Homework Help: Div v and curl v

1. Sep 19, 2011

Rubik

1. The problem statement, all variables and given/known data
Find the div v and curl v of v = (x2 + y2 + z2)-3/2(xi + yj + zk)
2. Relevant equations

div v = $\nabla$ $\cdot$ v and $\nabla$ $\times$ v

3. The attempt at a solution
I am just confused and drawing a blank in basic algebra

Is it right to expand v like this

v = x-3 + y-3 + z-3(xi + yj + zk)
= (z-3xi + z-3yj + z-2k)

is the right vector for v?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 19, 2011

issacnewton

$$(x^2+y^2+z^2)^{-3/2}\neq (x^{-3}+y^{-3}+z^{-3})$$

3. Sep 19, 2011

Lancelot59

There's your problem. Also, that's not really the right approach. Do you know what del represents?

4. Sep 21, 2011

Rubik

The del operator so instead I should be using ($\partial$/$\partial$x)i + ($\partial$/$\partial$y)j + ($\partial$/$\partial$z)k??

5. Sep 21, 2011

lineintegral1

Correct. Apply this operator to your vector $v$. Are you having trouble understanding what this does to your vector?

And just as a correction to your OP, $div(v)=\triangledown \cdot v$ only; it is not equal to the dot product and the cross product. The curl is the cross product.

6. Sep 21, 2011

Rubik

Yes I am struggling to apply this to my vector v? Is it right that I have to apply the product rule when applying the del operator to my vector?

7. Sep 21, 2011

Rubik

So if differentiate with respect to x that is my ($\partial$/$\partial$x)i term is this remotely correct

u = (x2 + y2 + z2)-3/2
u' = -3(x2 + y2 + z2)-5/2
v = xi + yj + zk
v' = 1

then using the product rule

(x2 + y2 + z2)-3/2 - 3x(x2 + y2 + z2)-5/2(x + y + z)

Then I add in with respect to y and z or is the completely off track?

8. Sep 21, 2011

HallsofIvy

This is meaningless, u' is not defined for u a function of more than one variable.

Pretty much. Do you not know how to take a partial derivative? I don't see any reference to partial derivatives!

9. Sep 21, 2011

Rubik

Sorry what I meant was using the product rule to take the partial derivative with respect to x then y then z and then putting it all together and getting $\nabla$$\cdot$ v

That is
the first term = (x2 + y2 + z2)-3/2
then we take the partial derivative of the first tems with respect to x gives
= -3x(x2 + y2 + z2)-5/2

Now the second term is (xi + yj + zk)
and the partial derivative with respect to x is 1 so using the product rule the first and second term come together to give

-3x(x2 + y2 + z2)-5/2 + (x2 + y2 + z2)-3/2

And doing the same by taking the partial derivative with respect to y and z then gives

$\nabla$$\cdot$ v = (x + y + z)(x2 + y2 + z2)-5/2(-3x - 3y - 3z) + 3(x2 + y2 + z2)-3/2

Does that look right?

10. Sep 22, 2011

Lancelot59

Not as such no... I got the divergence to be zero. Using straight vector notation here we have:

$$\nabla \cdot \vec{v} = (\frac{\delta}{\delta x},\frac{\delta}{\delta y},\frac{\delta}{\delta z}) \cdot (x(x^2+y^2+z^2)^{-3/2},y(x^2+y^2+z^2)^{-3/2},z(x^2+y^2+z^2)^{-3/2})$$

Your partial derivative with respect to x was different. I got an x2 in front of the term with power (-5/2)

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