# Homework Help: Simple Partial Differentiation

1. Jan 6, 2010

### Chewy0087

1. The problem statement, all variables and given/known data
Hey >.< another simple problem;

find $$\frac{\partial u}{\partial y} , \frac{\partial v}{\partial y}, \frac{\partial w}{\partial y}$$

given that

u - v + 2w = x + 2z
2u + v - 2w = 2x - 2z
u - v + w = z - y

3. The attempt at a solution

I don't know, this question is really bugging me, i understand that the whole idea of partial differentiation is only to have to deal with the change of one variable, so why are you given the first two equations? if you assume that everything apart from y & u are constant then of course

$$\frac{\partial u}{\partial y} = -1$$

however this isn't the answer, i thought maybe you can't assume that they're constant, because u is dependant on them, however this becomes an insanely hard/impossible simultaneous equation if you try to solve for all of their changes!

could someone give me a hint? i think i'm missing the point of this question (done all of the other questions in the exercise so far)

and of course, once again thank you for help you give me! =D

p.s answers are 0, 2 & 1

Last edited: Jan 6, 2010
2. Jan 6, 2010

### Dick

Take d/dy of all three equations. You'll get a pretty simple set of three equations in the three unknowns that you are looking for. Since it looks like x,y and z are your independent variables, dx/dy=dz/dy=0.

3. Jan 6, 2010

### Chewy0087

Hm, but how do you know that x, y and z are the independant variables, just because they ask for u, v and w's change with respect to y? Also i suppose that that means I CAN assume like you said that dx/dy = dz/dy = 0, but can't assume that dw/dy = 0?

Why can i simply choose to do that? =P (sorry if that's a long explainy question, if it is just forget it and i'll try to find a link)

and thanks alot sir, the help is invaluable (trying to self-teach this, you can tell how it's going xP)

4. Jan 6, 2010

### mugaliens

http://en.wikipedia.org/wiki/Partial_differential" [Broken]may help.

If not, consider this: Partial derivatives force all variables except one to be constants at some determined point in order to see the behavioral (derivative aka slope) result of the variable in question. However, instead of assigning constants to the other variables, partials continue to solve for the variable in question mathematically, so as to provide for a derivative solution at all values of the other variables.

I'm not a mathematician, though, so I hope one comes along, soon...

Last edited by a moderator: May 4, 2017
5. Jan 6, 2010

### Dick

It is sort of implied. It should have been explicitly stated. If you are taking partial derivatives with respect to y, that implies there are other independent variables around. Since x,y,z are consecutive letters in the alphabet I would assume they are the independent variables and that u,v and w are to be regarded as functions of x,y and z. I.e. u(x,y,z), v(x,y,z) and w(x,y,z). As I said, I think they really should have stated that if they didn't. I'm just reading between the lines.

6. Jan 6, 2010

### Chewy0087

I see, it's just common sense really, i guess if i had more experience answering these questions it would have been more obvious to me...much obliged Dick (i managed to get the answer out)! =D ps i voted for you in the polls a couple of weeks ago :D

and i have read that link...it's not VERY helpful but i agree I need to read more into the theory, i'm just learning to compute right now (not very good)