--- title: "Derive cubic Bezier equation and properties" subject: "Computer Graphics" chapter: 5 key: "semester_5/computer_graphics:bezier-cubic-equation-properties" source: https://www.bctengineer.com/q/semester_5/computer_graphics/bezier-cubic-equation-properties site: BCT Engineering --- # Derive cubic Bezier equation and properties Computer Graphics · Chapter 5: Curve Modeling - Asked in: 2071 Shrawan, 8 marks - Video: [Bezier Curve Derivation in Computer Graphics | CG | Lec-43 | Bhanu pray](https://www.youtube.com/watch?v=2OSEyo9oS6E) — Education 4u - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/computer_graphics/bezier-cubic-equation-properties - Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-5 ## Question Derive the equation for cubic Bezier curve. Also write down its properties. ## Short answer Cubic Bezier: P(u) = (1-u)^3 P0 + 3u(1-u)^2 P1 + 3u^2(1-u) P2 + u^3 P3. It passes through P0 and P3, with end tangents 3(P1-P0) and 3(P3-P2). Properties: lies inside the convex hull, affine invariance, global control, symmetry, variation diminishing, and blending functions are positive and sum to 1. ## Answer #### Cubic Bezier equation A cubic Bezier curve has four control points P0, P1, P2, P3 (n = 3). The general Bezier curve is P(u) = Σ P_i · C(n,i) u^i (1−u)^(n−i), 0 ≤ u ≤ 1. For n = 3: C(3,0) = 1, C(3,1) = 3, C(3,2) = 3, C(3,3) = 1 **P(u) = (1−u)³ P0 + 3u(1−u)² P1 + 3u²(1−u) P2 + u³ P3** Blending functions: B0 = (1−u)³, B1 = 3u(1−u)², B2 = 3u²(1−u), B3 = u³ (they add to 1). Expanding in powers of u: P(u) = (−P0 + 3P1 − 3P2 + P3) u³ + (3P0 − 6P1 + 3P2) u² + (−3P0 + 3P1) u + P0 Matrix form: ```text | -1 3 -3 1 | | P0 | P(u) = [u³ u² u 1]| 3 -6 3 0 | | P1 | | -3 3 0 0 | | P2 | | 1 0 0 0 | | P3 | ``` Check at u = 0: P = P0. At u = 1: P = P3. #### Properties 1. Starts at P0 and ends at P3; other points are not on the curve. 2. Tangent at the ends: P'(0) = 3(P1 − P0), P'(1) = 3(P3 − P2). 3. Lies inside the convex hull of the control points. 4. Blending functions are positive and sum to 1. 5. Affine invariance. 6. Global control: moving a point changes the whole curve. 7. Symmetric: reversing the point order gives the same curve. 8. Variation diminishing. 9. Adjacent sections have C⁰ continuity if P3 of the first = P0 of the second, and C¹ if also the points around the join are collinear with equal spacing. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is the cubic Bezier equation? A. (1-u)^3 P0 + 3u(1-u)^2 P1 + 3u^2(1-u) P2 + u^3 P3 B. (1-u)^3 P0 + u(1-u)^2 P1 + u^2(1-u) P2 + u^3 P3 C. (1-u)^2 P0 + 2u(1-u) P1 + u^2 P2 D. u^3 P0 + 3u^2(1-u) P1 + 3u(1-u)^2 P2 + (1-u)^3 P3 + P0 Answer: A. The binomial coefficients C(3,i)=1,3,3,1 multiply u^i(1-u)^(3-i). (Key point: Cubic Bezier equation) 2. What are the coefficients C(3,i) used in the cubic Bezier? A. 1, 2, 2, 1 B. 1, 3, 6, 1 C. 1, 3, 3, 1 D. 3, 3, 3, 3 Answer: C. C(3,0)=1, C(3,1)=3, C(3,2)=3, C(3,3)=1. (Key point: Binomial coefficients) 3. What are the end tangents of a cubic Bezier curve? A. P'(0)=P1-P0 and P'(1)=P3-P2 B. P'(0)=3(P2-P0) and P'(1)=3(P3-P1) C. P'(0)=3(P0-P1) and P'(1)=3(P2-P3) D. P'(0)=3(P1-P0) and P'(1)=3(P3-P2) Answer: D. The tangent at each end points along the first or last polygon edge, scaled by n=3. (Key point: End tangents) 4. Moving one control point of a Bezier curve: A. Changes only the part near that point B. Changes the whole curve (global control) C. Changes nothing unless it is an end point D. Moves the curve without changing its shape Answer: B. Every blending function is non-zero over the interval, so Bezier curves have global, not local, control. (Key point: Control type) 5. Where does a Bezier curve lie? A. Always on the control polygon B. Outside the control polygon C. Anywhere in the plane D. Inside the convex hull of its control points Answer: D. Blending functions are non-negative and sum to 1, so every point is a convex combination of control points. 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