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

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Homework Statement


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.

Homework Equations

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.
 
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masterchiefo said:

Homework Statement


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 line that lies on that z(x,y) function.

Homework Equations

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.
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?
 
SteamKing said:
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?
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.
 
masterchiefo said:

Homework Statement


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.

Homework Equations

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.

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)##.
 
Ray Vickson said:
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)##.
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.
 
masterchiefo said:
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.
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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