Bézier paths | Year 12 | Ascent Library

01 / LEARN

The idea, step by step

Blend the controls

A quadratic Bézier path is $B = {(1 - t)}^{2} P_{0} + 2 t (1 - t) P_{1} + t^{2} P_{2}$ , $0 \leq t \leq 1$ . It starts at the first control and ends at the last. The middle control influences the bend but is not usually on the curve.

The three weights are nonnegative and sum to 1 on this interval, so the curve remains inside the control triangle. Evaluate weights before combining corresponding coordinates.

Endpoint tangents

Quadratic endpoint derivatives are $B^{'} (0) = 2 (P_{1} - P_{0})$ and $B^{'} (1) = 2 (P_{2} - P_{1})$ . A nonzero derivative gives the tangent direction.

A cubic uses four controls with weights ${(1 - t)}^{3}$ , $3 t {(1 - t)}^{2}$ , $3 t^{2} (1 - t)$ and $t^{3}$ . Its endpoint derivatives are $3 (P_{1} - P_{0})$ and $3 (P_{3} - P_{2})$ . Matching tangent direction differs from matching the complete velocity vector.

A common trap

The path passes through every control.

Instead: Only the first and last controls are guaranteed endpoints.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Find the midpoint-parameter point and initial tangent direction for quadratic controls (0, 0), (2, 3), (4, 0).

At t=1/2, weights are 1/4, 1/2, 1/4.
The weighted point is (2, 1.5).
The initial derivative is (4, 6), parallel to (2, 3).

Point (2, 1.5); initial direction (2, 3).

Example 2

Worked example

Find endpoint derivatives for cubic controls (0, 0), (1, 2), (3, 2), (4, 0).

Initial derivative is 3 times the first control displacement.
Final derivative is 3 times the last control displacement.
Apply the cubic formulas.

(3, 6)

and

(3, - 6)

Check your understanding

Find B (1/2) for quadratic controls (0, 0), (1, 4), (3, 0).

Show answer

$(1.25, 2)$

Use weights 1/4, 1/2, 1/4.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find endpoints for quadratic controls (1, 2), (4, 5), (6, 1).
2 marks
Find B (1/2) for quadratic controls (0, 0), (2, 4), (6, 0).
2 marks
Find endpoint derivatives for quadratic controls (1, 1), (2, 4), (5, 2).
2 marks
Expand the quadratic rule with controls (0, 0), (2, 3), (4, 0).
2 marks
For a cubic with P 0=(0, 0), P 1=(2, 0), find its initial derivative.
2 marks
Show quadratic Bézier weights sum to 1.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find endpoints for quadratic controls (1, 2), (4, 5), (6, 1).
2 marks
At t=0 only the first weight remains.
At t=1 only the last remains.
Start (1, 2), end (6, 1).
Find B (1/2) for quadratic controls (0, 0), (2, 4), (6, 0).
2 marks
Use weights 1/4, 1/2, 1/4.
The coordinates are 1+1.5 and 2.
$(2.5, 2)$
Find endpoint derivatives for quadratic controls (1, 1), (2, 4), (5, 2).
2 marks
Compute twice each end control displacement.
$2 (1, 3)$ and $2 (3, - 2)$ .
$(2, 6)$ and $(6, - 4)$ .
Expand the quadratic rule with controls (0, 0), (2, 3), (4, 0).
2 marks
The x coordinate is $4 t (1 - t) + 4 t^{2} = 4 t$ .
The y coordinate is $6 t (1 - t)$ .
$B = (4 t, 6 t - 6 t^{2})$ .
For a cubic with P 0=(0, 0), P 1=(2, 0), find its initial derivative.
2 marks
Use $3 (P_{1} - P_{0})$ .
Multiply (2, 0) by 3.
$(6, 0)$
Show quadratic Bézier weights sum to 1.
2 marks
Expand every product.
All t terms cancel, leaving 1.
${(1 - t)}^{2} + 2 t (1 - t) + t^{2} = 1$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics topic 6.3–6.5: polynomial and trigonometric parameterisations, related coordinate rates, motion, tangent directions and arc length. This is school Vector calculus, not vector-field calculus.

Official curriculum source ↗

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