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Math Help - Points of inflection in a implicite curve

  1. #1
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    Points of inflection in a implicite curve

    Hi!

    You can say like finding the eventual points of inflection of a curve given in implicit form f (x, y)=0 and that not it can or is difficult to explicit in form y=y (x) or x=x (y),
    how for example
    x^5 + y^5 + x^2*y - x*y^2 = 0 ?


    P.S. I hope it is the correct subforum ...
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  2. #2
    MHF Contributor chisigma's Avatar
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    Given a function y(x) expressed in the form f(x,y)=0 , if certain conditions are satisfied, the derivative exists and is...

    \displaystyle y^{'} (x) = - \frac{f_{x}^{'} (x,y)}{f_{y}^{'} (x,y)} (1)

    Now if further conditions are satisfied it exists also the second derivative and is...

    \displaystyle y^{''} (x) = - \frac{f_{xx}^{''}\ f_{y}^{' 2} - 2 f_{xy}^{''}\ f_{x}^{'}\ f_{y}^{'} + f_{yy}^{''}\ f_{x}^{'2}}{f_{y}^{'3}} (2)

    At this point the inflection points of y(x) are found in standard way imposing that y^{''} (x)=0 ...

    Kind regards

    \chi \sigma
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  3. #3
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    chisigma, I am confused as to why you have both the f' that is normally used with ordinary derivatives and f_x and f_y for partial derivatives.

    I would say that if f(x,y)= 0, then f_x+ f_y\frac{dy}{dx}= 0 so that \frac{dy}{dx}= -\frac{f_x}{f_y}.

    Then, using the quotient rule, \frac{d^2y}{dx^2}=-\frac{f_{xx}f_y- f_xf_{yy}}{(f_y)^2}. That will be 0 when the numerator is 0: when f_{xx}f_y= f_xf_{yy}.
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  4. #4
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    Thanks for your explanation.
    Therefore x,y is a point of inflection when
    simply fxx*fy = fx*fyy ?
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    ... I would say that if f(x,y)= 0, then f_x+ f_y\frac{dy}{dx}= 0 so that \frac{dy}{dx}= -\frac{f_x}{f_y}. Then, using the quotient rule, \frac{d^2y}{dx^2}=-\frac{f_{xx}f_y- f_xf_{yy}}{(f_y)^2}. That will be 0 when the numerator is 0: when f_{xx}f_y= f_xf_{yy}.
    The matter is a little different: starting from...

    f(x,y(x))=0 (1)

    ... deriving both terms respect to x You obtain...

    \displaystyle  f_{x} + f_{y} \ y^{'} = 0 \rightarrow y^{'} = - \frac{f_{x}}{f_{y}} (2)

    Now if You derive both terms of (2) respect to x You obtain...

    \displaystyle f_{xx} + 2\ f_{xy}\ y^{'} + f_{yy}\ y^{'2} + f_{y}\ y^{''} = 0 (3)

    ... that setting y^{'} = - \frac{f_{x}}{f_{y}} becomes...

    \displaystyle y^{''} = - \frac{f_{xx}\ f_{y}^{2} - 2\ f_{xy}\ f_{x}\ f_{y} + f_{yy}\ f_{x}^{2}}{f_{y}^{3}} (4)

    Now the point of inflexion of y(x), if they exist, are the points in the x y plane for which the relations...

    f = 0

    f_{xx}\ f_{y}^{2} - 2\ f_{xy}\ f_{x}\ f_{y} + f_{yy}\ f_{x}^{2} =0 (5)

    ... are both satisfied with the condition that f_{y} \ne 0...

    Kind regards

    \chi \sigma
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  6. #6
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    Thanks, that was a lot clearer.
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  7. #7
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    You've stated necessary but not sufficient conditions for points to be points of inflection. Technically, the second derivative being zero does not imply that you have a point of inflection. You also need the lowest-order non-zero derivative to be of odd order. So, check out the first derivative at any candidate points. If it's nonzero at a candidate point, then it's a point of inflection. If the first derivative is zero at a candidate point, then check the third derivative (oh, joy!). Alternatively, a point of inflection is a point where the second derivative changes sign. So, if you find all the zeros of the second derivative, you should be able, by plugging in appropriate x-values, to see whether the second derivative changes sign at its zeros or not.
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  8. #8
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    A doubt about the points of inflection of a curve

    Thank you very much for your explanations about point of inflection.
    But I have a doubt...
    For the quintic
    F = y(20^4 - 60y^2 + 39) + 20x^2y(5x^2 + 6y^2 - 7) + x = 0
    I have 13 points of inflection or 14?
    The 14 point is infinite orizzontal point ?
     
    I have the graphic of this curve
    F = 0
    and its intersections with the curve
    FxxFy^2-2FxyFxFy+FyyFx^2 = 0 ,
    but I don't know how to send you (please, without the http:\!) 
    Help me, please!
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