Question of partial differentiation in polar coordinates

[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.

I am missing something, please help.

 

[adi0803199]

The above reasoning is wrong.

You can now put in 45^.<\[\theta\]<90 , you'll get r>x and \[\partial\]r/\[\partial\]x=x/r

.
which will be equal to 1/2 for theta of 60 degrees which means that r=2x which implies that x/r=1/2.This implies that \[\partial\]r/\[\partial\]x=x/r=0.5 which is right and satisfies the condition.

 

[Alan0354]

The above reasoning is wrong..
which will be equal to 1/2 for theta of 60 degrees which means that r=2x which implies that x/r=1/2.This implies that \[\partial\]r/\[\partial\]x=x/r=0.5 which is right and satisfies the condition.

Thanks for the reply, but in my example, \[\frac{\partial {r}}{\partial {x}}=2\]

 

[albbg]

if x=r*cos(theta) ==> r=x/cos(theta) ==> dr/dx=1/cos(theta)

 

[Alan0354]

if x=r*cos(theta) ==> r=x/cos(theta) ==> dr/dx=1/cos(theta)

I follow what you are doing, problem is the book and other articles showed exactly how I derived in my first post that \[\frac{\partial{r}}{\partial{x}}=\frac {x}{r}\]. That is really my question. You can see from my first post that \[\frac{\partial{y}}{\partial{x}}=0\]


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

 

[albbg]

Yes, my derivative is wrong. This is due to the fact the x, y and r are related one each other by the equation:

r^2=x^2+y^2

so we cannot derive only one of the two x=r*cos(theta) or y=r*sin(theta) as I did.

We have r=(x^2+y^2)^(1/2) then deriving with respect to x:

dr/dx=2*x/[2*(x^2+y^2)^(1/2)]=x/[(x^2+y^2)^(1/2)], substituting the polar form:

dr/dx=r*cos(theta)/r=cos(theta) that is the same result you have.

 

[Alan0354]

Yes, my derivative is wrong. This is due to the fact the x, y and r are related one each other by the equation:

r^2=x^2+y^2

so we cannot derive only one of the two x=r*cos(theta) or y=r*sin(theta) as I did.

We have r=(x^2+y^2)^(1/2) then deriving with respect to x:

dr/dx=2*x/[2*(x^2+y^2)^(1/2)]=x/[(x^2+y^2)^(1/2)], substituting the polar form:

dr/dx=r*cos(theta)/r=cos(theta) that is the same result you have.

Thanks

Now what does that leave me.....or us? You agree with me, but neither one of us agree with the book and some articles!!!

 

[Alan0354]

Now I am officially lost!!! I since posted this question in two different math forums, people there both asked and clarified, and then 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 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

 

[albbg]

Thanks

Now what does that leave me.....or us? You agree with me, but neither one of us agree with the book and some articles!!!

I think my answer agreed with the articles. dr/dx=cos(theta)

instead dr/dx=1/cos(theta) is wrong, as I said this is due to the fact that you cannot forget that x,y and r are all related one each other.

The same is in your example you just take into accout only one of the equations. Isn't it ?

 

[adi0803199]

I think that there must be some printing mistake in the book from what I see.

 

[Ratch]

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.

I am missing something, please help.

What on earth are you doing? Assuming "r" is a constant, any derivative of r will be zero.



Ratch