vector calculus - show field is conservative

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May 12th 2008, 04:11 AM
hunkydory19

vector calculus - show field is conservative

If F = $-2\frac{xi + yj + zk}{(x^2 + y^2 + z^2)^2}$

Show by calculating $\nabla x F$ that vector field F is conservative.

I know that $\nabla x F = Curl F = 0$

And then I do the determinant with i, j and k across the top row, and $\frac{\partial}{\partial x}, \frac{\partial}{\partial y} and \frac{\partial}{\partial z}$ across the second row.

But I'm confused as to what I put along the 3rd row since F is a fraction...

Would i split it up into $\frac{-2x}{(x^2 + y^2 + z^2)^2}, \frac{-2y}{(x^2 + y^2 + z^2)^2}$ and $\frac{-2z}{(x^2 + y^2 + z^2)^2}$

Thanks in advance!
May 12th 2008, 04:27 AM
mr fantastic

Quote:

Originally Posted by hunkydory19

If F = $-2\frac{xi + yj + zk}{(x^2 + y^2 + z^2)^2}$

Show by calculating $\nabla x F$ that vector field F is conservative.

I know that $\nabla x F = Curl F = 0$

And then I do the determinant with i, j and k across the top row, and $\frac{\partial}{\partial x}, \frac{\partial}{\partial y} and \frac{\partial}{\partial z}$ across the second row.

But I'm confused as to what I put along the 3rd row since F is a fraction...

Would i split it up into $\frac{-2x}{(x^2 + y^2 + z^2)^2}, \frac{-2y}{(x^2 + y^2 + z^2)^2}$ and $\frac{-2z}{(x^2 + y^2 + z^2)^2}$

Thanks in advance!

Yes.

The third row gets the components of F. Whatever form they may be in. Including the form of a fraction.
May 12th 2008, 05:13 AM
hunkydory19

Thanks for that mr fantastic, thats what I tried originally but then got stuck at the differentiation part...

Calculating the determinant, (I'll just do the i bit to save time) I got:

$\frac{\partial}{\partial y} \frac{-2z}{(x^2 + y^2 + z^2)^2} - \frac{\partial}{\partial z} \frac{-2y}{(x^2 + y^2 + z^2)^2} i$

But I dont understand how to partially differentiate these...

So for the first partial diff, $\frac{\partial}{\partial y} \frac{-2z}{(x^2 + y^2 + z^2)^2}$

I know you would treat the z's and x's as constants, so would it just go to 0?

And hence all of the terms go to 0?

Thanks again!
May 12th 2008, 05:24 AM
mr fantastic

Quote:

Originally Posted by hunkydory19

[snip]

But I dont understand how to partially differentiate these...

So for the first partial diff, $\frac{\partial}{\partial y} \frac{-2z}{(x^2 + y^2 + z^2)^2}$

I know you would treat the z's and x's as constants, so would it just go to 0?

And hence all of the terms go to 0?

Thanks again!

No. For the particular example you give, note that y is in the denominator. I'd just use the quotient rule. For example:

$\frac{\partial}{\partial y} \frac{-2z}{(x^2 + y^2 + z^2)^2} = \frac{(0)(x^2 + y^2 + z^2)^2 - [2(x^2 + y^2 + z^2)(2y)](-2z)}{[(x^2 + y^2 + z^2)^2]^2} = ......$