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Vector Calculus: Identity

  1. Apr 29, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]\widetilde{F}[/itex](r)=F1(r)i+F2(r)j+F3(r)k
    [itex]\hat{r}[/itex]=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:
    [itex]\nabla[/itex][itex]\cdot[/itex]F = [itex]\hat{r}[/itex][itex]\cdot[/itex]dF/dr.



    2. Relevant equations



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

    Then, Start on the RHS.
    [itex]\hat{r}[/itex][itex]\cdot[/itex]dF/dr
    =[itex]\hat{r}[/itex][itex]\cdot[/itex](dF1/dr,dF2/dr,dF3/dr)
    =((x,y,z)/r)[itex]\cdot[/itex](dF1/dr,dF2/dr,dF3/dr)
    Now, i use the chain rule here.
    =((x,y,z)/r)[itex]\cdot[/itex](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?
    Thanks in advanced!
     
  2. jcsd
  3. Apr 29, 2013 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

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

    Secondly, there is a dot in between, which means that you should get a scalar and not a vector like you have written.
     
  4. Apr 30, 2013 #3
    ops! So my definition is wrong! That's why i can't do it further more!
    I can prove it now!
    thanks CompuChip
     
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