Multivariable Calc: ∇ ∙ (r^3 * ȓ )

[toboldlygo]

Homework Statement

Evaluate: ∇ • (r^3 * ȓ ) (del dot (r^3 times vector r)), where r = sqrt(x^2+y^2+z^2) and vector r = (x, y, z)

The Attempt at a Solution

[/B]
So, taking the partial derivative of the x component, I got 2r^3/2. Doing the same thing for the y, z components I got a similar answer. This gives me a solution of ∇ • (r^3 * ȓ ) = 6r^3/2. However, according to my professor, the actual answer is just 6r^3. I don't know where I'm going wrong in my calculations, and any and all help would be appreciated. Thank you!

 

[Dr. Courtney]

Try the del operator in spherical coordinates.

 

[toboldlygo]

Try the del operator in spherical coordinates.

I thought about doing that, but my professor always uses r = ||r|| where r is a vector (x, y, z), and I've used this same definition of r previously in the same worksheet to get the right answer. However, if I were to do that, how would I set that problem up?

 

[Ray Vickson]

Homework Statement

Evaluate: ∇ • (r^3 * ȓ ) (del dot (r^3 times vector r)), where r = sqrt(x^2+y^2+z^2) and vector r = (x, y, z)

The Attempt at a Solution

[/B]
So, taking the partial derivative of the x component, I got 2r^3/2. Doing the same thing for the y, z components I got a similar answer. This gives me a solution of ∇ • (r^3 * ȓ ) = 6r^3/2. However, according to my professor, the actual answer is just 6r^3. I don't know where I'm going wrong in my calculations, and any and all help would be appreciated. Thank you!

You have the wrong derivatives:
$$\frac{\partial}{\partial x} x r^3 = \frac{\partial}{\partial x} x (x^2+y^2+z^2)^{3/2} \neq 2 r^{3/2} = 2 (x^2+y^2+z^2)^{3/4}$$

 

[toboldlygo]

You have the wrong derivatives:
$$\frac{\partial}{\partial x} x r^3 = \frac{\partial}{\partial x} x (x^2+y^2+z^2)^{3/2} \neq 2 r^{3/2} = 2 (x^2+y^2+z^2)^{3/4}$$

I'm sorry, but how did you get that? I get to $(x^2+y^2+z^2)^{1/2}(4x^2+y^2+z^2)$ and I don't understand how you simplified that to $2(x^2+y^2+z^2)^{3/4}$

 

[Ray Vickson]

I'm sorry, but how did you get that? I get to $(x^2+y^2+z^2)^{1/2}(4x^2+y^2+z^2)$ and I don't understand how you simplified that to $2(x^2+y^2+z^2)^{3/4}$

No: that is what YOU wrote, and I am saying it is wrong. You wrote $2 r^{3/2}$ and that equals $2 (x^2+y^2+z^2)^{3/4}$, because $r = (x^2+y^2+z^2)^{1/2}$.

 

[toboldlygo]

No: that is what YOU wrote, and I am saying it is wrong. You wrote $2 r^{3/2}$ and that equals $2 (x^2+y^2+z^2)^{3/4}$, because $r = (x^2+y^2+z^2)^{1/2}$.

Wow. I completely misread that haha. Sorry about that. So, I probably simplified $(x^2+y^2+z^2)^{1/2}(4x^2+y^2+z^2)$ incorrectly, but I think that's the correct derivative. If so, is there a way to manipulate the expression so I get 2r?

 

[Ray Vickson]

Wow. I completely misread that haha. Sorry about that. So, I probably simplified $(x^2+y^2+z^2)^{1/2}(4x^2+y^2+z^2)$ incorrectly, but I think that's the correct derivative. If so, is there a way to manipulate the expression so I get 2r?

No, but you don't need to do that.

 

[ehild]

Wow. I completely misread that haha. Sorry about that. So, I probably simplified $(x^2+y^2+z^2)^{1/2}(4x^2+y^2+z^2)$ incorrectly, but I think that's the correct derivative. If so, is there a way to manipulate the expression so I get 2r?

You got the derivative with respect to x. But you need the divergence of the given function. How is divergence defined? What product of the vector-vector function with the nabla operator?

 

[toboldlygo]

@Ray Vickson, @ehild: I actually did the whole problem instead of just one part and got the right answer. Who'dathunk? I guess I got caught up in just the first part. This is what happens when the right answer is directly underneath the problem haha. Thanks for helping me out!