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Solving response surface for 0

  1. Oct 5, 2011 #1
    Hi,

    I've done a response surface analysis, resulting in an equation of the form:

    P= A + B*X + C*X^2 + D*Y + E*Y^2 + F*X*Y

    Where A,B,C,D,E, and F are known values.

    I want to rearrange the equation so I can get a polynomial describing when P=0, but I'm stuck.

    Any suggestions?

    Thanks,

    Jono
     
  2. jcsd
  3. Oct 5, 2011 #2

    Mute

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    Homework Helper

    Your equation is a quadratic form. You generally will not be able to solve it in the form y = f(x). Such an equation (with P = 0) describes a conic section (ellipses, hypberbolas, etc).

    See this wikipedia page for details about how to identify which conic section your equation gives you.
     
  4. Oct 7, 2011 #3
    I see... So I guess I need to re-think how I've done things.

    I have a bunch of bacteria growth rates at different temperatures and water levels, and want to whack a curve on it that describes the boundary between growth and no growth for x (temp) and y(water). I've been playing around with response surfaces and binary logistic regressions, but I can't figure out how to get the resulting equations to give me an y=f(x) that describes the boundary.

    Thanks for your help, it has been very useful.

    J.
     
  5. Oct 7, 2011 #4

    Mute

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    Why does the boundary have to be a function? Couldn't there be regions of growth and non-growth that are bounded by curves that can't be described by a form y = f(x)?

    If you plot P(x,y) = 0 (e.g., using the contour-plot in Mathematica), what does the curve look like given your numbers A...E? Depending on your constants, it could be possible that the resulting curve could be described by a function for x > 0 and y > 0. For example, the curve

    (x-2)^2 + y^2 = 4

    is a conic section, and obviously a circle. You can't solve it exactly as y = f(x), although you can split it into two functions: [itex]y = \pm \sqrt{4-(x-2)^2}[/itex]. In this case, if you only care about x > 0, y > 0, you only need the [itex]y = + \sqrt{4-(x-2)^2}[/itex] solution.

    Depending on your parameters, something similar could happen for your case, but in general it won't. However, again, I'm not sure why your boundary must be a function y = f(x)?
     
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