# Divergent and rotational

• Apr 24th 2009, 11:29 AM
Apprentice123
Divergent and rotational
• Apr 24th 2009, 11:41 AM
Calculus26
The divergence is correct

Not sure what you have in mind for rotF

but it is 5z^2 i +3xz^2 j +4 y^4x k

The middle row of the determinant are the operators d/dx d/dy and d/dz
and the last row are the x , y , and z components of F not the parial derivatives
• Apr 24th 2009, 11:45 AM
Apprentice123
In the middle row is the partial derivatives (div) ?
In the last line components i, j and k of F ?

I solved this way not correct ?

• Apr 24th 2009, 11:52 AM
Calculus26
No the middle row is just the operators d/dx etc

Write F =f i + g j + h k

then the last row is f - g - h

Then rotF = (dh/dy-dg/dz)i - (dh/dx-df/dz) j + (dg/dx-df/dy)k
• Apr 24th 2009, 12:12 PM
Apprentice123
Thank you very much. Look this is correct

$F(x,y,z) = ln(x)i + e^{xyz}j + arctg(\frac{x}{z})k$

$div = x^{-1} + xze^{xyz} + \frac{x}{(x^2+y^2)}$

$rot = -xye^{xyz}i + \frac{z}{x^2+z^2}j + yze^{xyz}k$
• Apr 24th 2009, 12:23 PM
Calculus26
Yes except the last term in your calculation of divF should be

-x/(x^2+z^2)
• Apr 24th 2009, 12:58 PM
Apprentice123
Thanks

How do I calculate : $\bigtriangledown . (F x G)$

$F(x,y,z) = 2xi + j + 4yk$
$G(x,y,z) = xi + yj - zk$
• Apr 24th 2009, 01:22 PM
Calculus26
Compute FxG then compute the divergence of the result
• Apr 24th 2009, 01:57 PM
Calculus26
you can also use del*(FxG) = G*(del x F)-F*(del x G)
• Apr 24th 2009, 01:59 PM
Apprentice123
The symbol $\bigtriangledown$ means divergent ?

If I have: $\bigtriangledown . ( \bigtriangledown x F)$ and have F how can I calculate ?

How can I verify that the vector projection $r$ have the property:
1) $\bigtriangledown ||r|| = \frac{r}{||r||}$

• Apr 24th 2009, 02:13 PM
Calculus26
http://www.mathhelpforum.com/math-he...8739824f-1.gif is an operator and means d/dx i +d/dy j +d/dz k where d/dx d/dy d/dz ar the partial derivative operators

http://www.mathhelpforum.com/math-he...8739824f-1.gif f is the gradient where f is a scalar valued function

http://www.mathhelpforum.com/math-he...8739824f-1.gif X F is the curl or rotation where F is a vector field

http://www.mathhelpforum.com/math-he...8739824f-1.gif*F is the divergence where F is a vector field * means dot product

From the previous identity http://www.mathhelpforum.com/math-he...8739824f-1.gif* (http://www.mathhelpforum.com/math-he...8739824f-1.gif x F) = 0

r= xi +yj +zk

||r|| = (x^2 +y^2 +z^2)^(1/2)

Compute http://www.mathhelpforum.com/math-he...8739824f-1.gif ||r|| and the result follows directly
• Apr 24th 2009, 03:13 PM
Apprentice123
Thanks, you're helping me a lot

check my learning:

1) $\bigtriangledown . (FxG)$
$F(x,y,z) = yzi + xzj + xyk$
$G(x,y,z) = xyj + xyzk$

My solution:

$FxG = (x^2yz^2 - x^2y^2)i + (-xy^2z^2)j + (xy^2z)k$
$div = (2xyz^2 - 2xy^2)i + (-2xyz^2)j + (xy^2)k$

2) $\bigtriangledown . (\bigtriangledown x F)$
$F(x,y,z) = sen(x)i + cos(x-y)j + zk$

My solution:

$rotF = 0i - 0j + (-sen(x-y))k$
$div(rotF) = 0i + 0j + 0k$

3) Check that: $\bigtriangledown ||r|| = \frac{r}{||r||}$
$r = xi + yj + zk$

My solution:

$div||r|| = \frac{x}{\sqrt{x^2+y^2+z^2}} + \frac{y}{\sqrt{x^2+y^2+z^2}} + \frac{z}{\sqrt{x^2+y^2+z^2}}$

$\bigtriangledown ||r|| = \frac{r}{||r||}$ is correct

• Apr 24th 2009, 04:01 PM
Calculus26
Looks good you have learned well.

Note

http://www.mathhelpforum.com/math-he...f0ae7419-1.gif is always 0 regardless of what F is. so you don't have to calculate but it was a good exercise to verify this.
• Apr 24th 2009, 05:24 PM
Apprentice123
thanks. good Note