Bézier paths | Year 12 | Ascent Library

01 / LEARN

Understand the method.

THE CORE RULE

\begin{aligned} B (t) & = {(1 - t)}^{2} P_{0} \\ + 2 t (1 - t) P_{1} \\ + t^{2} P_{2} \\ 0 & \leq t \leq 1 \end{aligned}

A quadratic Bézier curve blends three controls. It starts at the first and ends at the last; the middle control shapes the bend.

Blend three controls to make a quadratic path

A quadratic Bézier path is a weighted combination of three control points. Evaluate the weights at the required parameter value, then combine corresponding coordinates.

For $0 \leq t \leq 1$ , the weights are nonnegative and sum to $1$ . The path stays in the control triangle, starts at $P_{0}$ and ends at $P_{2}$ . It generally does not pass through $P_{1}$ .

\begin{aligned} {(1 - t)}^{2} + 2 t (1 - t) \\ + t^{2} = 1 \end{aligned}

A quadratic Bézier curve with controls P0=(0,0), P1=(2,3), P2=(4,0), a dashed control polygon, midpoint-parameter point (2,3/2), and endpoint tangent arrows following the control edges.

Read tangent directions from the controls

A nonzero endpoint derivative gives the oriented tangent direction. For a quadratic, the initial direction follows the first control-polygon edge; the final direction follows the last edge.

A positive scalar multiple gives the same oriented tangent direction, but it does not give the same derivative magnitude. If neighbouring controls coincide, the corresponding endpoint derivative is zero and cannot alone supply a tangent direction.

\begin{aligned} B^{'} (0) & = 2 (P_{1} - P_{0}) \\ B^{'} (1) & = 2 (P_{2} - P_{1}) \end{aligned}

Use four controls for a cubic path

A cubic Bézier path blends four control points. Its endpoint derivatives use the first and last control edges with a factor of $3$ , instead of the quadratic factor of $2$ .

To join paths, first match their endpoint positions. Match nonzero oriented tangent directions for a common travel direction; match the complete derivative vectors when equal parameter rates are required. A dimensionless parameter derivative is not automatically a physical velocity.

\begin{aligned} C (t) & = {(1 - t)}^{3} P_{0} \\ + 3 t {(1 - t)}^{2} P_{1} \\ + 3 t^{2} (1 - t) P_{2} \\ + t^{3} P_{3} \\ C^{'} (0) & = 3 (P_{1} - P_{0}) \\ C^{'} (1) & = 3 (P_{3} - P_{2}) \end{aligned}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Evaluate a quadratic and its initial direction

A quadratic Bézier curve has controls $P_{0} = (0, 0)$ , $P_{1} = (2, 3)$ and $P_{2} = (4, 0)$ . Find $B (\frac{1}{2})$ and the initial tangent direction.

Worked solution

1
Evaluate the three weights.
$\begin{aligned} {(1 - \frac{1}{2})}^{2} & = \frac{1}{4} \\ 2 (\frac{1}{2}) (1 - \frac{1}{2}) & = \frac{1}{2} \\ {(\frac{1}{2})}^{2} & = \frac{1}{4} \end{aligned}$
The weights sum to $1$ and must be applied to the controls in their original order.
2
Combine the corresponding coordinates.
$\begin{aligned} B (\frac{1}{2}) & = \frac{1}{4} (0, 0) \\ + \frac{1}{2} (2, 3) \\ + \frac{1}{4} (4, 0) \\ = (2, \frac{3}{2}) \end{aligned}$
The curve point lies below the middle control $(2, 3)$ ; the control is not a point the curve is required to visit.
3
Use the first control displacement.
$\begin{aligned} B^{'} (0) & = 2 (\begin{array}{cccccccccc} 2 - 0 \\ 3 - 0 \end{array}) \\ = (\begin{array}{cccccccccc} 4 \\ 6 \end{array}) \end{aligned}$
This nonzero derivative is a positive multiple of $(2, 3)$ , so both vectors give the same initial oriented tangent direction.

Answer

$B (\frac{1}{2}) = (2, \frac{3}{2})$ ; the initial tangent direction is $(2, 3)$ , with derivative $(4, 6)$ .

EXAMPLE 02

Find the endpoint derivatives of a cubic

A cubic Bézier curve has controls $(0, 0)$ , $(1, 2)$ , $(3, 2)$ and $(4, 0)$ , in that order. Find its endpoint derivatives and its point at $t = \frac{1}{2}$ .

Worked solution

1
Apply the cubic initial-derivative formula.
$\begin{aligned} C^{'} (0) & = 3 (\begin{array}{cccccccccc} 1 - 0 \\ 2 - 0 \end{array}) \\ = (\begin{array}{cccccccccc} 3 \\ 6 \end{array}) \end{aligned}$
The first control edge determines the initial direction, scaled by the cubic degree.
2
Apply the cubic final-derivative formula.
$\begin{aligned} C^{'} (1) & = 3 (\begin{array}{cccccccccc} 4 - 3 \\ 0 - 2 \end{array}) \\ = (\begin{array}{cccccccccc} 3 \\ - 6 \end{array}) \end{aligned}$
Subtract the penultimate control from the final control to preserve the direction of increasing parameter.
3
Evaluate using the midpoint weights.
$\begin{aligned} C (\frac{1}{2}) & = \frac{1}{8} (0, 0) \\ + \frac{3}{8} (1, 2) \\ + \frac{3}{8} (3, 2) \\ + \frac{1}{8} (4, 0) \\ = (2, \frac{3}{2}) \end{aligned}$
The cubic weights at one half are $\frac{1}{8}$ , $\frac{3}{8}$ , $\frac{3}{8}$ and $\frac{1}{8}$ . They sum to $1$ .

Answer

$C^{'} (0) = (3, 6)$ , $C^{'} (1) = (3, - 6)$ and $C (\frac{1}{2}) = (2, \frac{3}{2})$ .

EXAMPLE 03

Choose a control for a smooth join

A drawing path uses quadratic controls $(0, 0)$ , $(2, 3)$ and $(4, 0)$ . A second quadratic starts at $Q_{0} = (4, 0)$ and ends at $Q_{2} = (8, 0)$ . Choose $Q_{1}$ so its initial derivative equals the first curve's final derivative.

Worked solution

1
Find the derivative arriving at the join.
$\begin{aligned} B^{'} (1) & = 2 (\begin{array}{cccccccccc} 4 - 2 \\ 0 - 3 \end{array}) \\ = (\begin{array}{cccccccccc} 4 \\ - 6 \end{array}) \end{aligned}$
The first curve arrives in the direction from its middle control to its endpoint.
2
Equate the next curve's initial derivative.
$2 (Q_{1} - (\begin{array}{cccccccccc} 4 \\ 0 \end{array})) = (\begin{array}{cccccccccc} 4 \\ - 6 \end{array})$
Matching the entire derivative fixes both its direction and magnitude for the two curve parameters.
3
Solve for the middle control and verify.
$\begin{array}{c} \begin{aligned} Q_{1} & = (\begin{array}{cccccccccc} 4 \\ 0 \end{array}) + (\begin{array}{cccccccccc} 2 \\ - 3 \end{array}) \\ = (\begin{array}{cccccccccc} 6 \\ - 3 \end{array}) \end{aligned} \\ 2 ((\begin{array}{cccccccccc} 6 \\ - 3 \end{array}) - (\begin{array}{cccccccccc} 4 \\ 0 \end{array})) = (\begin{array}{cccccccccc} 4 \\ - 6 \end{array}) \end{array}$
The shared endpoint already matches position. This control also makes the one-sided derivative vectors equal.

Answer

Choose $Q_{1} = (6, - 3)$ . The curves share $(4, 0)$ and both have derivative $(4, - 6)$ at the join.

03 / YOUR TURN

Try it yourself.

Calculate, explain, then apply.

CHECK 01

Calculate a weighted point

Find $B (\frac{1}{2})$ for quadratic controls $(0, 0)$ , $(1, 4)$ and $(3, 0)$ .

Show answer & reasoning

$(\frac{5}{4}, 2)$ .

Use weights $\frac{1}{4}$ , $\frac{1}{2}$ , $\frac{1}{4}$ . The horizontal coordinate is $0 + \frac{1}{2} + \frac{3}{4} = \frac{5}{4}$ and the vertical coordinate is $0 + 2 + 0 = 2$ .

CHECK 02

Interpret the middle control

A quadratic has controls $(0, 0)$ , $(2, 3)$ and $(4, 0)$ . A learner says the path must pass through $(2, 3)$ because it is a control point. Explain the error.

Show answer & reasoning

The middle control shapes the path; it is not a required point on it. Here the curve has maximum height $\frac{3}{2}$ , below $3$ .

The vertical coordinate is $y = 6 t (1 - t)$ , whose maximum on $0 \leq t \leq 1$ is $\frac{3}{2}$ at $t = \frac{1}{2}$ . Only the first and last controls are guaranteed endpoints.

CHECK 03

Apply a join between different degrees

A cubic ends at $(6, 0)$ with penultimate control $(5, 2)$ . A quadratic begins at $Q_{0} = (6, 0)$ . Choose its next control $Q_{1}$ so the endpoint derivatives match.

Show answer & reasoning

$Q_{1} = (\frac{15}{2}, - 3)$ .

The arriving cubic derivative is $3 ((6, 0) - (5, 2)) = (3, - 6)$ . Set $2 (Q_{1} - (6, 0)) = (3, - 6)$ , so $Q_{1} = (6, 0) + (\frac{3}{2}, - 3)$ . The degree factors differ.

Keep practising

6 worksheet questions · 12 marks

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Read the questions online

Find endpoints for quadratic controls $(1, 2)$ , $(4, 5)$ , $(6, 1)$ .
2 marks
Find $B (\frac{1}{2})$ 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 $B (\frac{1}{2})$ for quadratic controls $(0, 0)$ , $(2, 4)$ , $(6, 0)$ .
2 marks
Use weights $\frac{1}{4}$ , $\frac{1}{2}$ , $\frac{1}{4}$ .
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 school Vector calculus: polynomial and trigonometric parameterisations, related coordinate rates, Bézier paths, componentwise differentiation and integration, motion, tangent directions, circular motion and arc length. Reasoning and proofs use these school topics; vector-field calculus is outside this chapter.

Official curriculum source ↗

In this lesson

Understand the method.

Blend three controls to make a quadratic path

Read tangent directions from the controls

Use four controls for a cubic path

See the method in action.

Evaluate a quadratic and its initial direction

Evaluate the three weights.

Combine the corresponding coordinates.

Use the first control displacement.

Find the endpoint derivatives of a cubic

Apply the cubic initial-derivative formula.

Apply the cubic final-derivative formula.

Evaluate using the midpoint weights.

Choose a control for a smooth join

Find the derivative arriving at the join.

Equate the next curve's initial derivative.

Solve for the middle control and verify.

Try it yourself.

Calculate a weighted point

Interpret the middle control

Apply a join between different degrees

Keep practising