Vector calculus identity proof.

rock.freak667
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Homework Statement



Let f(x,y,z) be a function of three variables and G(x,y,z) be a vector field defined in 3D space. Prove the identity:

div(fG)= f*div(G)+G*grad(f)

Homework Equations



For F=Pi +Qj+Rk

div(F)=dF/dx + dQ/dy + dR/dz

grad(F)=dF/dx i + dQ/dy j + dR/dz k

The Attempt at a Solution




My problem starts with how do I find fG? Because I am thinking that f is a vector with i,j,k components and so is G. So fG should be the dot product of f and G, which gives a scalar, and one can't get the divergence of a scalar :confused: (Since this is an identity, I know somewhere I am missing some elementary fact)
 
on Phys.org
No, [itex]f(x,y,z)[/itex] is a scalar function, [itex]\textbf{G}(x,y,z)[/itex] is a vector field.

So, [tex]f\textbf{G}=f(x,y,z)G_x(x,y,z)\textbf{i}+f(x,y,z)G_y(x,y,z)\textbf{j}+f(x,y,z)G_z(x,y,z)\textbf{k}[/tex]
 
gabbagabbahey said:
No, [itex]f(x,y,z)[/itex] is a scalar function, [itex]\textbf{G}(x,y,z)[/itex] is a vector field.

So, [tex]f\textbf{G}=f(x,y,z)G_x(x,y,z)\textbf{i}+f(x,y,z)G_y(x,y,z)\textbf{j}+f(x,y,z)G_z(x,y,z)\textbf{k}[/tex]

:biggrin: I got it now!
 

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