Calcular \left|\begin{array}{ll}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \varphi} \\\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \varphi}\end{array}\right|, se x=r \cos \varphi e y=r \operatorname{sen} \varphi.

Passo 1

O enunciado define a transformação de coordenadas polares para cartesianas por
x = r\cos\varphi, \qquad y = r\sin\varphi
e pede o determinante da matriz Jacobiana
J = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \varphi} \\\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \varphi} \end{vmatrix}.
Para resolver, precisamos calcular as derivadas parciais de x e y em relação a r e a \varphi.

Passo 2

Calculamos cada derivada parcial:
\frac{\partial x}{\partial r} = \frac{\partial}{\partial r}\bigl(r\cos\varphi\bigr) = \cos\varphi,
\frac{\partial x}{\partial \varphi} = \frac{\partial}{\partial \varphi}\bigl(r\cos\varphi\bigr) = -r\sin\varphi,
\frac{\partial y}{\partial r} = \frac{\partial}{\partial r}\bigl(r\sin\varphi\bigr) = \sin\varphi,
\frac{\partial y}{\partial \varphi} = \frac{\partial}{\partial \varphi}\bigl(r\sin\varphi\bigr) = r\cos\varphi.

Passo 3

Substituindo na definição de J e usando a fórmula do determinante de matriz 2\times2,
J = \begin{vmatrix} \cos\varphi & -r\sin\varphi \\[6pt] \sin\varphi & r\cos\varphi \end{vmatrix} = \cos\varphi\,(r\cos\varphi)\;-\;(-r\sin\varphi)\,(\sin\varphi).]

Passo 4

Agora basta agrupar e usar a identidade trigonométrica \cos^2\varphi + \sin^2\varphi = 1:
J = r\cos^2\varphi + r\sin^2\varphi = r\bigl(\cos^2\varphi + \sin^2\varphi\bigr) = r.

Resposta

r

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