# Simple PDE Question

• Jan 19th 2013, 01:39 PM
Hal2001
Simple PDE Question
I ran across the following in a book, but I don't quite understand why we have fx+p fz=0 (and similarly with y). A little guidance would be greatly appreciated.
Attachment 26615
• Jan 19th 2013, 06:18 PM
TheEmptySet
Re: Simple PDE Question
Quote:

Originally Posted by Hal2001
I ran across the following in a book, but I don't quite understand why we have fx+p fz=0 (and similarly with y). A little guidance would be greatly appreciated.
Attachment 26615

The assumptions is that z is a function of both x and y, that is

$\displaystyle z(x,y) \implies f(x,y,z(x,y),a,b)=0$

Now just use the chain rule

$\displaystyle \frac{\partial }{\partial }f(x,y,z(x,y),a,b)= \frac{\partial f}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial f}{\partial a}\frac{\partial a}{\partial x}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial x}=0$

This reduces to

$\displaystyle \frac{\partial }{\partial x}f(x,y,z(x,y),a,b)= \frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial x} =0$
• Jan 19th 2013, 06:26 PM
Hal2001
Re: Simple PDE Question
:) got it thanks. Didn't realize at the time that they were defining z as a function of x and y. But it makes sense now.