This should be an easy partial derivative

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Homework Help Overview

The discussion revolves around the computation of partial derivatives in polar coordinates, specifically the relationships between the variables \( r \), \( x \), and \( \theta \). The original poster expresses confusion over obtaining two different expressions for the partial derivative of \( r \) with respect to \( x \) from related equations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to compute the partial derivatives of \( r \) with respect to \( x \) using two different equations and questions the validity of the resulting expressions. Some participants suggest that both expressions can be correct under different conditions, prompting further exploration of the assumptions involved.

Discussion Status

Participants are engaged in clarifying the original poster's confusion regarding the partial derivatives. There is a recognition of the need to consider the context of the variables involved, and some guidance has been provided regarding the proper notation for derivatives.

Contextual Notes

The original poster is also attempting to transform coordinates of the strain tensor from Cartesian to polar, indicating a potential complexity in the problem setup. There is mention of a desire to check the correctness of their approach against established resources.

redstone
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Homework Statement


Homework Equations


This should be easy, I don't know what I've done wrong...

polar coordinates
[tex]x=r cos(\theta)[/tex]
[tex]y=r sin(\theta)[/tex]
[tex]r^2=x^2+y^2[/tex]

The Attempt at a Solution


so with [tex]x=r cos(\theta)[/tex]
[tex]\partial{x}/\partial{r}=cos(\theta)[/tex]
[tex]\partial{x}/\partial{r}=x/r[/tex]
thus the inverse
[tex]\partial{r}/\partial{x}=r/x[/tex]

similarly with [tex]r=x/cos(\theta)[/tex]
partial of r wrt x
I get
[tex]\partial{r}/\partial{x}=1/cos(\theta)[/tex]
[tex] \partial{r}/\partial{x} = r/x[/tex]

now doing same thing on the r^2 equation
partial of r wrt x
[tex] 2r\partial{r}/\partial{x}=2x[/tex]
[tex] \partial{r}/\partial{x}=x/r[/tex]

what the heck? why am I getting two different (and inverse) answers from these two related equations?
 
Last edited:
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Welcome to PF!

Hi redstone ! Welcome to PF! :smile:

(have a theta: θ and a curly d: ∂ :wink:)
redstone said:
what the heck? why am I getting two different (and inverse) answers from these two related equations?

Both are correct! :biggrin:

∂r/∂x means the change in r if you change x, keeping all the other variables constant.

If you define r = x/cosθ, then the other variable is θ …

draw a diagram, and you find that if θ is fixed, then increasing x by dx does increase r by r/x dx …

but if you keep y fixed, increasing x by dx increases r by x/r dx :wink:
 
Ha! That's been frustrating me for two days and I was right the whole time!
Thanks!

As a related question, is it absurd to write the derivative with partials like this:
r = x/cosθ
∂r = ∂x/cosθ - xsinθ ∂θ ?

(I'm trying to transform coordinates of the strain tensor from Cartesian to polar, and trying to remember the math I need... so if you happen to know an online source that has the transform all worked out, I'd be interested in checking my math to it when I'm done)

Thanks again for the help!
 
Hi redstone! :smile:

You'd write it with d rather than ∂:

dr = dx/cosθ - xsinθ/cos2θ dθ

But whyever would you want to express as a function of x and θ? :confused:
 
oops, silly math mistake on my part there...:blushing:

But my mistake aside, I don't really want to express it like that, just wanted to check the proper use of partials, and whether it was OK to write that expression with a partial. I guess not, you need to use the full derivative.

Thanks a lot for your help.
:cool:
 

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