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.

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- Oct 11th 2009, 01:59 PMjpezzeParametric Cubic Curve Middle Coordinates
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. - Oct 14th 2009, 07:56 AMMedia_Man
First, write the general formula...

$\displaystyle x(t)=at^3+bt^2+ct+d$

$\displaystyle x'(t)=3at^2+2bt+c$

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

$\displaystyle s=x(0)=d$

$\displaystyle t=x'(0)=c$

$\displaystyle u=x(1)=a+b+c+d$

$\displaystyle v=x'(1)=3a+2b+c$

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

$\displaystyle a=2s+t-2u+v$

$\displaystyle b=-3s-2t+3u-v$

$\displaystyle c=t$

$\displaystyle d=s$

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

Plug and chug.