
Re: Equation to find Highest Point on a curve!
Posted:
Sep 12, 2013 2:41 AM


On Wed, 11 Sep 2013, mervynmccrabbe@gmail.com wrote:
> x^4 + y^4 + A(x^2)  A(y^2) + 2(x^2)(y^2)  Bxy + C = 0 > > If I have correctly evaluated > dy/dx to be = [4(x^3) + 2Ax + 4x(y^2)  By]
You haven't. Letting y' = dy/dx, 4x^3 + 4y^3 y' + 2ax  2ayy' + 4xy^2 + 4x^2 yy'  by  bxy' = 0
4y^3 y'  2ayy' + 4x^2 yy'  bxy' + 4x^3 + 2ax + 4xy^2  by = 0
y' = (4x^3 + 2ax + 4xy^2  by)/(4y^3  2ay + 4x^2 y  bx)
> 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.
What? To find extreme values of y, set y' = 0. That gives you two equations to solve for x and y with the additional requirement that 4y^3  2ay + 4x^2 y  bx /= 0
> I've tried completing squares etc but can not get rid of composite XY > terms. > Complete the square on 4x^3 + 2ax + 4xy^2  by = 0 to get y in terms of x and then substitue y = f(x) into the first equation and solve for x.
That's a heafty task. Next, get values for y and check to see if they satisfy the addional requirement and then if they're maximal, miminal or points of inflection.
> If I could eliminate X and get a generalised YOnly equation > then I could manage the rest. > > Any help would be appreciated > > Mervyn Mc Crabbe >

