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

# Cubic Bezier point and tangent calculation

Computer Aided Design · Chapter 3: Mathematical Representation of Curves

- Asked in: year not given, 6 marks
- Video: [S 2.42 Numerical on Bezier Curve to Find Parametric Equation and Degree of Polynomials](https://www.youtube.com/watch?v=a5XRogfZvAE) — 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-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/computer_aided_design/chapter-3

## Question

A cubic Bezier curve has control points B0 = (0, 0), B1 = (1, 3), B2 = (4, 4) and B3 = (5, 0). Calculate the points on the curve at u = 0.25 and u = 0.5 and the tangent vectors at the two ends. Verify the u = 0.5 point by the de Casteljau method.

## Short answer

Using Bernstein polynomials: P(0.25)=(1.0625,1.828), P(0.5)=(2.5,2.625). End tangents: P'(0)=3(B1-B0)=(3,9), P'(1)=3(B3-B2)=(3,-12).

## Answer

#### Equation
$$
\mathbf P(u)=(1-u)^3\mathbf B_0+3u(1-u)^2\mathbf B_1+3u^2(1-u)\mathbf B_2+u^3\mathbf B_3
$$

#### At u = 0.5
Weights: $0.125,\ 0.375,\ 0.375,\ 0.125$.

$$
\begin{aligned}
x&=0.125(0)+0.375(1)+0.375(4)+0.125(5)=0.375+1.5+0.625=2.5 \\
y&=0.125(0)+0.375(3)+0.375(4)+0.125(0)=1.125+1.5=2.625
\end{aligned}
$$

#### At u = 0.25
Weights: $(0.75)^3=0.421875$; $3(0.25)(0.75)^2=0.421875$; $3(0.25)^2(0.75)=0.140625$; $(0.25)^3=0.015625$.

$$
\begin{aligned}
x&=0.421875(1)+0.140625(4)+0.015625(5)=0.421875+0.5625+0.078125=1.0625 \\
y&=0.421875(3)+0.140625(4)=1.265625+0.5625=1.828125
\end{aligned}
$$

#### End tangents
$$
\mathbf P'(0)=3(\mathbf B_1-\mathbf B_0)=3(1,3)=(3,\ 9),\qquad
\mathbf P'(1)=3(\mathbf B_3-\mathbf B_2)=3(1,-4)=(3,\ -12)
$$

#### De Casteljau check at u = 0.5
Repeated mid-points:

| Level | Points |
|---|---|
| 0 | (0,0), (1,3), (4,4), (5,0) |
| 1 | (0.5,1.5), (2.5,3.5), (4.5,2) |
| 2 | (1.5,2.5), (3.5,2.75) |
| 3 | (2.5, 2.625) |

The result $(2.5,\ 2.625)$ equals the formula value.

**Answer: P(0.25) = (1.0625, 1.828); P(0.5) = (2.5, 2.625); P'(0) = (3, 9); P'(1) = (3, -12).**

## Check yourself

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

1. What is P(0.5) for B0..B3 = (0,0),(1,3),(4,4),(5,0)?
   A. (2.5, 2.625)
   B. (2.5, 2.0)
   C. (2.0, 2.625)
   D. (1.0625, 1.828)
   Answer: A. Weights 0.125, 0.375, 0.375, 0.125 give (2.5, 2.625). (Key point: Midpoint)

2. What is P'(0)?
   A. (1, 3)
   B. (3, 9)
   C. (3, -12)
   D. (1, 9)
   Answer: B. 3(B1 - B0) = 3(1,3) = (3,9). (Key point: Start tangent)

3. How does de Casteljau compute the point at u = 0.5?
   A. Multiplying Bernstein functions by derivatives
   B. Averaging end points only
   C. Repeated mid-points of the polygon until one point remains
   D. Solving a cubic equation
   Answer: C. At u = 0.5 each level takes mid-points; the final point is on the curve. (Key point: De Casteljau)

4. What is P'(1)?
   A. (3, 12)
   B. (-3, 12)
   C. (1, -4)
   D. (3, -12)
   Answer: D. 3(B3 - B2) = 3(1,-4) = (3,-12). (Key point: End tangent)

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