Solving Polar Co-ordinate Paradox: Finding \frac {\partial r} {\partial x}

[dirtydog]

Hi I am having a bit of difficulty working with plane polar co-ordinates.
We have:
r^2 = x^2 + z^2
x = r cos \theta
z=r sin \theta
I wish to find \frac {\partial r} {\partial x}

Using r^2 = x^2 + z^2
We have:

\frac {\partial (r^2)} {\partial x} = \frac {\partial (x^2)} {\partial x} + \frac {\partial (z^2)} {\partial x}
Thus 2r\frac {\partial r} {\partial x} = 2x
\frac {\partial r} {\partial x} = \frac {x} {r} = \frac {r cos \theta} {r}
Therefore \frac {\partial r} {\partial x} = cos \theta

But if we find \frac {\partial r} {\partial x} using x = r cos \theta
We have:
r = \frac {x} {cos \theta}
Therefore \frac {\partial r} {\partial x} = \frac {1} {cos \theta}

What is going on here? Which answer is wrong and why?

 

[Tide]

Ooops! In your last step you \theta is NOT a constant! You need to differentiate it wrt x as well.

 

[shmoe]

But if we find \frac {\partial r} {\partial x} using x = r cos \theta
We have:
r = \frac {x} {cos \theta}
Therefore \frac {\partial r} {\partial x} = \frac {1} {cos \theta}

Hi, when you differentiated with respect to x here, you treated \theta as a constant. It's not, it's dependant on x, so you have to use the quotient rule and you'll have a \frac {\partial \theta} {\partial x} appear via the chain rule. Use {\cos \theta}=\frac{x}{\sqrt{x^2+z^2}} to work out \frac {\partial \theta} {\partial x}.

eidt-Tide was quicker on the draw!

 

[NeutronStar]

I'm not sure, but I think your mistake is in the following procedure:

r = \frac {x} {cos \theta} \rightarrow \frac {\partial r} {\partial x} = \frac {1} {cos \theta}

I believe that \theta is dependent on x. In other words \theta = \theta (x) so you can't just treat cos \theta as a constant when taking the derivative with respect to x.

 

[NeutronStar]

Whoops! I guess I'm pretty late on that one!

This is the first time I used the equation formatting and it took me a while to format the post. (ha ha)

At least, I know now that I was on the right track.

 

[Tide]

Way to go, NS!

 

[dirtydog]

Thanks for your help guys!