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.
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.
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.