Re: Functions of several variables

On Sat, 9 Feb 2008 21:43:08 -0800 (PST), mathwiz <mathwiz@xxxxxxxxxx>
wrote:

Hi guys, we're just getting in to functions of several variables, and
I'm having trouble with one of the problems. Here it is:

Find a function f(x,y,z) whose level surface f=1 is the graph of the
function g(x,y)=x+2y

First, I set f=1, then if f(x,y,z) is a function, and g(x,y) is a
function, where g = f when f = 1, then the equation x+2y=z+1 should
represent the surface of f when z=0. So moving the variables over, I
end up with f(x,y,z)=x+2y-z-1, am I doing this correct?

No.

If you take your proposed solution, then set f = 1 and solve for z,
you get z = x + 2y - 2, which doesn't match the requirements.

Instead, try this ...

Let w = f(x,y,z)

You want to the 2 equations

w = 1

z = x + 2y

to be equivalent, where w is some function of x,y,z.

Forcing the right hand sides to be zero, you want the 2 equations

w - 1 = 0

z - x - 2y = 0

to be equivalent.

It suffices to set the left hand sides equal and solve for w, yielding

w = z - x - 2y + 1

Thus, one solution is

f(x,y,z) = z - x - 2y + 1

quasi
.