# Equation of a line that lies on a z=f(x,y) function

1. Oct 12, 2015

### masterchiefo

• Member warned about posting with no effort shown
1. The problem statement, all variables and given/known data
Hello,
I have this function z(x,y) = sin(2*x)*cos(2*y)*e-(x^2+y^2)/6

I need to find an equation of any curve line that lies on that z(x,y) function.

2. Relevant equations

3. The attempt at a solution
Sorry, I really have no idea how to proceed on this, been looking on my note book and google and have not find anything.

2. Oct 12, 2015

### SteamKing

Staff Emeritus
What do you mean a line "that lies on that ... function"?

Are you talking about a line which is tangent to the function's surface?

3. Oct 12, 2015

### masterchiefo

Sorry edited the post, its a curve line.

To clarify, the curve line has to be part of the z(x,y) function. The curve line has to be completely on the function. If I didn't clarify anything, let me know.

4. Oct 12, 2015

### Ray Vickson

A curve $(x(t),y(t),z(t))$ lying in the surface must project down to a curve on the $x\,y$ plane, obtained by just ignoring the $z$-component $z(t)$.

5. Oct 12, 2015

### masterchiefo

how do I proceed on this? what are the steps I have to take.

Do I take a random number of x and y and plot it in my equation to find z?
sorry I have never done this before.

6. Oct 12, 2015

### Staff: Mentor

It might be that you need to find level curves on this surface. As already mentioned, the graph of $z = f(x, y) = \sin(2x) \cos(2y) e^{-(x^2+y^2)/6}$ represents a surface in three dimensions. If you set z to some specific value, that defines a curve that lies on the surface. On level curve is when z = 0, or $0 = \sin(2x) \cos(2y) e^{-(x^2+y^2)/6}$. Geometrically, this level curve is the intersection of the surface with the x-y plane.