Determine \frac{d z}{d t} \mathrm{e} \frac{d z}{d s}, onde:
z=\operatorname{sen}(\theta) \cos (\varphi), \theta=\pi t s, \varphi=\sqrt{t+s}.

Passo 1

O enunciado define \displaystyle z=\sin(\theta)\,\cos(\varphi),
com \displaystyle \theta=\pi\,t\,s e \displaystyle \varphi=\sqrt{t+s},
e pede as derivadas parciais \frac{\partial z}{\partial t} e \frac{\partial z}{\partial s}.
O conteúdo aqui é aplicação da regra da cadeia para funções compostas de duas variáveis.

Passo 2

Calcular as derivadas parciais de \theta e \varphi em relação a t:
\frac{\partial \theta}{\partial t} = \pi\,s, \qquad \frac{\partial \varphi}{\partial t} = \frac{1}{2\sqrt{t+s}}.

Passo 3

Aplicar a regra da cadeia para \frac{\partial z}{\partial t}:
\frac{\partial z}{\partial t} = \frac{\partial}{\partial t}\bigl[\sin(\theta)\cos(\varphi)\bigr] = \cos(\theta)\cos(\varphi)\,\frac{\partial \theta}{\partial t} +\sin(\theta)\,\bigl(-\sin(\varphi)\bigr)\,\frac{\partial \varphi}{\partial t}.
Substituindo as derivadas de \theta e \varphi:
\frac{\partial z}{\partial t} = \pi s\,\cos(\theta)\cos(\varphi) - \frac{\sin(\theta)\,\sin(\varphi)}{2\sqrt{t+s}}.

Passo 4

Calcular as derivadas parciais de \theta e \varphi em relação a s:
\frac{\partial \theta}{\partial s} = \pi\,t, \qquad \frac{\partial \varphi}{\partial s} = \frac{1}{2\sqrt{t+s}}.

Passo 5

Aplicar a regra da cadeia para \frac{\partial z}{\partial s}:
\frac{\partial z}{\partial s} = \cos(\theta)\cos(\varphi)\,\frac{\partial \theta}{\partial s} +\sin(\theta)\,\bigl(-\sin(\varphi)\bigr)\,\frac{\partial \varphi}{\partial s} = \pi t\,\cos(\theta)\cos(\varphi) - \frac{\sin(\theta)\,\sin(\varphi)}{2\sqrt{t+s}}.

Resposta

\frac{dz}{dt} = \pi s\,\cos\bigl(\pi t s\bigr)\cos\bigl(\sqrt{t+s}\bigr) - \frac{\sin\bigl(\pi t s\bigr)\,\sin\bigl(\sqrt{t+s}\bigr)}{2\sqrt{t+s}},
\frac{dz}{ds} = \pi t\,\cos\bigl(\pi t s\bigr)\cos\bigl(\sqrt{t+s}\bigr) - \frac{\sin\bigl(\pi t s\bigr)\,\sin\bigl(\sqrt{t+s}\bigr)}{2\sqrt{t+s}}.

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