An animation demonstrates a car driving along a road that is given by a Bezier curve along with the subsequent control points:
Xk

0

5

40

50

Yk

0

40

5

15

The animation finals 10 seconds and the key frames are to be calculated at 1 second intervals. Compute the position of the car on the road at the beginning of the 6^{th} second of the animation.
Solution: By using similar process in the previous exercise we can compute the blending functions as:
1. B_{03} = 3!/(0! x (30)!) u^{0}(1  u)^{(30)} = 1u^{0}(1  u)^{3} = (1  u)^{3}
2. B_{13} = 3!/(1! x (31)!) u^{1}(1  u)^{(31)} = 3u^{1}(1  u)^{2} = 3u(1  u)^{2}
3. B_{23} = 3!/(2! x (32)!) u^{2}(1  u)^{(32)} = 3u^{2}(1  u)^{1} = 3u^{2}(1  u)
4. B_{33} = 3!/(3! x (33)!) u^{3}(1  u)^{(33)} = 1u^{3}(1  u)^{0} = u^{3}
The function x(u) is equivalent to x(u) = ∑x_{k}B_{k}; here k=0,1,2,3
x(u) = ∑ x_{k}B_{k} = x_{0}B_{03} + x_{1}B_{13} + x_{2}B_{23 }+ x_{2}B_{33}
= (0)(1  u)^{3 } + 5 [3u(1  u)^{2}] + 40 [3u^{2}(1  u)] + 50 u^{3}
= 15u(1  u)^{2} + 120u^{2}(1  u) + 50u^{3}
As the same y(u) = y_{0}B_{03} + y_{1}B_{13}+ y_{2}B_{23}+ y_{2}B_{33}
= (0)(1  u)^{3} + 40 [3u(1  u)^{2}] + 5 [3u^{2}(1  u)] + 15 u^{3}
= 120u(1  u)^{2} + 15u^{2}(1  u) + 15u^{3}
At the beginning of the sixth second of the animation, that is, when u=0.6, we can utilize these equations to work out such x(0.6) = 29.52 and y(0.6) = 16.92.
The way of the car looks like as: