Question Se z=e^{\frac{x}{y}} \operatorname{sen}\left(\frac{x}{y}\right)+e^{\frac{y}{x}} \cos \left(\frac{y}{x}\right), mostre que x \frac{\delta z}{\delta x}+y \frac{\delta z}{\delta y}=0

Passo 1

O enunciado dá
z = e^{\frac{x}{y}}\sin\!\bigl(\tfrac{x}{y}\bigr)\;+\;e^{\frac{y}{x}}\cos\!\bigl(\tfrac{y}{x}\bigr)
e pede para mostrar que
x\frac{\partial z}{\partial x} \;+\;y\frac{\partial z}{\partial y} \;=\;0.
O conteúdo aqui é cálculo de derivadas parciais combinadas com a regra da cadeia.

Passo 2

Para simplificar, defina
u = \frac{x}{y}, \qquad v = \frac{y}{x}.
Então
z = e^{u}\sin(u) \;+\; e^{v}\cos(v).
Precisamos de
\frac{\partial u}{\partial x} = \frac{1}{y}, \quad \frac{\partial u}{\partial y} = -\frac{x}{y^2}, \quad \frac{\partial v}{\partial x} = -\frac{y}{x^2}, \quad \frac{\partial v}{\partial y} = \frac{1}{x}.

Passo 3

Calculemos \displaystyle \frac{\partial z}{\partial x} usando a regra da cadeia:
\frac{\partial z}{\partial x} = \frac{d}{du}\bigl(e^u\sin u\bigr)\cdot\frac{\partial u}{\partial x} + \frac{d}{dv}\bigl(e^v\cos v\bigr)\cdot\frac{\partial v}{\partial x}.
Agora,
\frac{d}{du}\bigl(e^u\sin u\bigr) = e^u\sin u + e^u\cos u = e^u\bigl(\sin u + \cos u\bigr),
\frac{d}{dv}\bigl(e^v\cos v\bigr) = e^v\cos v - e^v\sin v = e^v\bigl(\cos v - \sin v\bigr).
Logo
\frac{\partial z}{\partial x} = e^u(\sin u + \cos u)\,\frac{1}{y} \;+\;e^v(\cos v - \sin v)\,\Bigl(-\frac{y}{x^2}\Bigr).

Passo 4

Multiplicando por x, obtemos
x\frac{\partial z}{\partial x} = \frac{x}{y}\,e^u(\sin u + \cos u) \;-\;\frac{xy}{x^2}\,e^v(\cos v - \sin v) = \frac{x}{y}e^u(\sin u + \cos u) \;-\;\frac{y}{x}e^v(\cos v - \sin v).

Passo 5

Agora \displaystyle \frac{\partial z}{\partial y}:
\frac{\partial z}{\partial y} = e^u(\sin u + \cos u)\,\frac{\partial u}{\partial y} + e^v(\cos v - \sin v)\,\frac{\partial v}{\partial y},
isto é
\frac{\partial z}{\partial y} = e^u(\sin u + \cos u)\,\Bigl(-\frac{x}{y^2}\Bigr) \;+\;e^v(\cos v - \sin v)\,\frac{1}{x}.

Passo 6

Multiplicando por y, fica
y\frac{\partial z}{\partial y} = -\frac{xy}{y^2}\,e^u(\sin u + \cos u) \;+\;\frac{y}{x}\,e^v(\cos v - \sin v) = -\frac{x}{y}e^u(\sin u + \cos u) \;+\;\frac{y}{x}e^v(\cos v - \sin v).

Passo 7

Somando as duas partes:
x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = \Bigl(\tfrac{x}{y}e^u(\sin u + \cos u)-\tfrac{y}{x}e^v(\cos v-\sin v)\Bigr) + \Bigl(-\tfrac{x}{y}e^u(\sin u + \cos u)+\tfrac{y}{x}e^v(\cos v-\sin v)\Bigr) = 0.

Resposta

x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = 0.

Exercícios relacionados