# Calc III: Use of variables in functions

1. Aug 11, 2008

### PhysicsHelp12

If I'm given an equation eg. 3x^2-2xy=3z+1 (just a random eg.)

Is it a correct assumption to think of the x,y and z in the equation as corresponding

to x,y and z as I'd think of them on the cartesian plane...They're not just dummy variables

but in this case actually represent something ...like they arent going to switch them around

on me and use y in place of z and x in place of y...

because x=sqrt(y) (right side of parabola) isnt the same as y=sqrt(x) ....so is this

assumption that x and y and z are used on purpose --and arent just 'dummy variables'

2. Aug 12, 2008

### HallsofIvy

I'm not sure I understand your question. In a specific application you are either told what "x, y, z" mean or you assign them yourself (and it would be a good idea to say what "x, y, z" mean at the beginning). If, in a problem, one person derives y= x2 and another x= y2, they really have the same thing- one graph has just swapped the x,y coordinates of the other. No, they are not just "dummy variables"- but swapping them around would just give the same graph in a different coordinate system.

If, for example, I have a problem that says "A farmer has 400 yards of fencing. He wants to fence a rectangular pasture in which he can use a river bank as one side (and so needs no fencing). What are the dimensions that will give the largest area?"

It is my choice whether I use x to mean the length of the side along the river bankand y the side perpendicular to it or vice-versa. If I do choose x to be the length of the side along the river bank, my equations become, "length of fencing used: x+ 2y= 400, area= xy" and I get x= 200 yards, y= 100 yards. If I choose it the other way around, y is the length of the side along the river bank, the equations become y+ 2x= 400, area= xy and I get y= 200 yards, x= 100 yards, but those are exactly the same answer.