Parametric Cubic Curve Middle Coordinates

**jpezze** · October 11th 2009, 01:59 PM

If you have two end points of x0,y0,z0 and x1, y1, z1 respectively of a parametric cubic curve and their known derivatives how do you find its middle coordinates x0.5, y0.5, z0.5

Any help that would be great thanks.

**Media_Man** · October 14th 2009, 07:56 AM

First, write the general formula...

$x(t)=at^3+bt^2+ct+d$
$x'(t)=3at^2+2bt+c$

For vector variables $x,a,b,c$

$s=x(0)=d$
$t=x'(0)=c$
$u=x(1)=a+b+c+d$
$v=x'(1)=3a+2b+c$

Given vectors $s,t,u,v$ , you should be able to easily solve this system for $a,b,c,d$ ...

$a=2s+t-2u+v$
$b=-3s-2t+3u-v$
$c=t$
$d=s$

All of these are vectors, so you now have your general parametric equation $x(t)=at^3+bt^2+ct+d$

Plug and chug.

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