Why Is Partial Differentiation Different in Polar Coordinates?

[yungman]

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

 

[mfb]

If you calculate the partial derivatives at the end of your last equation, you'll see that the result is zero.

 

[yungman]

So $\frac{\partial y}{\partial x}=0$ for Polar coordinates.

Then I am even more confused!

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

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

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, $\frac{\partial r}{\partial x}=\frac{x}{r}=0.5$ which does not agree with the example I gave.

Please help, I've been stuck for over a day.

Thanks

 

[mfb]

Let's just use an example where $\theta =60^o$

If you fix θ, x and y get dependent, but then you are no longer in the general polar coordinates.

 

[yungman]

If you fix θ, x and y get dependent, but then you are no longer in the general polar coordinates.

Thanks for your patience, I really don't get this. Even if I don't fix the $\theta$,
$$\frac{\partial r}{\partial x}=\cos\theta\;\hbox { as }\;\frac{\partial y}{\partial x}=0$$

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

Here I am not fixing $\theta$, I just use $\frac{\partial y}{\partial x}=0$ only.

I think I am serious missing something.

 

[yungman]

Now I am officially lost! I since posted this question in two different math forums, people there both asked and clarified, then it's been 12 hours with no response just like here!

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 textbook 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