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Thread: Polar coordinates

  1. #1
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    Polar coordinates

    This is a post by nomad609.

    Quote Originally Posted by nomad609 View Post
    Hi all, my question is essentially the title.


    I would like to find the polar coordinates for the Laplace of a particular scalar function.


    My question is attached here, if me attaching the image did not work then I will try and drag and drop the image.


    Here is the link of the question: Laplace from a scalar function - Album on Imgur









    Now my working out for this is here, although I am not entirely sure it is correct.


    Again here is the link and the image?


    Link: My working out - Album on Imgur


    Image:























    I would really appreciate if anyone can help me out here.


    -nomad609
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  2. #2
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    Re: Polar coordinates

    I have found this: https://en.wikibooks.org/wiki/Advanc...e%27s_Equation

    Although I do not feel it helps in the context of my question.

    Does anyone have any other ideas to help me out please?
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    Re: Polar coordinates

    Anyone have a clue :P
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    Re: Polar coordinates

    Use the formulas connecting Cartesian and polar coordinates- x= r cos(\theta), y= r sin(\theta), r= \sqrt{x^2+ y^2}= (x^2+ y^2)^{1/2], and y= arctan\left(\frac{y}{x}\right)- and the chain rule. It is straight forward but tedious.

    \frac{\partial f}{\partial x}= \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}

    \frac{\partial r}{\partial x}= \frac{\partial (x^2+ y^2)^{1/2}}{\partial x}= (1/2)(x^2+ y^2)^{-1/2}(2x)= \frac{x}{(x^2+ y^2)^{1/2}}= \frac{r cos(\theta)}{r}= cos(\theta)

    \frac{\partial \theta}{\partial x}= \frac{\partial arctan(y/x)}{\partial x}= \frac{1}{1+ \frac{y^2}{x^2}}\left(-\frac{y}{x^2}\right)= -\frac{y}{x^2+ y^2}= -\frac{r sin(\theta)}{r^2}= -\frac{1}{r} sin(\theta)

    So \frac{\partial f}{\partial x}= cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}

    Now do it again:
    \frac{\partial^2 f}{\partial x^2}= \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)= \frac{\partial}{\partial x}\left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)
    = cos(\theta)\frac{\partial\left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)}{\partial r}+ \frac{1}{r}sin(\theta)\frac{\partial \left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)}{\partial \theta}
    = cos(\theta)\left(cos(\theta)\frac{\partial^2 f}{\partial r^2}+ \frac{1}{r^2}sin(\theta)\frac{\partial f}{\partial \theta}- \frac{1}{r}sin(\theta)\frac{\partial^2f}{\partial r\partial \theta}\right)+  \frac{1}{r}sin(\theta)\left(-sin(\theta)\frac{\partial f}{\partial r}+ cos(\theta)\frac{\partial^2 f}{\partial r\partial \theta}- \frac{1}{r}cos(\theta)\frac{\partial f}{\partial \theta}- \frac{1}{r}sin(\theta)\frac{\partial^2 f}{\partial \theta^2}\right)

    Well, I said it was tedious! I will leave the rest for you to try.

    Notice that this contains a cos^2(\theta)\frac{\partial^2 f}{\partial r^2}. The second derivative with respect to y will contain sin^2(\theta)\frac{\partial^2 f}{\partial r^2} so they add to \frac{\partial^2 f}{\partial r^2}. Similar things happen with the other derivatives.
    Last edited by HallsofIvy; Oct 30th 2016 at 03:54 AM.
    Thanks from nomad609
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    Re: Polar coordinates

    Quote Originally Posted by HallsofIvy View Post
    Use the formulas connecting Cartesian and polar coordinates- x= r cos(\theta), y= r sin(\theta), r= \sqrt{x^2+ y^2}= (x^2+ y^2)^{1/2], and y= arctan\left(\frac{y}{x}\right)- and the chain rule. It is straight forward but tedious.

    \frac{\partial f}{\partial x}= \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}

    \frac{\partial r}{\partial x}= \frac{\partial (x^2+ y^2)^{1/2}}{\partial x}= (1/2)(x^2+ y^2)^{-1/2}(2x)= \frac{x}{(x^2+ y^2)^{1/2}}= \frac{r cos(\theta)}{r}= cos(\theta)

    \frac{\partial \theta}{\partial x}= \frac{\partial arctan(y/x)}{\partial x}= \frac{1}{1+ \frac{y^2}{x^2}}\left(-\frac{y}{x^2}\right)= -\frac{y}{x^2+ y^2}= -\frac{r sin(\theta)}{r^2}= -\frac{1}{r} sin(\theta)

    So \frac{\partial f}{\partial x}= cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}

    Now do it again:
    \frac{\partial^2 f}{\partial x^2}= \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)= \frac{\partial}{\partial x}\left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)
    = cos(\theta)\frac{\partial\left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)}{\partial r}+ \frac{1}{r}sin(\theta)\frac{\partial \left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)}{\partial \theta}
    = cos(\theta)\left(cos(\theta)\frac{\partial^2 f}{\partial r^2}+ \frac{1}{r^2}sin(\theta)\frac{\partial f}{\partial \theta}- \frac{1}{r}sin(\theta)\frac{\partial^2f}{\partial r\partial \theta}\right)+  \frac{1}{r}sin(\theta)\left(-sin(\theta)\frac{\partial f}{\partial r}+ cos(\theta)\frac{\partial^2 f}{\partial r\partial \theta}- \frac{1}{r}cos(\theta)\frac{\partial f}{\partial \theta}- \frac{1}{r}sin(\theta)\frac{\partial^2 f}{\partial \theta^2}\right)

    Well, I said it was tedious! I will leave the rest for you to try.

    Notice that this contains a cos^2(\theta)\frac{\partial^2 f}{\partial r^2}. The second derivative with respect to y will contain sin^2(\theta)\frac{\partial^2 f}{\partial r^2} so they add to \frac{\partial^2 f}{\partial r^2}. Similar things happen with the other derivatives.
    Wow thank you somuch for this, I will try and finish this off. And if it okay, show you my working out.

    Really appreciate this, thank you.

    -nomad609
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  6. #6
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    Re: Polar coordinates

    Quote Originally Posted by HallsofIvy View Post
    Use the formulas connecting Cartesian and polar coordinates- x= r cos(\theta), y= r sin(\theta), r= \sqrt{x^2+ y^2}= (x^2+ y^2)^{1/2], and y= arctan\left(\frac{y}{x}\right)- and the chain rule. It is straight forward but tedious.

    \frac{\partial f}{\partial x}= \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}

    \frac{\partial r}{\partial x}= \frac{\partial (x^2+ y^2)^{1/2}}{\partial x}= (1/2)(x^2+ y^2)^{-1/2}(2x)= \frac{x}{(x^2+ y^2)^{1/2}}= \frac{r cos(\theta)}{r}= cos(\theta)

    \frac{\partial \theta}{\partial x}= \frac{\partial arctan(y/x)}{\partial x}= \frac{1}{1+ \frac{y^2}{x^2}}\left(-\frac{y}{x^2}\right)= -\frac{y}{x^2+ y^2}= -\frac{r sin(\theta)}{r^2}= -\frac{1}{r} sin(\theta)

    So \frac{\partial f}{\partial x}= cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}

    Now do it again:
    \frac{\partial^2 f}{\partial x^2}= \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)= \frac{\partial}{\partial x}\left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)
    = cos(\theta)\frac{\partial\left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)}{\partial r}+ \frac{1}{r}sin(\theta)\frac{\partial \left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)}{\partial \theta}
    = cos(\theta)\left(cos(\theta)\frac{\partial^2 f}{\partial r^2}+ \frac{1}{r^2}sin(\theta)\frac{\partial f}{\partial \theta}- \frac{1}{r}sin(\theta)\frac{\partial^2f}{\partial r\partial \theta}\right)+  \frac{1}{r}sin(\theta)\left(-sin(\theta)\frac{\partial f}{\partial r}+ cos(\theta)\frac{\partial^2 f}{\partial r\partial \theta}- \frac{1}{r}cos(\theta)\frac{\partial f}{\partial \theta}- \frac{1}{r}sin(\theta)\frac{\partial^2 f}{\partial \theta^2}\right)

    Well, I said it was tedious! I will leave the rest for you to try.

    Notice that this contains a cos^2(\theta)\frac{\partial^2 f}{\partial r^2}. The second derivative with respect to y will contain sin^2(\theta)\frac{\partial^2 f}{\partial r^2} so they add to \frac{\partial^2 f}{\partial r^2}. Similar things happen with the other derivatives.


    I think I have solved it

    Although please, if you don't mind explaining how do you get the bolded part? This is the y=arctan(y/x) part this. Should the "y" on the LHS not be theta? by trig. I think this is a typo maybe?

    If you could just clarify this, that would be awesome.

    But thanks again for the help.
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  7. #7
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    Re: Polar coordinates

    Yes, that was a typo. \theta= arctan(y/x).
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  8. #8
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    Re: Polar coordinates

    Quote Originally Posted by HallsofIvy View Post
    Yes, that was a typo. \theta= arctan(y/x).
    Thank you bro

    I realised after you typed it later on

    -nomad609
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  9. #9
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    Re: Polar coordinates

    Quote Originally Posted by HallsofIvy View Post
    Yes, that was a typo. \theta= arctan(y/x).
    Hey m8 can I ask that if you have time could you please look at my worked solutions that I have done, I am very close to the answer I feel my final stage is similar to the form that I wanted it to be in but not exact.

    I have uploaded my solutions here: First Image - Album on Imgur

    Thank you again for your help, your work helped me start my attempted (Attached solution)

    -nomad609
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