Find div v and curl v of v Vector

[Rubik]

Homework Statement

Find the div v and curl v of v = (x² + y² + z²)^-3/2(xi + yj + zk)

Homework Equations

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

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?

 

[issacnewton]

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

 

[Lancelot59]

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

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

 

[Rubik]

The del operator so instead I should be using (\partial/\partialx)i + (\partial/\partialy)j + (\partial/\partialz)k??

 

[lineintegral1]

The del operator so instead I should be using (\partial/\partialx)i + (\partial/\partialy)j + (\partial/\partialz)k??

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.

 

[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?

 

[Rubik]

So if differentiate with respect to x that is my (\partial/\partialx)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?

 

[HallsofIvy]

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

u = (x² + y² + z²)^-3/2
u' = -3(x² + y² + z²)^-5/2

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

[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?

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

 

[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?

 

[Lancelot59]

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 vThat 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?

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)