# Homework Help: Div and curl

1. May 14, 2008

### ineedmunchies

1. The problem statement, all variables and given/known data
F = $$\frac{r}{r}$$

Find divF
and curl F

2. Relevant equations
r = x$$\widehat{i}$$ + y$$\widehat{j}$$ + z$$\widehat{k}$$

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

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

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

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?

2. May 14, 2008

### 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.

3. May 14, 2008

### 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! )

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

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

4. May 14, 2008

### ineedmunchies

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

5. May 14, 2008

### tiny-tim

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. )

6. May 14, 2008

### 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.