Finding divF and curl F for F = r/r

[ineedmunchies]

Homework Statement

F = \frac{<b>r</b>}{r}

Find divF
and curl F

Homework Equations

r = x\widehat{i} + y\widehat{j} + z\widehat{k}

r = \sqrt{(x^{2} + y^{2} + z^{2})}

The Attempt at a Solution

F = \frac{x}{(\sqrt{x^{2} + y^{2} + z^{2}})}\widehat{i} + \frac{y}{(\sqrt{x^{2} + y^{2} + z^{2}})}\widehat{j} + \frac{z}{(\sqrt{x^{2} + y^{2} + z^{2}})}\widehat{k}


div F = \frac{\partial}{\partial x} (x(x^{2} + y^{2} + z^{2})^{\frac{-1}{2}} + \frac{\partial}{\partial y}(y(x^{2} + y^{2} + z^{2})^{\frac{-1}{2}} + \frac{\partial}{\partial z}(z(x^{2} + y^{2} + z^{2})^{\frac{-1}{2}}

Take the partial derivative of x first.

let u = x^{2} + y^{2} + z^{2}
and a = xu^{2}

\frac{\partial u}{\partial x} = 2x
\frac{\partial a}{\partial u} = \frac{-1}{2}xu^{\frac{-3}{2}}

\frac{\partial a}{\partial x} = -x^{2}u^{\frac{-3}{2}}
= -x^{2}(\frac{1}{\sqrt{(x^{2} + y^{2} + z^{2})}})^{3}

The other derivatives would give similar answers, and the final answer would be

-\frac{x^{2}}{r^{3}}-\frac{y^{2}}{r^{3}}-\frac{z^{2}}{r^{3}}

This is apparently the incorrect answer, can anybody help?

 

[ineedmunchies]

The three parts that didnt come up are partial derivatives; du/dx da/du and da/dx

And the r as the numerator in the first equation is meant to be in bold to denote that it is a vector.

 

[tiny-tim]

Hi ineedmunchies!

(here's √ and ² and ∂ for you to copy-and-paste
oh … and tags like B don't work in LaTeX
hmm … nice LaTeX apart from that, though! )

let u = x^{2} + y^{2} + z^{2}
and a = xu^{2}

\frac{\partial u}{\partial x} = 2x
\frac{\partial a}{\partial u} = \frac{-1}{2}xu^{\frac{-3}{2}}

ooooh … this is horrible … can't look … need air …

Just use r² = x² + y² + z², so 2r∂r/∂x = … ? ​

 

[ineedmunchies]

hmmm sorry where are you getting 2r∂r/∂x from?

 

[tiny-tim]

hmmm sorry where are you getting 2r∂r/∂x from?

Differentiating r² with respect to x.

(Chain rule: ∂r²/∂x = dr²/dr ∂r/∂x = 2r ∂r/∂x.)

(Can you read r² on your computer? if not, it's supposed to be r^2. )

 

[ineedmunchies]

ah actually never mind, I've worked it out. I had left out part of the product rule in my differentiation. I would post the full worked solution but its quite long and the forums seem to be running slow for me today.