Differential x, cylindrical coordinates

[fluidistic]

1. Homework Statement +attempt at solution+equations
In Cartesian coordinates, x translate into $$x=r \cos \theta$$ into cylindrical coordinates, $$y=r \sin \theta$$ and $$z=z$$ .


However $$dx=\cos \theta dr - r \sin \theta d\theta$$. This is what I don't understand.

Since x is a function of both $$\theta$$ and $$r$$, I can write $$x=f(\theta , r)$$.
It's not really clear to me what dx represent, it's not a derivative with respect to any variable. But I can write $$\frac{dx}{dt}$$ for an arbitrary variable t. And this is worth $$\frac{\partial f(r,\theta)}{\partial r} \frac{\partial r}{\partial t}+ \frac{\partial f(r, \theta)}{\partial \theta} \frac{\partial \theta}{\partial t}$$.
Multiplying by dt or $$\partial t$$ I get a non sense result $$(dx=2 \partial f(r, \theta )$$. So I'm doing something wrong.

Also I realize that $$dx =\frac{ \partial (r \cos \theta)}{r}dr + \frac{ \partial (r \cos \theta)}{\theta} d\theta$$ but I don't understand why. Can someone tell me what I should relearn? I've Boas mathematical book.
Thanking you.

 

[Kreizhn]

Hey,

I'm not sure where the problem is coming from, but here's my attempt to help:

I think in multivariable calculus, one normally defines the total differential of a function right? In particular, you wrote $x = f(\theta, r)$ so when you write $dx$ you want to use the chain rule

$$\begin{align*} dx &= \frac{dx}{dr} dr + \frac{dx}{d\theta}d\theta \\ &= \frac{\partial f}{\partial r}dr + \frac{\partial f}{\partial \theta} d\theta & \text{ since } x = f(\theta,r) \end{align*}$$

The true meaning of the dx is very hard to explain and depends on what you're doing. In measure theory, it usually represents the measure and in this case how it relates to a different coordinate basis. However, I'm not super fluent on the measure theoretical aspects of this formalism.

What is probably more useful is that dx is a http://en.wikipedia.org/wiki/Differential_form" . To properly explain what a differential form is would require a great deal of time and geometry.

Both explanations are quite complicated. Perhaps if you could give some insight as to where you are seeing this occur I might be able to give a better response.

 

[fluidistic]

I appreciate your reply.
But I think there's a problem, you wrote $$dx &= \frac{dx}{dr} dr + \frac{dx}{d\theta}d\theta \\ &= \frac{\partial f}{\partial r}dr + \frac{\partial f}{\partial \theta} d\theta & \text{ since } x = f(\theta,r)$$.
Since $$\frac{dr}{dr}=1$$ and $$\frac{d\theta}{d\theta}=1$$, you are saying that $$dx=2dx$$. This is true only if dx=0, or I'm wrong?!

Edit: Nevermind, I just found on the Internet your explanation. Just replace a dr by $$\partial r$$ and you don't get dx=2dx.
Thanks for your help, problem solved.