Find the divergence of the function

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Discussion Overview

The discussion revolves around finding the divergence of a function, specifically in the context of vector calculus. Participants explore the gradient, curl, and divergence of vector fields, with a focus on spherical and cylindrical coordinates. The conversation includes homework-related queries and attempts at solutions.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a function V and requests the divergence, curl, and gradient.
  • Another participant asserts that the formulas used are incorrect and emphasizes the need for proper notation.
  • There is a discussion about the necessity of unit vectors in expressing the gradient and the nature of curl and divergence.
  • Participants express confusion over the correct application of formulas and the use of unit vectors in different coordinate systems.
  • Some participants suggest using Wolfram Alpha to verify calculations related to gradients and divergences.
  • There are multiple attempts to clarify the divergence of specific vector fields, with varying degrees of success and correctness noted.
  • One participant questions the correctness of their formula and seeks confirmation from others.
  • Discussions include the importance of understanding the mathematical identities related to curl and gradient operations.
  • Participants also discuss the implications of using different coordinate systems and the need for careful application of divergence formulas.

Areas of Agreement / Disagreement

Participants generally express uncertainty about the correctness of their formulas and calculations. There is no consensus on the final answers, and multiple competing views on the correct approach and notation remain throughout the discussion.

Contextual Notes

Some participants mention confusion regarding the transition between different coordinate systems and the corresponding formulas for divergence and gradient. There are unresolved mathematical steps and assumptions regarding the use of unit vectors in spherical and cylindrical coordinates.

Who May Find This Useful

Students and individuals interested in vector calculus, particularly those working with divergence and gradient in various coordinate systems.

DODGEVIPER13
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Homework Statement


Let V = (sin(theta)cos(phi))/r Determine:
(a) ∇V
(b) ∇ x ∇V
(c) ∇∇V


Homework Equations





The Attempt at a Solution


Uploaded
 

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Your formulas are all wrong. You need to get the right formulas as you finally did with your gradient exercises.
 
I uploaded 2 parts of the answer
 

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DODGEVIPER13 said:
I uploaded 2 parts of the answer

No time to check it thoroughly now, but 2 cooents:
1. you're using i j k again when you shouldn't.
2. curl(grad f) = 0 always. The curl of the gradient of a function is always zero. (Not all functions have gradients).
3 your writing is hard for me to read but it looks like you wrote grad when you meant div.
 
Is my formula even close to correct? If I take off I,j, and k and does it make it a curl when you take the cross product of a gradient?
 
Sorry that was one long run on sentence. I meant to say will it be ok if I take off I,j, and k?
 
DODGEVIPER13 said:
Sorry that was one long run on sentence. I meant to say will it be ok if I take off I,j, and k?

No. The gradient of a function is a vector. Therefore you need the three unit vectors. Do't use i j k unless you're in an xyz cartesian coordinate system. Use the ones I gave you previously.

The curl can only be taken of a vector. There is no such thing as the curl of a scalar.

It's a mathematical identity that curl (grad f) = 0 for any f. It's very good to remember that identity.
 
Ok is my formula ok? I'm going to go look again but I believe Wikipedia had that listed I can add unit vectors that's no problem.
 
DODGEVIPER13 said:
Ok is my formula ok? I'm going to go look again but I believe Wikipedia had that listed I can add unit vectors that's no problem.

If wikipedia listed it I'm sure it was right. But they must have had the unit vectors too?
(Now I know how you did so well with the gradients, right?) :smile:
 
  • #10
Ugggg that is all Wikipedia lists I'm going to go to my instructors web page and check there
 
  • #11
##\hat{i},\hat{j},\hat{k}## are just labels for unit vectors. They traditionally represent the unit vectors in Cartesian coordinates but you can call them whatever you want, they're just labels. Generally one would write ##\hat{e}_1,\hat{e}_2,\hat{e}_3## to represent unit vectors in arbitrary coordinates. If you are using spherical coordinates, then you need to use the correct formulas for ##\nabla f, \nabla \cdot V, \nabla \times V, \nabla^2 f## in spherical coordinates. You can find them easily online; for example ##\nabla = \hat{e}_r\partial_r + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta} + \frac{1}{r\sin\theta} \hat{e}_{\varphi}\partial_{\varphi}##.
 
  • #12
DODGEVIPER13 said:
Ugggg that is all Wikipedia lists I'm going to go to my instructors web page and check there

Never mind, I screwed you up & forgot you're now doing divergences, not gradients.

Divergence of a vector is a scalar so you don't attach the unit vectors.

Sorry!
 
  • #13
actually I screwed up the problem never said anything about divergence! That being said I corrected what I did and have uploaded it hopefully its ok!
 

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  • #14
DODGEVIPER13 said:
actually I screwed up the problem never said anything about divergence! That being said I corrected what I did and have uploaded it hopefully its ok!

Oh, OK. I think I mistook this for another of your threads whichI think did have div's in them.

I am too lazy to check all your math steps but for curl[grad(f)] the answer must be zero so I'm afraid something went awry there.

Why don't you run grad(f) thru wolfram alpha to make sure you got that part right.
 
  • #15
ok how do I use wolfram for gradients?
 
  • #16
DODGEVIPER13 said:
ok how do I use wolfram for gradients?

Stand by, I'll give it a shot, haven't done that myself yet ...

EDIT: OK, punch in " gradient of sin(theta)*cos(phi)/r " in their top window.

They don't include the unit vectors. They give you the r, theta and phi components in that order, separated by commas, instead. Dumb, but better than screwing up solving it ourselves!
 
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  • #17
ok uploaded something else I can could use some assistance with
 

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  • #18
DODGEVIPER13 said:
ok uploaded something else I can could use some assistance with

Can't read that. What happened with grad f?

Back in 1 hr.
 
  • #19
Evaluate the divergence of the following:
(A) A=xyUx+y^2Uy-xzUz
(B) B=pz^2Up+psin^2(phi)Uphi+2pzsin^2(phi)Uz
(C) C=rUr+rcos^2(theta)Uphi
 
  • #20
The previous post was in reference to my uploaded answer that could not be read.
 
  • #21
(A) y+2y-x
(B) z^2+sin(2phi)+2psin^2(phi)
(C) 1
These are my answers
 
  • #22
DODGEVIPER13 said:
(A) y+2y-x
(B) z^2+sin(2phi)+2psin^2(phi)
(C) 1
These are my answers

Unfortunately, B and C are incorrect. They are tricky so work carefully. And check your results with wolfram.

For example, B=pz2Up + psin2(phi)Uphi + 2pzsin2(phi)Uz

so what is ∂Bρ/∂ρ + Bρ/ρ? Not z2.
 
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  • #23
div B=pz^2+psin^2(phi)+2pzsin^2(phi) I entered that in wolfram it didnt work what syntax should I use.
 
  • #24
DODGEVIPER13 said:
div B=pz^2+psin^2(phi)+2pzsin^2(phi) I entered that in wolfram it didnt work what syntax should I use.

look at what I came up with for the gradient and improvise! Play around with it for a while. I pointed out that it gave the gradient in comma-separated chunks, omitting the unit vectors. Wolfram will have examples of how to enter a vector to get the divergence.

I would suggest using r instead of rho.

Answer my post 22.
 
  • #25
Bp/p
 
  • #26
Whoops I meant what is Bp/p was that in the formula
 
  • #27
div (rho(z)^2,(rho)sin^2(phi),2(rho)zsin^2(phi)) tried this on wolfram worked but gave the answer in Cartesian
 
  • #28
DODGEVIPER13 said:
div (rho(z)^2,(rho)sin^2(phi),2(rho)zsin^2(phi)) tried this on wolfram worked but gave the answer in Cartesian

Did you look carefully? When I ran grad V3 it gave it in cartesian but also in spherical.
 
  • #29
DODGEVIPER13 said:
Whoops I meant what is Bp/p was that in the formula

sure was! ∂Bρ/∂ρ + Bρ/ρ is the Bρ part of div B.
 
  • #30
Hmm Bp/p isn't a derivative I am failing to see what I can do with this. I am going to go on a long shot here and guess that the unit vectors cancel as they multiply together, and p cancels the other p and leaves me with z^2 again? Then z^2+z^2 heh that can't be right, what am doing wrong anymore hints?
 

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