QUESTÃO 2 [valor: 3,0 pontos] Resolva cada item a seguir.
Sabendo que frac{\mathrm{d} z}{\mathrm{dt}}=frac{partial f}{partial x} cdot frac{\mathrm{d} x}{\mathrm{dt}}+frac{partial f}{partial y} cdot frac{\mathrm{d} y}{\mathrm{d} t}, calcule frac{\mathrm{dz}}{\mathrm{dt}} para \mathrm{z}=frac{x}{y^{'}}, x(t)=e^{-t} e y(t)=ln t.

Passo 1

O enunciado pede para usar a regra da cadeia
\frac{dz}{dt} = \frac{\partial f}{\partial x}\,\frac{dx}{dt} + \frac{\partial f}{\partial y}\,\frac{dy}{dt}
no caso em que z=f(x,y)=\frac{x}{y}, com x(t)=e^{-t} e y(t)=\ln t. Vamos precisar calcular as derivadas parciais de f e as derivadas de x(t) e y(t).

Passo 2

Derivadas parciais de f(x,y)=\frac{x}{y}:
\frac{\partial f}{\partial x} = \frac{1}{y}, \qquad \frac{\partial f}{\partial y} = -\frac{x}{y^2}.

Passo 3

Derivadas de x(t) e y(t):
x(t)=e^{-t} \quad\Longrightarrow\quad \frac{dx}{dt} = -e^{-t},
y(t)=\ln t \quad\Longrightarrow\quad \frac{dy}{dt} = \frac{1}{t}.

Passo 4

Aplicando a regra da cadeia:
\frac{dz}{dt} = \frac{\partial f}{\partial x}\,\frac{dx}{dt} + \frac{\partial f}{\partial y}\,\frac{dy}{dt} = \frac{1}{y}\bigl(-e^{-t}\bigr) + \Bigl(-\frac{x}{y^2}\Bigr)\frac{1}{t}.

Passo 5

Substituindo x=e^{-t} e y=\ln t:
\frac{dz}{dt} = -\frac{e^{-t}}{\ln t} - \frac{e^{-t}}{t\,(\ln t)^2} = -\,\frac{e^{-t}\bigl(t\ln t +1\bigr)}{t\,(\ln t)^2}.

Resposta

\frac{dz}{dt} = -\frac{e^{-t}}{\ln t} - \frac{e^{-t}}{t\,(\ln t)^2} = -\,\frac{e^{-t}\bigl(t\ln t +1\bigr)}{t\,(\ln t)^2}.

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