Tutorial Solutions
Parametric Cubic Curves

1. A curve of the desired type will look something like the following:
   
   The general form is given by:

   x(t) = T A_x
   y(t) = T A_y
   z(t) = T A_z
   

   where

   T = [ t^3 t^2 t 1 ]
   A = [ a3 ]
       [ a2 ] 
       [ a1 ]
       [ a0 ]
   

   We know the following four constraints hold for x (and analogous constraints hold for y and z):

   t=0:   P0_x = [ 0     0   0  1 ] A_x
   t=1/3: P1_x = [ 1/27 1/9 1/3 1 ] A_x
   t=2/3: P2_x = [ 8/27 4/9 2/9 1 ] A_x
   t=1:   P3_x = [ 1     1   1  1 ] A_x
   

   Writing this in terms of a vector equation gives:

   G_x = B A_x
   

   Solving for A_x gives:

   A_x = B_inv G_x
   

   Thus,

   x(t) = T A_x
   x(t) = T B_inv G_x
   x(t) = T M G_x
     ( M = B_inv )
   

   where

           [ -9  27 -27  9 ]
   M = 0.5 [ 18 -45  36 -9 ]
           [-11  18  -9  2 ]
           [  2   0   0  0 ]
   

   The basis functions are given by T M:

     f1(t) = -4.5t^3 + 9t^2 - 5.5t + 1
     f2(t) = -13.5t^3 - 22.5t^2 + 9t
     f3(t) = -13.5t^3 + 18t^2 - 4.5 t
     f4(t) = 4.5t^3 - 4.5t^2 + t
   

   These look as follows:
   
   
   

2. A(0) = (0,0)
   A(1) = (1,1)
   B(0) = (1,1)
   B(1) = (3,6)
   

   A(1) = B(0), therefore we have atleast C0 (and G0) continuity.

   A'(1) = (1,2t) evaluated for t=1, giving (1,2)
   B'(0) = (2,3t^2 + 4) evaluated for t=0, givine (2,4)
   

   We see that we do not have C1 continuity.

   dy/dx for A(1) = 2/1 = 2
   dy/dx for B(0) = 4/2 = 2
   

   We therefore have G1 continuity.