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• Oct 30th 2013, 10:04 AM
Alan0354
For Polar coordinates, $r^2=x^2+y^2$ and $x=r\cos \theta$, $y=r\sin\theta$

$r^2=x^2+y^2\Rightarrow\; r\frac{\partial {r}}{\partial {x}}=x+y\frac{\partial {y}}{\partial {x}}$
$\Rightarrow\; \frac{\partial {r}}{\partial {x}}=\frac{x}{r}+\frac{y}{r}\frac{\partial {y}}{\partial {x}}$

$\frac{\partial {y}}{\partial {x}}=0$, then $\frac{\partial {r}}{\partial {x}}=\frac{x}{r}$

$\frac{\partial {r}}{\partial {x}}=\cos\theta$ as $\frac{\partial {y}}{\partial {x}}=0$

You can now put in $90^o\;>\;\theta\;>45^o$, you'll get $r>x$ and $\frac{\partial {r}}{\partial {x}}>1$

Let's just use an example where ##\theta =60^o##, so to every unit change of $x$, $r$will change for 2 unit. So $\frac{\partial {r}}{\partial {x}}=2$

The same reasoning, $r=2x$. So using this example, $\frac{\partial {r}}{\partial {x}}=\cos 60^o=0.5$ which does not agree with the example I gave.

• Oct 30th 2013, 10:36 AM
HallsofIvy
I am puzzled as to what your question is . Your title refers to a "PDE in polar coordinates" but there is no PDE, just relations between Cartesian Coordinates and Polar Coordinates. In any case, from $x= r cos(\theta)$, $r= \dfrac{x}{cos(\theta)}$ so $\dfrac{\partial r}{\partial x}= \dfrac{1}{cos(\theta)}$, not $cos(\theta)$. Similarly, $\dfrac{\partial r}{\partial x}= \dfrac{r}{x}= \dfrac{r}{rcos(\theta)}= \dfrac{1}{cos(\theta)}$.
• Oct 30th 2013, 12:44 PM
Alan0354
Quote:

Originally Posted by HallsofIvy
I am puzzled as to what your question is . Your title refers to a "PDE in polar coordinates" but there is no PDE, just relations between Cartesian Coordinates and Polar Coordinates. In any case, from $x= r cos(\theta)$, $r= \dfrac{x}{cos(\theta)}$ so $\dfrac{\partial r}{\partial x}= \dfrac{1}{cos(\theta)}$, not $cos(\theta)$. Similarly, $\dfrac{\partial r}{\partial x}= \dfrac{r}{x}= \dfrac{r}{rcos(\theta)}= \dfrac{1}{cos(\theta)}$.

Thanks for the reply, but according to the derivation shown in my first post, $\frac{\partial {r}}{\partial {x}}=\frac {x}{r}=\cos\theta$
This is verified by more than one book.
• Oct 30th 2013, 01:12 PM
SlipEternal
Quote:

Originally Posted by Alan0354
For Polar coordinates, $r^2=x^2+y^2$ and $x=r\cos \theta$, $y=r\sin\theta$

$r^2=x^2+y^2\Rightarrow\; r\frac{\partial {r}}{\partial {x}}=x+y\frac{\partial {y}}{\partial {x}}$
$\Rightarrow\; \frac{\partial {r}}{\partial {x}}=\frac{x}{r}+\frac{y}{r}\frac{\partial {y}}{\partial {x}}$

$\frac{\partial {y}}{\partial {x}}=0$, then $\frac{\partial {r}}{\partial {x}}=\frac{x}{r}$

$\frac{\partial {r}}{\partial {x}}=\cos\theta$ as $\frac{\partial {y}}{\partial {x}}=0$

You can now put in $90^o\;>\;\theta\;>45^o$, you'll get $r>x$ and $\frac{\partial {r}}{\partial {x}}>1$

Let's just use an example where ##\theta =60^o##, so to every unit change of $x$, $r$will change for 2 unit. So $\frac{\partial {r}}{\partial {x}}=2$

The same reasoning, $r=2x$. So using this example, $\frac{\partial {r}}{\partial {x}}=\cos 60^o=0.5$ which does not agree with the example I gave.

$\frac{\partial {r}}{\partial {x}}=\cos\theta$ if $\frac{\partial {y}}{\partial {x}}=0$. If $r=2x=\sqrt{x^2+y^2}$, then $y = \pm x\sqrt{3}$, so $\frac{\partial {y}}{\partial {x}} = \pm \sqrt{3} \neq 0$.
• Oct 30th 2013, 01:40 PM
Alan0354
Quote:

Originally Posted by SlipEternal
$\frac{\partial {r}}{\partial {x}}=\cos\theta$ if $\frac{\partial {y}}{\partial {x}}=0$. If $r=2x=\sqrt{x^2+y^2}$, then $y = \pm x\sqrt{3}$, so $\frac{\partial {y}}{\partial {x}} = \pm \sqrt{3} \neq 0$.

Thanks for the reply. This is just so confusing.

$\frac{\partial y}{\partial x}=\frac{\partial y(r,\theta)}{\partial r}\frac{\partial r}{\partial x(r,\theta)}+\frac{\partial y(r,\theta)}{\partial \theta}\frac{\partial \theta}{\partial x(r,\theta)}=\cos\theta \frac{\partial y(r,\theta)}{\partial r}-\frac{\sin\theta}{r}\frac{\partial y(r,\theta)}{\partial \theta }= 0$

So what is correct?
• Oct 30th 2013, 03:28 PM
topsquark
Just a thought...are we to assume that you have a DEq that is being integrated over a region? And that you are trying to convert that region from rectangular coordinates to polar coordinates? In that case x and y are independent (except on the boundaries) and thus $\frac{\partial y}{\partial x} = 0$.

It might help to post the entire problem.

-Dan
• Oct 30th 2013, 04:01 PM
Alan0354
Quote:

Originally Posted by topsquark
Just a thought...are we to assume that you have a DEq that is being integrated over a region? And that you are trying to convert that region from rectangular coordinates to polar coordinates? In that case x and y are independent (except on the boundaries) and thus $\frac{\partial y}{\partial x} = 0$.

It might help to post the entire problem.

-Dan

Thanks for the reply. This is the complete question. I just find this that I cannot explain. It is just simple Polar coordinates and I listed the derivation according to more than one sources in the first post. Then I just use an example to show a different result that I cannot explain.

As you can see from the derivation, $\frac{\partial{r}}{\partial{x}}=\frac{x}{r}$ if $\frac{\partial{y}}{\partial{x}}=0$ as I showed my work. But that does not explain the example I gave when $45^o<\theta<90^o$.

And the plot even thickens when SlipEternal showed that $\frac{\partial{y}}{\partial{x}}\neq 0$
Thanks
• Oct 30th 2013, 04:24 PM
topsquark
Quote:

Originally Posted by Alan0354
Thanks for the reply. This is the complete question. I just find this that I cannot explain. It is just simple Polar coordinates and I listed the derivation according to more than one sources in the first post. Then I just use an example to show a different result that I cannot explain.

As you can see from the derivation, $\frac{\partial{r}}{\partial{x}}=\frac{x}{r}$ if $\frac{\partial{y}}{\partial{x}}=0$ as I showed my work. But that does not explain the example I gave when $45^o<\theta<90^o$.

And the plot even thickens when SlipEternal showed that $\frac{\partial{y}}{\partial{x}}\neq 0$
Thanks

Well, that's why I'm curious. First about the DEq tag in your thread title. Why is that there if you don't have a DEq you are working with? And your comment that you found that $\frac{\partial y}{\partial x} = 0$ "verified by more than one book" leads me to believe you are converting a domain in xy to a domain in $r \theta$. At the very least you don't seem to be converting an equation (differential or otherwise) in this thread.

This whole question is just confusing!

-Dan
• Oct 30th 2013, 04:51 PM
Alan0354
Quote:

Originally Posted by topsquark
Well, that's why I'm curious. First about the DEq tag in your thread title. Why is that there if you don't have a DEq you are working with? And your comment that you found that $\frac{\partial y}{\partial x} = 0$ "verified by more than one book" leads me to believe you are converting a domain in xy to a domain in $r \theta$. At the very least you don't seem to be converting an equation (differential or otherwise) in this thread.

This whole question is just confusing!

-Dan

I might have use the wrong tittle, but my question is still valid. I triple checked my post before I posted it. Maybe, you can change it to a better tittle for me.
Thanks
• Oct 31st 2013, 12:07 PM
Alan0354
Now I am officially lost!!! I since posted this question in two different math forums, people there both asked and clarified, then no response!!!

I have one book and at least one article derived the equations like in my first post, but obviously the example I gave does not agree with the first post and I triple checked my example. I don't think I did anything wrong.......apparently I have not get any suggestion otherwise from three forums!!! I am pretty sure I am missing something as the book I used is a text book used in San Jose State and I studied through 7 chapters and yet to find a single mistake until the question here.

Anyone has anything to say?

From my example,
$\frac{\partial{r}}{\partial{x}}=\frac{r}{x}=\frac{ 1}{\cos\theta}$
And this answer makes a lot more sense.

Thanks
• Oct 31st 2013, 12:35 PM
SlipEternal
Quote:

Originally Posted by Alan0354
Now I am officially lost!!! I since posted this question in two different math forums, people there both asked and clarified, then no response!!!

I have one book and at least one article derived the equations like in my first post, but obviously the example I gave does not agree with the first post and I triple checked my example. I don't think I did anything wrong.......apparently I have not get any suggestion otherwise from three forums!!! I am pretty sure I am missing something as the book I used is a text book used in San Jose State and I studied through 7 chapters and yet to find a single mistake until the question here.

Anyone has anything to say?

From my example,
$\frac{\partial{r}}{\partial{x}}=\frac{r}{x}=\frac{ 1}{\cos\theta}$
And this answer makes a lot more sense.

Thanks

The reason no one has a suggestion for you is because you are not asking any valid questions. If $\frac{\partial{y}}{\partial{x}}=0$, then $\frac{\partial{r}}{\partial{x}} = \cos \theta$. But, if $\frac{\partial{y}}{\partial{x}}\neq 0$, then $\frac{\partial{r}}{\partial{x}} \neq \cos \theta$.

Also, as HallsofIvy showed, $\frac{\partial{r}}{\partial{x}} = \dfrac{1}{\cos\theta}$, so setting those two equal, we get $\cos^2\theta = 1$, so $\frac{\partial{y}}{\partial{x}}=0$ only if $\theta = n\pi$ where $n$ is an integer. Hence, $\frac{\partial{y}}{\partial{x}}=0$ only if $\frac{\partial{r}}{\partial{x}} = \pm 1$.
• Oct 31st 2013, 01:46 PM
Alan0354
Quote:

Originally Posted by SlipEternal
The reason no one has a suggestion for you is because you are not asking any valid questions. If $\frac{\partial{y}}{\partial{x}}=0$, then $\frac{\partial{r}}{\partial{x}} = \cos \theta$. But, if $\frac{\partial{y}}{\partial{x}}\neq 0$, then $\frac{\partial{r}}{\partial{x}} \neq \cos \theta$.

Also, as HallsofIvy showed, $\frac{\partial{r}}{\partial{x}} = \dfrac{1}{\cos\theta}$, so setting those two equal, we get $\cos^2\theta = 1$, so $\frac{\partial{y}}{\partial{x}}=0$ only if $\theta = n\pi$ where $n$ is an integer. Hence, $\frac{\partial{y}}{\partial{x}}=0$ only if $\frac{\partial{r}}{\partial{x}} = \pm 1$.

But I posted in post #5 proof $\frac{\partial{y}}{\partial{x}}= 0$ for polar coordinates.

On top of that, you showed $\frac{\partial{y}}{\partial{x}}\neq 0$, which I agree with you. But the proof in post #5 say otherwise, so what am I missing?
• Oct 31st 2013, 02:08 PM
SlipEternal
Quote:

Originally Posted by Alan0354
But I posted in post #5 proof $\frac{\partial{y}}{\partial{x}}= 0$ for polar coordinates.

On top of that, you showed $\frac{\partial{y}}{\partial{x}}\neq 0$, which I agree with you. But the proof in post #5 say otherwise, so what am I missing?

In post 5, you assumed $\frac{\partial{x(r,\theta)}}{\partial{r}} = \cos\theta$, which means you were already assuming $\frac{\partial{y}}{\partial{x}} = 0$. You never proved it.
• Oct 31st 2013, 03:15 PM
SlipEternal
Quote:

Originally Posted by SlipEternal
In post 5, you assumed $\frac{\partial{x(r,\theta)}}{\partial{r}} = \cos\theta$, which means you were already assuming $\frac{\partial{y}}{\partial{x}} = 0$. You never proved it.

Oops, I wrote that upside down. I meant you assumed $\frac{\partial{r}}{\partial{x(r,\theta)}} = \cos\theta$
• Oct 31st 2013, 03:54 PM
Alan0354
Quote:

Originally Posted by SlipEternal
Oops, I wrote that upside down. I meant you assumed $\frac{\partial{r}}{\partial{x(r,\theta)}} = \cos\theta$

I think you find my problem. Let me work this out:

$r^2=x^2+y^2\;\Rightarrow\;\frac{\partial{r}}{\part ial{x}}=\frac {x}{r}+\frac{y}{r}\frac{\partial{y}}{\partial{x}}$

$x=r\cos\theta\;\Rightarrow \partial{x}=\partial{r}\cos\theta\;\Rightarrow\;\f rac{\partial{r}}{\partial{x}}=\frac{1}{\cos\theta} =\frac{r}{x}$

$\Rightarrow\;\frac{r}{x}=\frac {x}{r}+\frac{y}{r}\frac{\partial{y}}{\partial{x}}$

$\Rightarrow\;\frac{\partial{y}}{\partial{x}}=\left (\frac{r}{x}-\frac{x}{r}\right)\frac{r}{y}$

$y=r\sin\theta$ and $x=r\cos\theta$

$\Rightarrow\;\frac{\partial{y}}{\partial{x}}=\left (\frac{r}{x}-\frac{x}{r}\right)\frac{r}{y} = \frac{r^2}{r^2 \cos\theta \sin\theta}-\frac{\cos\theta}{\sin\theta}=\tan\theta$

I really appreciate your help, I know it must be my mistake, I don't believe for a moment the book is wrong. Just small little mistake cause me over a day!!! Now everything comes together.

Thanks
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