Ejercicios de cálculo vectorial - Respuesta del ejemplo 19

Sabemos que el producto escalar de las funciones vectoriales F y G viene dado por :

\[ \overrightarrow{F}·\overrightarrow{G}= F_x · G_x + F_y · G_y +F_z · G_z \]
y además, tenemos :

\( \displaystyle \left[ Grad \left(\overrightarrow{F}·\overrightarrow{G}\right)\right]_x = \frac{\partial}{\partial x}\left(F_x · G_x + F_y · G_y +F_z · G_z \right) \)


\( \displaystyle = \frac{\partial G_x}{\partial x} · F_x+ \frac{\partial F_x}{\partial x} · G_x+ \frac{\partial G_y}{\partial x} · F_y+ \frac{\partial F_y}{\partial x} · G_y + \frac{\partial G_z}{\partial x} · F_z+ \frac{\partial F_z}{\partial x} · G_z \)

Expresiones análogas, con derivadas respecto de y y de z, tenemos para:

\[ \displaystyle \left[ Grad \left(\overrightarrow{F}·\overrightarrow{G}\right)\right]_y \qquad ; \qquad \left[ Grad \left(\overrightarrow{F}·\overrightarrow{G}\right)\right]_z \]
Por otra parte, desarrollando la expresión :

\( \displaystyle \overrightarrow{F}\wedge rot \overrightarrow{G} = \left[F_y\left(\frac{\partial G_y}{\partial x} - \frac{\partial G_x}{\partial y}\right) - F_z\left(\frac{\partial G_x}{\partial z} - \frac{\partial G_z}{\partial x}\right) \right]\hat{i} + \)


\( \displaystyle \left[F_z\left(\frac{\partial G_z}{\partial y} - \frac{\partial G_y}{\partial z}\right) - F_x\left(\frac{\partial G_y}{\partial x} - \frac{\partial G_x}{\partial y}\right) \right]\hat{j} + \left[F_x\left(\frac{\partial G_x}{\partial z} - \frac{\partial G_z}{\partial y}\right) - \quad ... \right. \)


\( \displaystyle \left. ... \quad - F_y\left(\frac{\partial G_z}{\partial y} - \frac{\partial G_y}{\partial z}\right) \right]\hat{k} \)

Por razones de brevedad, demostraremos la expresión únicamente para la primera componente. De ese modo, la componente respecto a x, de \(\overrightarrow{G}\wedge rot \overrightarrow{F} \quad \) es :

\[ \left[\overrightarrow{G}\wedge rot \overrightarrow{F}\right]_x = \left[G_y\left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) - F_z\left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \right]\hat{i} \]
Análogamente, las primeras componentes para (F.grad)G y (G.grad)F son, respectivamente :

\[ \left[\left(\overrightarrow{F}· Grad \right)\overrightarrow{G}\right]_x = \frac{\partial G_x}{\partial x}· F_x + \frac{\partial G_x}{\partial y}· F_y + \frac{\partial G_x}{\partial z}· F_z \]

\[ \left[\left(\overrightarrow{G}· Grad \right)\overrightarrow{F}\right]_x = \frac{\partial F_x}{\partial x}· G_x + \frac{\partial F_x}{\partial y}· G_y + \frac{\partial F_x}{\partial z}· G_z \]
Así pues, escribiendo :

\[ \left[\overrightarrow{F}\wedge rot \overrightarrow{G}\right]_x + \left[\overrightarrow{G}\wedge rot \overrightarrow{F}\right]_x + \left[\left(\overrightarrow{F}· Grad \right)\overrightarrow{G}\right]_x + \left[\left(\overrightarrow{G}· Grad \right)\overrightarrow{F}\right]_x \]
y desarrollando, tenemos :

\( \displaystyle \left[F_y\left(\frac{\partial G_y}{\partial x} - \frac{\partial G_x}{\partial y}\right) - F_z\left(\frac{\partial G_x}{\partial z} - \frac{\partial G_z}{\partial x}\right) \right] + \)


\( \displaystyle \left[G_y\left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) - G_z\left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \right] + \)


\( \displaystyle \left( \frac{\partial F_x}{\partial x}· G_x + \frac{\partial F_x}{\partial y}· G_y + \frac{\partial F_x}{\partial z}· G_z\right) + \left( \frac{\partial G_x}{\partial x}· F_x + \frac{\partial G_x}{\partial y}· F_y + \frac{\partial G_x}{\partial z}· F_z\right) \)

Pero, simplificando :

\[ \displaystyle \frac{\partial F_x}{\partial x}· G_x + \frac{\partial F_x}{\partial y}· G_y + \frac{\partial F_x}{\partial z}· G_z + \frac{\partial G_x}{\partial x}· F_x + \frac{\partial G_x}{\partial y}· F_y + \frac{\partial G_x}{\partial z}· F_z = \left[Grad \left(\overrightarrow{F}· \overrightarrow{G}\right)\right]\]
y queda demostrado lo que nos proponíamos.