contestada

If f is a scalar field and F, G are vector fields, then f F, F · G, and F × G are defined by the following. (f F)(x, y, z) = f(x, y, z) F(x, y, z) (F · G)(x, y, z) = F(x, y, z) · G(x, y, z) (F × G)(x, y, z) = F(x, y, z) × G(x, y, z) Find an identical expression, assuming that the appropriate partial derivatives exist and are continuous.

Respuesta :

usamakld99

Answer:

Step-by-step explanation:

Consider curl [tex](fF)[/tex] where [tex]f[/tex] is a scalar function and F is a vector function

                  [tex]F=F_{1}i +F_{2}j +F_{3}k[/tex]

                     i         j          k                

[tex]curl(fF) = \frac{\partial}{\partial x}[/tex]      [tex]\frac{\partial}{\partial y}[/tex]      [tex]\frac{\partial}{\partial z}[/tex]        

                  [tex]fF_{1}[/tex]    [tex]fF_{2}[/tex]    [tex]fF_{3}[/tex]  

[tex]curl(fF)=i(\frac{\partial}{\partial y}(fF_{3})-\frac{\partial}{\partial z}(fF_{2}))-j(\frac{\partial}{\partial x}(fF_{3})-\frac{\partial}{\partial z}(fF_{1}))+k(\frac{\partial}{\partial x}(fF_{2})-\frac{\partial}{\partial y}(fF_{1}))[/tex]

[tex]=i(\frac{\partial}{\partial y}(F_{3})+\frac{\partial}{\partial y}(f)-\frac{\partial}{\partial z}(F_{2})-\frac{\partial}{\partial z}(f))-j(\frac{\partial}{\partial x}(F_{3})+\frac{\partial}{\partial x}(f)-\frac{\partial}{\partial z}(F_{1})-\frac{\partial}{\partial z}(f))+k(\frac{\partial}{\partial x}(F_{2})+\frac{\partial}{\partial x}(f)-\frac{\partial}{\partial y}(F_{1})-\frac{\partial}{\partial y}(f))[/tex]

[tex]curl(fF)=f(\Delta\times F)+(\Delta f)\times F[/tex]