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 dependent 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!