quasi
Posts:
11,060
Registered:
7/15/05


Re: Building an Equation to find (Maximum Y) ie Highest Point on a curve!
Posted:
Sep 12, 2013 3:03 AM


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 = 4x^3 + 2ax + 4x(y^2)  by
No, that's not correct.
The correct result is:
dy/dx =
(4x^3 + 2ax + 4x(y^2)  by)/(4y^3 + 2ay  4y(x^2) + bx)
However, it _is_ true that dy/dx = 0 implies
4x^3 + 2ax + 4x(y^2)  by = 0
>If I could eliminate X and get a generalised YOnly equation >then I could manage the rest.
Ok, you asked for it ...
The system of equations
x^4 + y^4 + a(x^2)  a(y^2) + 2(x^2)(y^2)  bxy + c = 0
4x^3 + 2ax + 4x(y^2)  by = 0
yields the yonly equation
(d_8)(y^8) + (d_6)(y^6) + (d_4)(y^4) + (d_2)(y^2) + (d_0) = 0
where
d_8 = 256b^2 + 1024a^2
d_6 = 192(b^2)a  1024ac  768(a^3)
d_4 = 240(a^4)  168(a^2)(b^2) + 1920(a^2)c  27(b^4) + 288(b^2)c + 256(c^2)
d_2 = 16(a^5)  4(a^3)(b^2) + 384(a^3)c  1280a(c^2) + 144a(b^2)c
d_0 = 256(c^3) + 16(a^4)c  128(a^2)(c^2)
quasi

