# level curves

• Jul 6th 2008, 07:50 PM
chris25
level curves
construct a function whose level curves are lines passing through the origin.

i don't know how to express this function...can anyone help? is the center have to be 0,0 but i don't know how to write this...
• Jul 6th 2008, 09:45 PM
mr fantastic
Quote:

Originally Posted by chris25
construct a function whose level curves are lines passing through the origin.

i don't know how to express this function...can anyone help? is the center have to be 0,0 but i don't know how to write this...

Contemplate the function $f(x, y) = \frac{2xy}{x^2 + y^2}$ as an example .......
• Jul 8th 2008, 10:21 AM
Mathstud28
Quote:

Originally Posted by mr fantastic
Contemplate the function $f(x, y) = \frac{2xy}{x^2 + y^2}$ as an example .......

Maybe I am misunderstanding the question, but isn't the OP looking for a function such that $f(x,y)=c$ is linear and passes through the origin? So if your function was correct we would have that

$cx^2+cy^2=2xy$ which is not even a function let alone one that is linear.
• Jul 8th 2008, 02:31 PM
mr fantastic
Quote:

Originally Posted by Mathstud28
Maybe I am misunderstanding the question, but isn't the OP looking for a function such that $f(x,y)=c$ is linear and passes through the origin? So if your function was correct we would have that

$cx^2+cy^2=2xy$ which is not even a function let alone one that is linear.

Solve $cy^2 - 2xy + cx^2 = 0$ using the quadratic formula. You get solutions of the form y = mx where m is a function of c. There is only a certain set of values of c for which these lines exist. No level curves exist for values of c lying outside of these values.

Note: c = 0 gives the level curves x = 0 and y = 0 (which obviously pass through the origin).

Clearly this example is one of an infinite number of possibilities. I can only hope the OP has not been discouraged from contemplating it as a solution to his/her problem.
• Jul 8th 2008, 06:57 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
Solve $cy^2 - 2xy + cx^2 = 0$ using the quadratic formula. You get solutions of the form y = mx where m is a function of c. There is only a certain set of values of c for which these lines exist. No level curves exist for values of c lying outside of these values.

Note: c = 0 gives the level curves x = 0 and y = 0 (which obviously pass through the origin).

Clearly this example is one of an infinite number of possibilities. I can only hope the OP has not been discouraged from contemplating it as a solution to his/her problem.

Of relevance is the general Cartesian equation of a conic, viz.

$ax^2 + 2{\color{red}c}xy + {\color{red}b}y^2 + 2dx + 2ey + f = 0$.

(I have switched the standard notation slightly [see red] to be consistent with the example I gave).

It defines a hyperbola, a parabola, an ellipse, a circle or a pair of lines depending on the value of the invariants of the curve.

A little bit of research should show origins of the example I gave.
• Jul 9th 2008, 06:07 AM
jme44
I know there can be many different answers. But what exactly do you do to go about getting one.

is F(x,y) = 2x +3y
• Jul 9th 2008, 02:36 PM
mr fantastic
Quote:

Originally Posted by jme44
I know there can be many different answers. But what exactly do you do to go about getting one.

is F(x,y) = 2x +3y

Post #5 gives the idea I started with for constructing my example.

The level curves of the surface you've given (a plane) are lines, but the lines do not all pass through the origin. Therefore it fails to answer the question asked by the OP.