
Re: Building an Equation to find (Maximum Y) ie Highest Point on a curve!
Posted:
Sep 11, 2013 12:20 PM


Am 11.09.2013 17:28, schrieb mervynmccrabbe@gmail.com: > x^4 + y^4 + A(x^2)  A(y^2) + 2(x^2)(y^2)  Bxy + C = 0
I get: 4x^3 + 4y^3*y' + 2Ax  2Ayy' + 4xy^2 +4x^2yy'  By  Bxy' = 0 where y' = dy/dx. Then:
(*): y'(4y^3  2Ay + 4x^2y  Bx) = 4x^3  2Ax 4xy^2 + By. Then (**): y' = (4x^3  2Ax 4xy^2 + By) / (4y^3  2Ay + 4x^2y  Bx)
> > If I have correctly evaluated > > dy/dx to be = [4(x^3) + 2Ax + 4x(y^2)  By]
No. You must differentiate your equation w.r.t. x, treating y as a function of x applying the chainrule correctly. You then end up with equation (*) and  if you want  equation (**) comes next. But you don't need it. Just set y' = 0 in (*) since you are only interested in extrema. Then you get the equation:
4(x^3) + 2Ax + 4x(y^2)  By = 0 as a necessary condition for y' = 0.
> > and if it is proper to set this then equal to zero > to give a new equation that could be merged with the original to get rid > of the cumbersome XY terms  then that i failed to do.
Multiply your new equation by x and add it to the original one, that gets rid of the xyterm. I get:
(***) 5x^4 + y^4 + 3Ax^2  Ay^2 + 6x^2y^2 + C = 0
> > I've tried completing squares etc but can not get rid of composite XY terms. > > If I could eliminate X and get a generalised YOnly equation > then I could manage the rest.
Can you handle (***)? This is a quadratic equation in X=x^2 and Y=y^2.
> > Any help would be appreciated > > Mervyn Mc Crabbe >
 Thomas Nordhaus

