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Gradient in polar coordinate

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove this equation attachment.php?attachmentid=37039&stc=1&d=1310225650.jpg

    2. Relevant equations
    attachment.php?attachmentid=37040&stc=1&d=1310225650.jpg


    3. The attempt at a solution
    I almost get the answer. But I don't know why all of the sin and cos are in reciprocal form.


    attachment.php?attachmentid=37041&stc=1&d=1310225650.jpg
     

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  2. jcsd
  3. Jul 9, 2011 #2

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    Hi athrun200! :smile:

    You seem to understand how to do multivariate calculus.
    I didn't check all your equations, but they seem right.

    However, you end up in an expression with i and j, when you should be ending up in an expression with [itex]\boldsymbol{e_{r}}[/itex] and [itex]\boldsymbol{e_{\theta}}[/itex].

    Perhaps you should make a substitution like that?
     
  4. Jul 9, 2011 #3
    In fact, the first term on my last step is very similar to [itex]\boldsymbol{e_{r}}[/itex] and same for the second term. Can I really convert them to [itex]\boldsymbol{e_{\theta}}[/itex] by substitution ?

    Sorry that I can't think of any substitution which is useful.
     
  5. Jul 9, 2011 #4

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    You're making a transition from cartesian coordinates to polar coordinates.

    In your relevant equations section you have defined the relation between the unit vectors of the two sets of coordinates.
    What you need is the inverse relation.

    And in your attempt you switch the coordinates from cartesian to polar, which is good, but you also need to switch the unit vectors from cartesian to polar.


    Edit: I just checked a little bit of your equations, and I do not understand how you got [itex]\frac {\partial r} {\partial x} = \frac 1 {\cos \theta}[/itex].
    That looks wrong.
     
  6. Jul 9, 2011 #5
    attachment.php?attachmentid=37055&stc=1&d=1310256726.jpg

    Do you mean this?
    But the final result still have a factor 2.
     

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  7. Jul 9, 2011 #6

    I obtain it like this. (I am also not sure about it.)
    attachment.php?attachmentid=37056&stc=1&d=1310257777.jpg
     

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  8. Jul 10, 2011 #7

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    Ah well, that is not right.
    If you take the partial derivative on the right side, you need to apply the product rule, meaning you also have to differentiate theta.
    If you do that, you'll not find it so easy to calculate dr/dx.

    Instead I suggest you fill in [itex]r = \sqrt{x^2+y^2}[/itex], take the derivative, and simplify any way you like afterward.
    You'll get a different result for [itex]\frac {\partial r} {\partial x}[/itex] then, which is also why your end result is off by a factor 2.
    If you correct this and the similar mistakes, you'll find the proper end result.
     
    Last edited: Jul 10, 2011
  9. Jul 10, 2011 #8
    I get the answer! Thx!
     
  10. Jul 10, 2011 #9

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    Good! And thx for your thx. :smile:
     
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