Is ##\frac{dx}{dr}=\frac{r}{x}## Correct in Polar Coordinates?

[yungman]

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

Then is it true $\frac{dx}{dr}=\frac{r}{x}$?

I am not sure this is correct as
r^2=x^2+y^2\;\Rightarrow \; 2r=2x\frac{dx}{dr}+2y\frac{dy}{dr}

 

[UltrafastPED]

If you take the differential of both sides you get:

2r dr = 2x dx + 2y dy which can be rewritten as:

dr = x/r dx + y/r dy

Converting this to derivative form we get:

dr/dx = x/r + y/r dy/dx

dr/dy = x/r dx/dy + y/rSo your first result is missing the dependence of y upon x.

 

[yungman]

If you take the differential of both sides you get:

2r dr = 2x dx + 2y dy which can be rewritten as:

dr = x/r dx + y/r dy

Converting this to derivative form we get:

dr/dx = x/r + y/r dy/dx

dr/dy = x/r dx/dy + y/r


So your first result is missing the dependence of y upon x.

I forgot to say this is x,y coordinates and r is the radius. So x and y are independent.

 

[UltrafastPED]

Are they? You seem to have created a functional relationship between them! It looks like a circle to me, but with a variable radius.

Perhaps you could describe the meaning of your starting equation, and what you are planning to accomplish.

 

[yungman]

This is just polar coordinates where $r^2=x^2+y^2$. So x and y are independent variables.

So, from what you show:$2rdr=2xdx+2ydy$
\Rightarrow\; 2r=2x\frac{dx}{dr}+2y\frac{dy}{dr}
\Rightarrow\;\frac{dx}{dr}=\frac{r}{z}-\frac{y}{x}\frac{dy}{dr}


2rdr=2xdx+2ydy\;\Rightarrow r\frac{dr}{dx}=2x+2y\frac{dy}{dx}=2x\;\Rightarrow\;\frac{dr}{dx}=\frac {x}{r}
BUT
\frac{dx}{dr}\neq\frac{r}{x}

Am I correct?

 

[UltrafastPED]

But x = r cos(theta), y = r sin(theta) ... you have not included these relationships!

Thus x and y are no longer independent ... try plugging these into the differential dr = x/r dx + y/r dy, and see what you get.

 

[yungman]

I am confused. This is the second questionable thing in this book. I have a questionable formula of this book in another thread right now! Here is the scanned page of what I have been asking:

 

[vanhees71]

What's the problem with the scanned page, except the equation
\theta=\arctan(y/x),
which has to be used with very much care. Better is (for y \neq 0)
\theta=\mathrm{sign} y \arccos \left(\frac{x}{\sqrt{x^2+y^2}} \right )
For y=0 you can set
\theta=\begin{cases} 0 &amp; \text{for} \quad x=r&gt;0,\\<br /> \pi &amp;\text{for} \quad x=-r&lt;0.<br /> \end{cases}
The polar coordinates are indetermined at the origin (coordinate singularity).

So, what is considered here is the transformation between cartesian and polar coordinates of the Euclidean plane,
x=r \cos \theta, \quad y=r \sin \theta.
It's also important to distinguish partial and total derivatives. You have to decide which two variables you take as the independent ones, i.e., \frac{\partial}{\partial x} \equiv \partial_x means to take the partial derivative with respect to x while y is kept constant, i.e., indeed
r=\sqrt{x^2+y^2} \; \Rightarrow \; \partial_x r=\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}
etc.
For \theta I'd rather use
\cos \theta=\frac{x}{r}.
Then you get
-\sin \theta \partial_x \theta=\frac{1}{r}-\frac{x}{r^2} \partial_x r=\frac{1}{r}-\frac{x^2}{r^3}=\frac{r^2-x^2}{r^3}=\frac{y^2}{r^3}=\frac{r^2 \sin^2 \theta}{r^3}.
From this
\partial_x \theta = -\frac{y}{r^2},
etc.

I've not checked the other formulae from your book, but at least these to examples are perfectly fine.

 

[UltrafastPED]

You can type ∂ instead of d by using the Quick Symbols of Advance editing mode.

This certainly makes a difference in how to interpret a statement!

 

[yungman]

Yes, my bad to use "d" instead of $\partial$.

So $\frac{\partial y}{\partial x}=0$?

So back to my original question:

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

Then is it true $\frac{\partial x}{\partial r}=\frac{r}{x}$?

I am not sure this is correct as
r^2=x^2+y^2\;\Rightarrow \; 2r=2x\frac{\partial x}{\partial r}+2y\frac{\partial y}{\partial r}


Can I look at it this way, $x=r\cos\theta$
\frac{\partial x}{\partial r}=\frac{\partial (r\cos\theta)}{\partial r}=\cos\theta= \frac {x}{r}

But then
\frac{\partial x}{\partial r}=\frac{\partial r}{\partial x}

I am confused.

 

[UltrafastPED]

So $\frac{\partial y}{\partial x}=0$?

Yes.

So back to my original question:

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

Yes.

Then is it true $\frac{\partial x}{\partial r}=\frac{r}{x}$?

See this old thread on Reciprocals of Derivatives:
https://www.physicsforums.com/showthread.php?t=317983

 

[yungman]

Yes.



Yes.



See this old thread on Reciprocals of Derivatives:
https://www.physicsforums.com/showthread.php?t=317983

You mean I got it right, hit the jackpot?!

Thanks

 

[UltrafastPED]

I am not sure this is correct as
r^2=x^2+y^2\;\Rightarrow \; 2r=2x\frac{\partial x}{\partial r}+2y\frac{\partial y}{\partial r}


Can I look at it this way, $x=r\cos\theta$
\frac{\partial x}{\partial r}=\frac{\partial (r\cos\theta)}{\partial r}=\cos\theta= \frac {x}{r}

But then
\frac{\partial x}{\partial r}=\frac{\partial r}{\partial x}

I am confused.

For an analysis of the above, refer to the link given above ...

 

[yungman]

I see that
\frac{\partial x}{\partial r}=\frac{1}{\left(\frac{\partial r}{\partial x}\right)}

So back to my original question:

r^2=x^2+y^2\;\Rightarrow \; 2r\frac{\partial r}{\partial x}=2x\;\Rightarrow\; \frac{\partial r}{\partial x}=\frac{x}{r}
\Rightarrow\;\frac{\partial x}{\partial r}=\frac{1}{\left(\frac{x}{r}\right)}=\frac{r}{x}


But if I let $x=r\cos\theta$
\frac{\partial x}{\partial r}=\frac{\partial (r\cos\theta)}{\partial r}=\cos\theta= \frac {x}{r}
I am confused, please help.

 

[yungman]

On top of the above question. Let's put in some real numbers. Let $\theta=60^0$
\Rightarrow \; \frac {r}{x}=\frac{1}{\cos 60}=2
This also to every change of x, the r change twice as much.
\Rightarrow\; \frac{\partial r}{\partial x}=\frac {r}{x}=2
and also, to every change of r, x only change half as much.
\Rightarrow \; \frac{\partial x}{\partial r}=\frac {x}{r}=0.5


but from the last post,
\Rightarrow\; \frac{\partial r}{\partial x}=\frac {x}{r}=\frac {1}{2}
and according to $\frac{\partial x}{\partial r}=\frac {1}{ \left(\frac{\partial r}{\partial x}\right)}$
\frac{\partial x}{\partial r}=\frac{r}{x}=2

The two is opposite. It just does not make sense.

 

[UltrafastPED]

Here is a nicely prepared lesson on partial derivatives and the chain rule which also addresses your OP in some detail: http://www2.imperial.ac.uk/~jdg/AECHAIN.PDFIn particular, section 2, part 3.

 

[D H]

I see that
\frac{\partial x}{\partial r}=\frac{1}{\left(\frac{\partial r}{\partial x}\right)}

No! It's a mistake to think of $\frac{\partial x}{\partial r}$ and $\frac{\partial r}{\partial x}$ as fractions. If they were fractions, then the product $\frac{\partial x}{\partial r}\frac{\partial r}{\partial x}$ would be one. But this is not true. Instead, with these variables $\frac{\partial x}{\partial r}\frac{\partial r}{\partial x} = \cos^2\theta = \bigl(\frac x r\bigr)^2$.

 

[vanhees71]

That cannot be true! You again forgot that for any partial derivative you must tell which other variable is hold fixed. In the context of coordinate transformations you consider either the pair (x,y) or the pair (r,\theta) as independent variables, i.e., the expression \partial r/\partial x means that you take the derivative of r with respect to x at a fixed value of y, while \partial x/\partial r is the derivative of x wrt. r at a fixed value of \theta.

It may become more clear for you, when you think about the geometric meaning either fixing y and varying x or fixing \theta and varying r at a given point on the plane. You'll see that you get completely different curves (coordinate lines) or even tangent vectors on that coordinate lines!

 

[D H]

That cannot be true!

I don't know if you were addressing me or yungman, vanhees71.

You again forgot that for any partial derivative you must tell which other variable is hold fixed.

This is very important, particularly if it's not clear from context. In this case it is better to write $\frac{\partial r}{\partial x}$ as $\left.\frac{\partial r}{\partial x}\right|_y$ and $\frac{\partial x}{\partial r}$ as $\left.\frac{\partial x}{\partial r}\right|_{\theta}$.

 

[yungman]

Here is a nicely prepared lesson on partial derivatives and the chain rule which also addresses your OP in some detail: http://www2.imperial.ac.uk/~jdg/AECHAIN.PDF

Thanks, this really help. This answer a lot of my question. The normal Calculus III books do not get into this and the PDE book take for granted people should know this. I bet the instructor must go through this in class, problem is I am a self studier and never really link it up this good. But I still have question, I'm going to ask in a separate post.

Thanks

 

[yungman]

On top of the above question. Let's put in some real numbers. Let $\theta=60^0$
\Rightarrow \; \frac {r}{x}=\frac{1}{\cos 60}=2
This also to every change of x, the r change twice as much.
\Rightarrow\; \frac{\partial r}{\partial x}=\frac {r}{x}=2
and also, to every change of r, x only change half as much.
\Rightarrow \; \frac{\partial x}{\partial r}=\frac {x}{r}=0.5


but from the last post,
\Rightarrow\; \frac{\partial r}{\partial x}=\frac {x}{r}=\frac {1}{2}

The two is opposite. It just does not make sense.

I still have not resolve this, it does not make sense.

 

[UltrafastPED]

Review message #16 - there is a reference there to an analysis of a similar problem in polar coordinates.

As others have pointed out, there is a notational problem with partial derivatives: you must determine the variables being held constant from context, as they are normally omitted (except in thermodynamics!).

Thus you cannot tell if you are to apply the complete chain rule for partial derivatives (close relative of the total derivative), or a restricted application. And if restricted, what are the restrictions?

 

[yungman]

I know it is talking about the exact problem and I know how to derive $\frac{\partial r}{\partial x}=\frac{x}{r}$

What my question is using common sense by setting $\theta=60^o$ where $x=r\cos 60=0.5r$.

Common sense will tell you that to every unit increase of x will cause 2 units increase of r. So
\left[\frac{\partial r}{\partial x}\right]_{\theta=60}=2=\frac{r}{x}

I am trying to understand why the discrepancy.

 

[D H]

I am trying to understand why the discrepancy.

You're holding the "wrong" variable constant.

Alternatively, $\left.\frac{\partial r}{\partial x}\right|_y \ne \left.\frac{\partial r}{\partial x}\right|_{\theta}$, as you just found.

 

[yungman]

Do you all mean
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}

So $\frac{\partial r}{\partial x}=\frac {x}{r}$ only hold true if y is a constant which is not my example that y change with x. In my example, there is a relation between y and x in form of $\tan \theta=\frac {y}{x}$. So $\frac{\partial r}{\partial x}\neq\frac{x}{r}$

 

[yungman]

I think I understand now, I just want to verify

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

$x(r,\theta)$ and$y(r,\theta)$ are not independent to each other like in rectangular.

In rectangular coordinates, $\frac{\partial y}{\partial x}=\frac{dy}{dx}=0$

But in Polar coordinates,
\frac{\partial r}{\partial x}=\cos\theta,\;\frac{\partial \theta}{\partial x}=-\frac{\sin\theta}{r}
\frac{\partial y(r,\theta)}{\partial x(r,\theta)}=\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)}=<br /> <br /> (\cos\theta) \frac{\partial y(r,\theta)}{\partial r}-\left(\frac{\sin\theta}{r}\right)\frac{\partial y(r,\theta)}{\partial \theta}

Thanks