Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Building an Equation to find (Maximum Y) ie Highest Point on a curve!
Replies: 14   Last Post: Sep 14, 2013 3:40 PM

 Search Thread: Advanced Search

 Messages: [ Previous | Next ]
 quasi Posts: 12,067 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
 Plain Text Reply

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 Y-Only 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 y-only 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

© The Math Forum at NCTM 1994-2018. All Rights Reserved.