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Math Help - level curves

  1. #1
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    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...
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  2. #2
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    Quote Originally Posted by chris25 View Post
    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 .......
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    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.
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  4. #4
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    Quote Originally Posted by Mathstud28 View Post
    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.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    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.
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  6. #6
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    Answer

    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
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  7. #7
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    Quote Originally Posted by jme44 View Post
    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.
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