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Div and curl

  • #1

Homework Statement


F = [tex]\frac{r}{r}[/tex]

Find divF
and curl F

Homework Equations


r = x[tex]\widehat{i}[/tex] + y[tex]\widehat{j}[/tex] + z[tex]\widehat{k}[/tex]

r = [tex]\sqrt{(x^{2} + y^{2} + z^{2})}[/tex]

The Attempt at a Solution




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


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

Take the partial derivative of x first.

let u = [tex]x^{2} + y^{2} + z^{2}[/tex]
and a = [tex] xu^{2}[/tex]

[tex]\frac{\partial u}{\partial x}[/tex] = 2x
[tex]\frac{\partial a}{\partial u}[/tex] = [tex]\frac{-1}{2}[/tex]x[tex]u^{\frac{-3}{2}}[/tex]

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

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

-[tex]\frac{x^{2}}{r^{3}}[/tex]-[tex]\frac{y^{2}}{r^{3}}[/tex]-[tex]\frac{z^{2}}{r^{3}}[/tex]

This is apparently the incorrect answer, can anybody help?
 

Answers and Replies

  • #2
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
tiny-tim
Science Advisor
Homework Helper
25,832
249
Hi ineedmunchies! :smile:

(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! :smile:)

let u = [tex]x^{2} + y^{2} + z^{2}[/tex]
and a = [tex] xu^{2}[/tex]

[tex]\frac{\partial u}{\partial x}[/tex] = 2x
[tex]\frac{\partial a}{\partial u}[/tex] = [tex]\frac{-1}{2}[/tex]x[tex]u^{\frac{-3}{2}}[/tex]
ooooh … this is horrible … can't look … need air:cry:

Just use r² = x² + y² + z², so 2r∂r/∂x = … ? :smile:
 
  • #4
hmmm sorry where are you getting 2r∂r/∂x from?
 
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
249
hmmm sorry where are you getting 2r∂r/∂x from?
Differentiating r² with respect to x. :smile:

(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. :redface:)
 
  • #6
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.
 

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