# Vector Calculus: Identity

1. Apr 29, 2013

### yy205001

1. The problem statement, all variables and given/known data
$\widetilde{F}$(r)=F1(r)i+F2(r)j+F3(r)k
$\hat{r}$=r/r
r(x,y,z)=xi+yj+zk, r=abs(r)=sqrt(x2+y2+z2)
(Hint: The chain rule will be helpful for this question.)

Show that:
$\nabla$$\cdot$F = $\hat{r}$$\cdot$dF/dr.

2. Relevant equations

3. The attempt at a solution
My attempt:
First, $\nabla$$\cdot$F=(dF1/dr,dF2/dr,dF3/dr)

Then, Start on the RHS.
$\hat{r}$$\cdot$dF/dr
=$\hat{r}$$\cdot$(dF1/dr,dF2/dr,dF3/dr)
=((x,y,z)/r)$\cdot$(dF1/dr,dF2/dr,dF3/dr)
Now, i use the chain rule here.
=((x,y,z)/r)$\cdot$(dF1/dx*dx/dr, dF2/dy*dy/dr, dF3/dz*dz/dr)

And i cant do further more here, can anyone help me on this?

2. Apr 29, 2013

### CompuChip

I'm pretty sure that's not how the gradient works.
First of all, the "definition" is
$$\vec\nabla = \left( \frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz} \right)$$

Secondly, there is a dot in between, which means that you should get a scalar and not a vector like you have written.

3. Apr 30, 2013

### yy205001

ops! So my definition is wrong! That's why i can't do it further more!
I can prove it now!
thanks CompuChip