## anonymous 4 years ago Prove the following vector identities letting: V = u i + v j + w k and ∇= ∂╱∂x i + ∂╱∂y j + ∂╱∂z k that: ∇∙(fV̅) = f (∇∙V̅) + V̅ ∙∇f

1. anonymous

What is fV ?

2. anonymous

f is a scalar

3. anonymous

v is the vector

4. anonymous

I am aware. I meant write down what it actually is.

5. anonymous

Explicitly

6. anonymous

f (u)i + f(v) j + f(w)k

7. anonymous

f is multiply u, v, and w, right?

8. anonymous

yes

9. anonymous

So what's the divergence of that?

10. anonymous

∂╱∂x i + ∂╱∂y j + ∂╱∂z k

11. anonymous

No, what is the divergence of the expression you wrote for fV ?

12. anonymous

13. anonymous

$f\vec{V} = fu\vec{i} + fv\vec{j} + fw\vec{k}$ what is $\vec{\nabla} \cdot (f\vec{V})$ ?

14. anonymous

d f(u)/dx i + d f(v)/dy j + d f(w)/dz k

15. anonymous

the i, j, and k are gone.

16. anonymous

Okay, so expand that out.

17. anonymous

so just d f(u)/dx + d f(v)/dy + d f(w)/dz

18. anonymous

That's right. Now show that that equals the identity that you were given above.

19. anonymous

well the right hand side of the equation is where the problems start

20. anonymous

$\frac{\partial (fu)}{\partial x} = \frac{ \partial f}{\partial x} u + f\frac{\partial u}{\partial x}$ etc....

21. anonymous

those are equal what u just wrote above?

22. anonymous

That's only the first term...

23. anonymous

oh i think i just clicked what ur saying

24. anonymous

okay..so the entire left side is now expanded

25. anonymous

okay....i think i see where this is going now

26. anonymous

if i expand the right side it's gonna come out the same way huh?

27. anonymous

Yeah but it would probably be better if you just grouped the terms on the left. $f \frac{\partial u}{\partial x} + f\frac{\partial v}{\partial y} + f\frac{\partial w}{\partial z} = f(\vec{\nabla} \cdot \vec{V})$ and so forth..

28. anonymous

very good

29. anonymous

thank you soo soo much