Differential x, cylindrical coordinates

In summary, the problem is that when you use the differential equation notation, you accidentally get an equation that says dx=2dx. This is only true if dx=0, which isn't the case.
  • #1
fluidistic
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1. Homework Statement +attempt at solution+equations
In Cartesian coordinates, x translate into [tex]x=r \cos \theta[/tex] into cylindrical coordinates, [tex]y=r \sin \theta[/tex] and [tex]z=z[/tex] .


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

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

Also I realize that [tex]dx =\frac{ \partial (r \cos \theta)}{r}dr + \frac{ \partial (r \cos \theta)}{\theta} d\theta[/tex] but I don't understand why. Can someone tell me what I should relearn? I've Boas mathematical book.
Thanking you.
 
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  • #2
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 [itex] x = f(\theta, r) [/itex] so when you write [itex] dx [/itex] you want to use the chain rule

[tex]\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*}
[/tex]

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" [Broken]. 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.
 
Last edited by a moderator:
  • #3
I appreciate your reply.
But I think there's a problem, you wrote [tex]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)[/tex].
Since [tex]\frac{dr}{dr}=1[/tex] and [tex]\frac{d\theta}{d\theta}=1[/tex], you are saying that [tex]dx=2dx[/tex]. 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 [tex]\partial r[/tex] and you don't get dx=2dx.
Thanks for your help, problem solved.
 
Last edited:

What is differential x in cylindrical coordinates?

Differential x in cylindrical coordinates refers to the infinitesimal change in the x-coordinate of a point in a three-dimensional space represented using cylindrical coordinates. It is a measure of the change in the distance from the origin along the x-axis.

How is differential x calculated in cylindrical coordinates?

In cylindrical coordinates, differential x can be calculated using the formula dx = dr * cos(theta) - r * sin(theta) * d(theta), where dr and d(theta) are the infinitesimal changes in the radial and angular coordinates, respectively.

What is the significance of differential x in cylindrical coordinates?

Differential x is an important measure in cylindrical coordinates as it helps in determining the rate of change of a function or a physical quantity in the x-direction. It is also used in the calculation of line and surface integrals in cylindrical coordinates.

Can differential x be negative in cylindrical coordinates?

Yes, differential x can be negative in cylindrical coordinates. This occurs when the point in consideration moves in the negative x-direction, resulting in a decrease in the x-coordinate. However, the magnitude of differential x remains positive.

What is the relationship between differential x and other differentials in cylindrical coordinates?

In cylindrical coordinates, differential x is related to the other differentials by the conversion equations dx = dr * cos(theta) - r * sin(theta) * d(theta) and dy = dr * sin(theta) + r * cos(theta) * d(theta). This allows for the conversion between different coordinate systems, such as cylindrical and Cartesian coordinates.

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