# PDE's - Implicit Differentiation Query

• Jan 14th 2011, 02:22 PM
bugatti79
PDE's - Implicit Differentiation Query
hi,
I am looking at a question in 'advanced engineering mathematics'.

if $\displaystyle z=x^2-y^2$ and $\displaystyle x=r cos \theta$, $\displaystyle y=r sin \theta$

1) Do we say that $\displaystyle z=u(x,y)$ or $\displaystyle z=u(x(r,\theta),y(r,\theta))$ or $\displaystyle z=u(r,\theta)$?

2)I know we can get $\displaystyle \frac{\partial z}{\partial r}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$and so on for $\displaystyle \frac{\partial z}{\partial \theta}$.
Now lets say we have $\displaystyle z=u(x(r,\theta),y(r,\theta))$

How do I get $\displaystyle \frac{\partial z}{\partial r}$ and $\displaystyle \frac{\partial z}{\partial \theta}$ in terms of $\displaystyle \frac{\partial u}{\partial x}$ and $\displaystyle \frac{\partial u}{\partial y}$?
Do I differentiate both sides of the equation?

Thanks
• Jan 14th 2011, 02:38 PM
dwsmith
Quote:

Originally Posted by bugatti79
hi,
I am looking at a question in 'advanced engineering mathematics'.

if $\displaystyle z=x^2-y^2$ and $\displaystyle x=r cos \theta$, $\displaystyle y=r sin \theta$

1) Do we say that $\displaystyle z=u(x,y)$ or $\displaystyle z=u(x(r,\theta),y(r,\theta))$ or $\displaystyle z=u(r,\theta)$?

2)I know we can get $\displaystyle \frac{\partial z}{\partial r}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$and so on for $\displaystyle \frac{\partial z}{\partial \theta}$.
Now lets say we have $\displaystyle z=u(x(r,\theta),y(r,\theta))$

How do I get $\displaystyle \frac{\partial z}{\partial r}$ and $\displaystyle \frac{\partial z}{\partial \theta}$ in terms of $\displaystyle \frac{\partial u}{\partial x}$ and $\displaystyle \frac{\partial u}{\partial y}$?
Do I differentiate both sides of the equation?

Thanks

Does the question say use a change of variables or in a section about a change of variables?
• Jan 14th 2011, 03:05 PM
bugatti79
Quote:

Originally Posted by dwsmith
Does the question say use a change of variables or in a section about a change of variables?

The question in in the section on 'change of variables' and specifically relates to query 1 in my post.

Query 2 is my own question. Im interested in getting del z del r and del z del theta in terms of del u del x and del u del y.

Thanks
• Jan 14th 2011, 05:04 PM
dwsmith
Quote:

Originally Posted by bugatti79
The question in in the section on 'change of variables' and specifically relates to query 1 in my post.

Query 2 is my own question. Im interested in getting del z del r and del z del theta in terms of del u del x and del u del y.

Thanks

I know how to do a change of variables like this

$\displaystyle u(x,y)=u(\xi\cos{\alpha}-\eta\sin{\alpha},\xi\sin{\alpha}+\eta\cos{\alpha}) =\omega(\xi,\eta)$
• Jan 15th 2011, 05:18 AM
bugatti79
Quote:

Originally Posted by dwsmith
I know how to do a change of variables like this

$\displaystyle u(x,y)=u(\xi\cos{\alpha}-\eta\sin{\alpha},\xi\sin{\alpha}+\eta\cos{\alpha}) =\omega(\xi,\eta)$

No, i beileve that is not the change of variable the book is referring to. Anyway, diferentiate the LHS and RHS of the first equation on post #1 I calculate
$\displaystyle \frac{\partial z}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}$ and

$\displaystyle \frac{\partial z}{\partial \theta}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta}$

I wonder is this correct?
• Jan 15th 2011, 05:38 AM
Jester
Yes, that's correct.
• Jan 15th 2011, 08:56 AM
HallsofIvy
[QUOTE=bugatti79;605264]hi,
I am looking at a question in 'advanced engineering mathematics'.

if $\displaystyle z=x^2-y^2$ and $\displaystyle x=r cos \theta$, $\displaystyle y=r sin \theta$

1) Do we say that $\displaystyle z=u(x,y)$ or $\displaystyle z=u(x(r,\theta),y(r,\theta))$ or $\displaystyle z=u(r,\theta)$?
Quote:

Any one of those three is correct- with the understanding that x and are functions of u and v (and vice-versa) they all mean the same thing.

Quote:

2)I know we can get $\displaystyle \frac{\partial z}{\partial r}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$and so on for $\displaystyle \frac{\partial z}{\partial \theta}$.
Now lets say we have $\displaystyle z=u(x(r,\theta),y(r,\theta))$

How do I get $\displaystyle \frac{\partial z}{\partial r}$ and $\displaystyle \frac{\partial z}{\partial \theta}$ in terms of $\displaystyle \frac{\partial u}{\partial x}$ and $\displaystyle \frac{\partial u}{\partial y}$?
Do I differentiate both sides of the equation?
If you know x and y as functions, of r and $\displaystyle \theta$, do the differentiations and put them into that equation. For example, if $\displaystyle x= r cos(\theta)$ and $\displaystyle y= r sin(\theta)$, then
$\displaystyle \frac{\partial z}{\partial r}= \frac{\partial z}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$
$\displaystyle = cos(\theta)\frac{\partial z}{\partial x}+ sin(\theta)\frac{\partial z}{\partial y}$.

Thanks