---
title: "Bezier curve: equation and properties"
subject: "Computer Aided Design"
chapter: 3
key: "semester_4/computer_aided_design:bezier-curve-properties"
source: https://www.bctengineer.com/q/semester_4/computer_aided_design/bezier-curve-properties
site: BCT Engineering
---

# Bezier curve: equation and properties

Computer Aided Design · Chapter 3: Mathematical Representation of Curves

- Asked in: year not given, 6 marks
- Video: [S 2.41 Bezier Curve | Characteristics and Parametric Equation of Bezier Curve](https://www.youtube.com/watch?v=kZuTiIby5aU) — Dr. Khizar Ahmed Pathan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/computer_aided_design/bezier-curve-properties
- Whole chapter: https://www.bctengineer.com/md/semester_4/computer_aided_design/chapter-3

## Question

Define a Bezier curve. Write the general equation with Bernstein basis functions and explain the main properties of Bezier curves. What are the limitations of Bezier curves?

## Short answer

A Bezier curve is a parametric curve defined by control points; it approximates the polygon. Equation: P(u)=Σ B_i J_{n,i}(u) with Bernstein basis J_{n,i}=C(n,i) u^i (1-u)^{n-i}. Properties: passes through first and last points, end tangents along polygon edges, convex hull, variation diminishing, global control. Limitations: degree rises with points, no local control, cannot represent exact conics.

## Answer

A **Bezier curve** is a parametric curve defined by a set of control points forming a **control polygon**. The curve approximates the polygon: it starts at the first and ends at the last control point, and is pulled towards the other points. It was developed by Pierre Bezier at Renault.

#### Equation
For $n+1$ control points $\mathbf B_0,\dots,\mathbf B_n$ (degree $n$):

$$
\mathbf P(u)=\sum_{i=0}^{n}\mathbf B_i\,J_{n,i}(u),\qquad 0\le u\le1
$$

with the Bernstein basis

$$
J_{n,i}(u)=\binom{n}{i}u^i(1-u)^{n-i}=\frac{n!}{i!(n-i)!}u^i(1-u)^{n-i}
$$

For a cubic ($n=3$): $J_{3,0}=(1-u)^3$, $J_{3,1}=3u(1-u)^2$, $J_{3,2}=3u^2(1-u)$, $J_{3,3}=u^3$.

#### Properties
1. The curve is of degree one less than the number of control points.
2. It passes through the first and last control points: $\mathbf P(0)=\mathbf B_0$, $\mathbf P(1)=\mathbf B_n$.
3. End tangents lie along the first and last polygon sides: $\mathbf P'(0)=n(\mathbf B_1-\mathbf B_0)$, $\mathbf P'(1)=n(\mathbf B_n-\mathbf B_{n-1})$.
4. **Convex hull property:** the curve lies inside the convex hull of its control points, because the basis functions are non-negative and sum to 1.
5. **Variation diminishing:** the curve does not cross any straight line more times than the control polygon does.
6. Symmetry: reversing the order of the points gives the same curve.
7. Invariant under affine transformation: transform the control points and redraw.
8. Global control: moving one control point changes the whole curve.

#### Limitations
- Degree rises with the number of points, so many points give a high-degree, costly curve.
- No local control.
- Cannot represent exact conics such as a circle (rational Bezier is required).
- Joining segments smoothly needs the three points around the joint to be collinear (tangent, $G^1$, continuity); for $C^1$ continuity between segments of equal degree they must also be equally spaced.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which is the Bernstein basis J(n,i)(u)?
   A. u^i (1+u)^n
   B. C(n,i) u^i (1-u)^(n-i)
   C. C(n,i) u^n
   D. (1-u)^i
   Answer: B. J_{n,i}(u) = C(n,i) u^i (1-u)^{n-i}. (Key point: Bernstein basis)

2. Why does the curve lie inside the convex hull?
   A. It passes through all points
   B. Degree is fixed
   C. Basis functions are non-negative and sum to 1
   D. It is symmetric
   Answer: C. Non-negative weights summing to 1 make each point a convex combination of control points. (Key point: Convex hull)

3. What is P'(0) of a Bezier curve of degree n?
   A. B1 - B0
   B. (B1 - B0)/n
   C. n(B0 - B1)
   D. n(B1 - B0)
   Answer: D. The tangent at the start is n times the first polygon side. (Key point: End tangent)

4. Which is a limitation of Bezier curves?
   A. No local control
   B. Does not pass through end points
   C. Not affine invariant
   D. Cannot be closed
   Answer: A. Moving one control point changes the whole curve, and degree grows with the number of points. (Key point: Limitation)

## Related in Computer Aided Design

- [Bezier surface equation and properties](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Abezier-surface-properties) — Computer Aided Design, chapter 4
- [B-spline curve versus Bezier curve](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Abspline-vs-bezier) — Computer Aided Design, chapter 3
- [Derive Hermite cubic curve and blending functions](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Ahermite-cubic-derivation) — Computer Aided Design, chapter 3
- [Cubic Bezier point and tangent calculation](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Abezier-curve-numerical) — Computer Aided Design, chapter 3
- [Convert Hermite data to Bezier cubic](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Ahermite-to-bezier-conversion) — Computer Aided Design, chapter 3
- [Biquadratic Bezier surface point calculation](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Abezier-surface-numerical) — Computer Aided Design, chapter 4

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- [Bezier properties; point at u=0.2](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Abezier-properties-point-u-0-2) — Computer Graphics, chapter 5, asked 2×