)

Passo 1

O enunciado define
z = x^2 + y^2, \quad x = \operatorname{cosh}(s)\cos(t), \quad y = \operatorname{sinh}(s)\sin(t),
e depois w = f(x,y,z). Vamos usar a regra da cadeia primeiro para \partial z/\partial t e \partial z/\partial s, depois para \partial w/\partial t, \partial w/\partial s e \partial w/\partial r.

Passo 2

Como \;z=x^2+y^2, vale
\frac{\partial z}{\partial t} =2x\,\frac{\partial x}{\partial t}+2y\,\frac{\partial y}{\partial t}.
Calculando
\frac{\partial x}{\partial t}=-\operatorname{cosh}(s)\sin(t), \quad \frac{\partial y}{\partial t}=\operatorname{sinh}(s)\cos(t),
substituímos e obtemos
\frac{\partial z}{\partial t} =2\operatorname{cosh}(s)\cos(t)\bigl(-\operatorname{cosh}(s)\sin(t)\bigr) +2\operatorname{sinh}(s)\sin(t)\bigl(\operatorname{sinh}(s)\cos(t)\bigr) =-2\sin(t)\cos(t) =-\sin(2t).
De modo similar,
\frac{\partial z}{\partial s} =2x\,\frac{\partial x}{\partial s}+2y\,\frac{\partial y}{\partial s},
com
\frac{\partial x}{\partial s}=\operatorname{sinh}(s)\cos(t), \quad \frac{\partial y}{\partial s}=\operatorname{cosh}(s)\sin(t),
logo
\frac{\partial z}{\partial s} =2\operatorname{cosh}(s)\cos(t)\,\operatorname{sinh}(s)\cos(t) +2\operatorname{sinh}(s)\sin(t)\,\operatorname{cosh}(s)\sin(t) =2\operatorname{cosh}(s)\operatorname{sinh}(s) =\operatorname{sinh}(2s).

Passo 3

Agora w=f(x,y,z) é função de três variáveis, cada uma dependendo de (s,t). A regra da cadeia diz:
\frac{\partial w}{\partial t} =f_x\,\frac{\partial x}{\partial t} +f_y\,\frac{\partial y}{\partial t} +f_z\,\frac{\partial z}{\partial t},
\frac{\partial w}{\partial s} =f_x\,\frac{\partial x}{\partial s} +f_y\,\frac{\partial y}{\partial s} +f_z\,\frac{\partial z}{\partial s}.
Substituindo as derivadas parciais de x,y,z do Passo 2, temos
\frac{\partial w}{\partial t} =f_x\bigl(-\operatorname{cosh}(s)\sin(t)\bigr) +f_y\bigl(\operatorname{sinh}(s)\cos(t)\bigr) +f_z\bigl(-\sin(2t)\bigr),
\frac{\partial w}{\partial s} =f_x\bigl(\operatorname{sinh}(s)\cos(t)\bigr) +f_y\bigl(\operatorname{cosh}(s)\sin(t)\bigr) +f_z\bigl(\operatorname{sinh}(2s)\bigr).

Passo 4

Como nem x nem y dependem de r, tem-se
\frac{\partial x}{\partial r} =\frac{\partial y}{\partial r} =0 \;\Longrightarrow\; \frac{\partial z}{\partial r} =2x\cdot0+2y\cdot0 =0.
Aplicando a cadeia em w,
\frac{\partial w}{\partial r} =f_x\,\frac{\partial x}{\partial r} +f_y\,\frac{\partial y}{\partial r} +f_z\,\frac{\partial z}{\partial r} =0.

Resposta

\frac{\partial z}{\partial t} = -\sin(2t), \quad \frac{\partial z}{\partial s} = \operatorname{sinh}(2s),
\frac{\partial w}{\partial t} =f_x\bigl(-\operatorname{cosh}(s)\sin(t)\bigr) +f_y\bigl(\operatorname{sinh}(s)\cos(t)\bigr) +f_z\bigl(-\sin(2t)\bigr),
\frac{\partial w}{\partial s} =f_x\bigl(\operatorname{sinh}(s)\cos(t)\bigr) +f_y\bigl(\operatorname{cosh}(s)\sin(t)\bigr) +f_z\bigl(\operatorname{sinh}(2s)\bigr), \quad \frac{\partial w}{\partial r}=0.

Exercícios relacionados