# Parametric Cubic Curve Middle Coordinates

• Oct 11th 2009, 01:59 PM
jpezze
Parametric 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 AM
Media_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.