 Solution

ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2015/geometric-modelling/2/e/solution

Rotating the geometry through

18 0^{\circ}

sends every point

p

- p

. To start the extension at the origin, also reverse the parameter:

E (t) = - F (1 - t), E (t) = (2 t^{2}, - 2 t + 2 t^{2}), 0 \leq t \leq 1.

Then

E (0) = (0, 0) = F (1)

and

E (1) = (2, 0)

. Merely using

- F (t)

would trace the same rotated image toward the origin, instead of away from the join.

The reversed and rotated de Boor points are

- D_{2}, - D_{1}, - D_{0}

, namely

(0, 1), (0, - 1), (4, 1)

. The first two are already

D_{1}, D_{2}

, so there is only one new de Boor point,

D_{3} = (4, 1)

. Using the same uniform basis matrix,

E (t) = \frac{1}{2} [t^{2} t 1] 1 - 2 1 - 2 21 100 004 1 - 1 1 .

Equivalently, the Bézier curve handles are

C = (0, 0)

- B = (0, - 1)

and

- A = (2, 0)

, which gives the same polynomials immediately.

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