This should be an easy partial derivative

In summary, the two equations give different answers depending on whether θ is fixed or if y is fixed.
  • #1
redstone
26
0

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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  • #2
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:
 
  • #3
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!
 
  • #4
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:
 
  • #5
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:
 

Related to This should be an easy partial derivative

What is a partial derivative?

A partial derivative is a mathematical concept used to measure the rate of change of a function with respect to one of its input variables, while holding all other variables constant.

Why is this partial derivative considered easy?

This partial derivative is considered easy because it only involves one input variable and does not require any complex mathematical operations.

What is the purpose of calculating a partial derivative?

The purpose of calculating a partial derivative is to better understand how changes in one input variable affect the output of a function, and to help solve optimization problems in fields such as economics, physics, and engineering.

Can a partial derivative be negative?

Yes, a partial derivative can be negative. A negative partial derivative indicates that as the input variable increases, the output value decreases.

What are some common notations for partial derivatives?

Some common notations for partial derivatives include: ∂f/∂x, fx, Dxf, and f1.

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