Question

THEORETICAL EXERCISES 6. Suppose a cubic Bézier polynomial is placed through (uo, vo) and (u3, v3) with guidepoints (u\, v1) and (u2, U2), respectively. Derive the parametric equations for u(t) and v(t) assuming that а. u(0) = uo, и(1) — из, u' (0) = u, - Uo, и'(1) — из — и2 and v'(0) = v1- Vo v'(1)= v3 - v2 v (0) = vo, v(1)= v3, b. Let f(i/3) ui, for i = 0, 1, 2, 3, and g(i/3) = Vi, for i = 0, 1, 2,3. Show that the Bernstein polynomial of degree three in t for f is u(t) and the Bernstein polynomial of degree three in t for g is v(t). (See Exercise 23 of Section 3.1.)

Answer 1

Blending function (0) The Bezier curve is obtained by using which is given by u(t)= È uz Jni(t) For a cubic Bezier polynomial, set n= 3 lie uch). È uz J.: (+) wwhere Ini- n! į - tri 4 (n-1)] Jogo (t) = (1-+)3 J3, (t) = 3 (1-t²t Jg.z(t) = 3+ (1-t) J3, 3(t) = t3 Also Ž Ini(t) = 1 In our question, given u (o) = lo V(o)= Vo Ull)= Uz VC) = V₂ U'(o) = U,.lo V'(o) = V, -vo U'(1) - Uz-uz V(1) = Uz - Uz Using forward difference formula 6= 0 3 (urus) Uz=1 3(Uz-42) which is similar to the polynomial ult) = lot + 3 (4,-4)t +3z-U2) t² + (4,4) (42-412) t² Similarly V(t) - Vot + 3 (0,-)t + 3 (03-V2)ť + (Wi-vo) (V3 - Vn) t

(6) Using the formular for Beanstein polynomial u(t)= ug (1-4)3 + 3uit (1-4)2 + 34gt? (1-) + Ugt and VCL) = Vo (1-)3 + 3vit (1-t)2+ 3vzť (1-t) + Vot3