Usando a regra da cadeia para z=f(x, y) e w=f(x, y, z), calcule \frac{d z}{d t} e \frac{d w}{d t} :
(k) w=\frac{x+y+z}{x^{2}+y^{2}+z^{2}}, x=\cos (t), y=\operatorname{sen}(t), z=e^{t}

Passo 1

Temos as funções
x=\cos(t),\quad y=\operatorname{sen}(t),\quad z=e^{t},
e definimos
w=\frac{x+y+z}{x^{2}+y^{2}+z^{2}}.
Queremos primeiro \frac{dz}{dt} e depois, aplicando a regra da cadeia para função de três variáveis,
\frac{dw}{dt} =\frac{\partial w}{\partial x}\,\frac{dx}{dt} +\frac{\partial w}{\partial y}\,\frac{dy}{dt} +\frac{\partial w}{\partial z}\,\frac{dz}{dt}.

Passo 2

Como z=e^{t}, temos diretamente
\frac{dz}{dt}=\frac{d}{dt}\bigl(e^{t}\bigr)=e^{t}.

Passo 3

Calculemos as derivadas parciais de
w(x,y,z)=\frac{N}{D}, \quad N=x+y+z,\; D=x^{2}+y^{2}+z^{2}.
Por quociente,
\frac{\partial w}{\partial x} =\frac{D\cdot\frac{\partial N}{\partial x}-N\cdot\frac{\partial D}{\partial x}}{D^{2}} =\frac{(x^{2}+y^{2}+z^{2})\cdot1-(x+y+z)\cdot2x}{(x^{2}+y^{2}+z^{2})^{2}} =\frac{D-2x\,N}{D^{2}},
e analogamente
\frac{\partial w}{\partial y} =\frac{D-2y\,N}{D^{2}}, \qquad \frac{\partial w}{\partial z} =\frac{D-2z\,N}{D^{2}}.

Passo 4

As derivadas de x e y em relação a t são
\frac{dx}{dt}=-\operatorname{sen}(t), \qquad \frac{dy}{dt}=\cos(t).
Além disso, do Passo 2 temos \tfrac{dz}{dt}=e^{t}.

Passo 5

Agora substituímos em
\frac{dw}{dt} =\frac{\partial w}{\partial x}\,\frac{dx}{dt} +\frac{\partial w}{\partial y}\,\frac{dy}{dt} +\frac{\partial w}{\partial z}\,\frac{dz}{dt}.
Definindo
N=\cos(t)+\operatorname{sen}(t)+e^{t}, \qquad D= \cos^{2}(t)+\operatorname{sen}^{2}(t)+e^{2t}=1+e^{2t},
fica
\frac{\partial w}{\partial x} =\frac{D-2\cos(t)\,N}{D^{2}}, \quad \frac{\partial w}{\partial y} =\frac{D-2\operatorname{sen}(t)\,N}{D^{2}}, \quad \frac{\partial w}{\partial z} =\frac{D-2e^{t}\,N}{D^{2}}.
Portanto
\frac{dw}{dt} =\frac{D-2\cos t\,N}{D^{2}}\bigl(-\operatorname{sen} t\bigr) +\frac{D-2\operatorname{sen} t\,N}{D^{2}}\bigl(\cos t\bigr) +\frac{D-2e^{t}\,N}{D^{2}}\bigl(e^{t}\bigr).

Resposta

\frac{dz}{dt}=e^{t}, \qquad \frac{dw}{dt} =\frac{-\operatorname{sen} t\,(D-2\cos t\,N)+\cos t\,(D-2\operatorname{sen} t\,N)+e^{t}\,(D-2e^{t}\,N)}{D^{2}},
com N=\cos t+\operatorname{sen} t+e^{t} e D=1+e^{2t}.

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