# cylindrical coordinates

• Oct 22nd 2010, 03:43 AM
tsang
cylindrical coordinates

The relation between three-dimentional cartesian coordinates $\displaystyle (X,Y,Z)$ and cylindrical coordinates $\displaystyle (\rho,\phi,z)$ is given by
$\displaystyle X=\rho cos\phi$
$\displaystyle Y=\rho sin\phi$
$\displaystyle Z=z$

Calculate the partial derivatives $\displaystyle \frac{\partial \rho}{\partial X}$, $\displaystyle \frac{\partial \phi}{\partial Y}$ and $\displaystyle \frac{\partial z}{\partial Z}$ in terms of $\displaystyle \rho$,$\displaystyle \phi$,$\displaystyle z$.

My current problem here is like a puzzle. I certainly can solve the question and find the answer. For example, I did find $\displaystyle \frac{\partial \rho}{\partial X}=cos\phi$.Same as solution
However, my problem occured when I want to calculate $\displaystyle \frac{\partial X}{\partial \rho}$,if I just use $\displaystyle X=\rho cos\phi$, I would get $\displaystyle cos\phi$ as well!
I think it is weird, because I thought $\displaystyle \frac{\partial X}{\partial \rho}$ supposed to equal to the reciprocal of $\displaystyle \frac{\partial \rho}{\partial X}$, which means it supposed to be $\displaystyle \frac{1}{cos \phi}$.
So pretty sure $\displaystyle \frac{\partial \rho}{\partial X}=cos\phi$ is the right answer(same as solution),but why the recipocal rule does not apply here? I'm so confused about this puzzle. Same as two other questions.(Headbang)
• Oct 22nd 2010, 04:44 AM
Ackbeet
What was your process by which you obtained

$\displaystyle \displaystyle\frac{\partial\rho}{\partial x}?$
• Oct 22nd 2010, 07:07 AM
HallsofIvy
Oddly enough, both $\displaystyle \frac{\partial \rho}{\partial X}= cos(\theta)$ and $\displaystyle \frac{\partial X}{\partial \rho}= cos(\theta)$ are correct.
Your "mistake" is in thinking that $\displaystyle \frac{\partial \rho}{\partial X}$ and $\displaystyle \frac{\partial X}{\partial \rho}$ must be reciprocals. For partial derivatives, that simply is not true.
• Oct 22nd 2010, 09:31 PM
tsang
Quote:

Originally Posted by HallsofIvy
Oddly enough, both $\displaystyle \frac{\partial \rho}{\partial X}= cos(\theta)$ and $\displaystyle \frac{\partial X}{\partial \rho}= cos(\theta)$ are correct.
Your "mistake" is in thinking that $\displaystyle \frac{\partial \rho}{\partial X}$ and $\displaystyle \frac{\partial X}{\partial \rho}$ must be reciprocals. For partial derivatives, that simply is not true.

Thanks a lot.
I felt the reason must be because their are partial derivatives, therefore they are not reciprocals.
But I wonder the reason, is there any mathematical way can explain that? Thanks a lot.