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

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Thomas Nordhaus

Posts: 433
Registered: 12/13/04
Re: Building an Equation to find (Maximum Y) ie Highest Point on
a curve!

Posted: Sep 11, 2013 12:20 PM
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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 chain-rule 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 xy-term. 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 Y-Only 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



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