[A83] Re: The shape of a cord
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[A83] Re: The shape of a cord
> You can't exactly do that...first of all l = 5, and (x1,
y1) and (x2,
> y2) have to be at points exactly 5 units away on this
parabola...
As I said, it's quite easy to solve it for any given
case. I've already done that. However, deriving a
parametric solution seems to go beyond my capabilities.
(Patience being the most important one...)
But for a starter, here's what I could come up with
so far (I didn't do anything with these during the
last three hours...):
1.
If we have three points (x1, y1), (x2, y2) and (x3, y3),
to obtain the general formula y=ax^2+bx+c the
following formulae can be used:
y1(x2-x2)+y2(x1*x3)+y3(x2-x1)
a=(-1)-----------------------------
(x3-x2)(x1-x3)(x2-x1)
y1(x3^2-x2^2)+y2(x1^2-x3^2)+y3(x2^2-x1^2)
b=-----------------------------------------
(x3-x2)(x1-x3)(x2-x1)
y1x2x3(x2-x3)+y2x1x3(x3-x1)+y3x1x2(x1-x2)
c=-----------------------------------------
(x3-x2)(x1-x3)(x2-x1)
Of course, if any two x values are identical,
this is not a one-var function, that's why the
formulae don't work. If you look at them, you
can see that no matter how you change the
order of the points, the values don't change.
(I hope there are no typos, though.)
2.
The length of a ax^2+bx+c parabola segment between
i and j (definite integral):
int(i..j) (1+2ax+b)dx=
=(2aj+b+1)^(3/2)/3a-(2ai+b+1)^(3/2)/3a=
=((2aj+b+1)^(3/2)-(2ai+b+1)^(3/2))/3a=l (not one)
(in my case i=x1 and j=x2)
Now, it's easy to plug known values in this
equation. However, to find the general solution,
we have to substitute a and b with those nasty
formulae above (after substituting x3 with
(x1+x2)/2, naturally), and solve for y3,
i. e. express y3 with x1, x2, y1, x2 and l.
That's the homework for today. :)
PG
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