In each of the following parts, find a numerical value for div \mathbf{D} at the point specified:
\mathbf{D}=2 r \sin \theta \cos \varphi \mathbf{a}_{r}+r \cos \theta \cos \varphi \mathbf{a}_{\theta}-r \sin \varphi \mathbf{a}_{\varphi} at P_{C}\left(r=1.5, \theta=30^{\circ}, \varphi=50^{\circ}\right).

Passo 1

O enunciado nos dá o campo vetorial em coordenadas esféricas
\mathbf{D}=2r\sin\theta\cos\varphi\,\mathbf{a}_r \;+\; r\cos\theta\cos\varphi\,\mathbf{a}_\theta \;-\; r\sin\varphi\,\mathbf{a}_\varphi
e pede \nabla\cdot\mathbf{D} no ponto P_C(r=1{,}5,\theta=30^\circ,\varphi=50^\circ). Vamos usar a fórmula do divergente em esféricas.

Passo 2

A fórmula do divergente em coordenadas esféricas é
\nabla\cdot \mathbf{D} =\frac{1}{r^2}\frac{\partial}{\partial r}\bigl(r^2D_r\bigr) +\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\bigl(\sin\theta\,D_\theta\bigr) +\frac{1}{r\sin\theta}\frac{\partial D_\varphi}{\partial\varphi}.

Passo 3

Identificamos
D_r=2r\sin\theta\cos\varphi,\quad D_\theta=r\cos\theta\cos\varphi,\quad D_\varphi=-r\sin\varphi.
Calculamos cada termo:
\frac{1}{r^2}\frac{\partial}{\partial r}\bigl(r^2D_r\bigr) =\frac{1}{r^2}\frac{\partial}{\partial r}\bigl(2r^3\sin\theta\cos\varphi\bigr) =\frac{6r^2\sin\theta\cos\varphi}{r^2} =6\sin\theta\cos\varphi.
\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\bigl(\sin\theta\,D_\theta\bigr) =\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\bigl(r\sin\theta\cos\theta\cos\varphi\bigr) =\frac{\cos\varphi\,( \cos^2\theta-\sin^2\theta)}{\sin\theta} =\frac{\cos\varphi\cos2\theta}{\sin\theta}.
\frac{1}{r\sin\theta}\frac{\partial D_\varphi}{\partial\varphi} =\frac{1}{r\sin\theta}\bigl(-r\cos\varphi\bigr) =-\frac{\cos\varphi}{\sin\theta}.

Passo 4

Somando os termos:
\nabla\cdot\mathbf{D} =6\sin\theta\cos\varphi +\frac{\cos\varphi\cos2\theta}{\sin\theta} -\frac{\cos\varphi}{\sin\theta} =6\sin\theta\cos\varphi +\frac{\cos\varphi(\cos2\theta-1)}{\sin\theta}.
Como \cos2\theta-1=-2\sin^2\theta, segue
\nabla\cdot\mathbf{D} =6\sin\theta\cos\varphi-2\sin\theta\cos\varphi =4\sin\theta\cos\varphi.

Passo 5

No ponto dado \theta=30^\circ, \varphi=50^\circ:
\nabla\cdot\mathbf{D} =4\sin30^\circ\cos50^\circ =4\cdot\tfrac12\cdot\cos50^\circ =2\cos50^\circ \approx2\cdot0{,}6428 \approx1{,}286.

Resposta

1,286

Exercícios relacionados