Did you know that a smooth curve can be built without first writing an equation of the form y=f(x)? Choose three points—two endpoints and one control point—and blend their coordinates using a changing parameter. The result is a quadratic Bézier curve. This simple construction explains how mathematical curves can bend smoothly while remaining controlled by a small number of points.
This Mathematics in Bukit Timah guide develops linear, quadratic and cubic Bézier curves, the de Casteljau algorithm, interpolation weights, convex hulls, endpoints, tangent directions, derivatives, curvature, subdivision and modelling limits. The shapes, points and invented garden-like diagrams are entirely fictional. No curve is a surveyed Bukit Timah path, a safe access route, a real school facility or an official design specification.
Quick answer: With points P₀,P₁,P₂, a quadratic Bézier curve is B(t)=(1−t)²P₀+2t(1−t)P₁+t²P₂, for 0≤t≤1. For P₀=(0,0), P₁=(2,4), P₂=(6,0), the midpoint-parameter point is B(1/2)=(2.5,2). The curve begins at P₀, ends at P₂, and is pulled toward P₁ without generally passing through it.
Article contents
- What is the design question?
- Bukit Timah is context rather than physical route data
- A point and a vector are different objects
- Linear interpolation is the simplest starting point
- Worked example 1: a linear blend
- The quadratic curve adds a third point
- The quadratic blending weights
- Worked example 2: weights at the midpoint
- Calculate the sample curve
- Worked example 3: the middle point
- The parameter is not automatically horizontal distance
- Worked example 4: quarter parameter
- Endpoint conditions are automatic
- Worked example 5: verify endpoints
- A useful geometric construction: de Casteljau
- First-stage interpolation formulas
- Worked example 6: de Casteljau at one-half
- Worked example 7: de Casteljau at one-quarter
- Why the control point attracts the curve
- The convex-hull property
- Worked example 8: an impossible curve position
- The convex hull is a region, not the path
- Differentiate the quadratic curve
- Worked example 9: starting tangent direction
- Worked example 10: ending tangent direction
- Tangent direction and tangent slope
- Worked example 11: initial and final slopes
- The midpoint tangent is horizontal
- Worked example 12: find the highest point
- A second derivative describes changing tangent
- Worked example 13: calculate the second derivative
- Geometric curvature involves both derivatives
- Worked example 14: curvature at the top
- Cubic Bézier curves use four control points
- The cubic weights also add to one
- Worked example 15: a symmetric cubic
- Endpoint tangent vectors of a cubic
- Worked example 16: cubic starting tangent
- The parameter midpoint need not bisect arc length
- Arc length uses an integral
- Worked example 17: compare two easy lengths
- Why the curve length cannot be read from the control polygon
- Subdivision gives two smaller Bézier curves
- Worked example 18: split at one-half
- Control polygons help estimate where a curve lies
- A cubic can be subdivided the same way
- Worked example 19: first row of cubic midpoint subdivision
- Bézier control points are not always interpolation nodes
- A Bézier curve can be fitted to data
- The same curve can have different parameter speeds
- Connecting two cubic segments smoothly
- Worked example 20: an obvious corner
- Tangent continuity and derivative continuity differ
- Worked example 21: choose a tangent-aligned join
- A quadratic has a polynomial coordinate rule
- The quadratic midpoint is not the chord midpoint
- Bernstein polynomials explain the general degree
- Worked example 22: identify cubic Bernstein weights
- Affine transformations preserve the construction
- Worked example 23: translate the sample quadratic
- Uniform scaling changes sizes predictably
- Worked example 24: scale by two
- Bézier curves are not always parabolic arches in real life
- A control point can create a degenerate curve
- Worked example 25: all controls collinear
- Numerical sampling can miss sharp features
- Comparing a chord and control-polygon length
- A spline is a larger family of connected pieces
- Engineering meaning requires constraints
- Common error 1: assume the curve passes through every control point
- Common error 2: treat t as x
- Common error 3: forget the binomial coefficient two
- Common error 4: mistake a control polygon for arc length
- Common error 5: assume equal parameter increments give equal distances
- Common error 6: use a zero denominator in tangent slope
- Common error 7: treat a geometry construction as a real path
- Primary preparation: fractions and line segments
- Lower Secondary: coordinates and quadratic functions
- Advanced Mathematics and course boundaries
- A twenty-minute family diagnostic
- Seven days of independent practice
- Practice questions with checked answers
- Frequently asked questions
- Official references and related Mathematics in Bukit Timah
What is the design question?
A learner may know how to draw straight line segments between given coordinates. But some problems ask for a smooth transition between endpoints.
Drawing a freehand curve can produce a shape, but it does not specify exact intermediate positions or tangent directions. A Bézier construction replaces the guess with a polynomial rule.
The user of that rule can calculate a point for every parameter t between zero and one.
Bukit Timah is context rather than physical route data
A family near Beauty World or Sixth Avenue may see a gently curved illustration and ask how designers construct such forms.
The curves below are mathematical diagrams only. No point specifies an actual walkway, property boundary, climbing route or infrastructure layout.
An attractive coordinate curve is not a certification that an object could or should be built with that exact geometry.
A point and a vector are different objects
A point P=(x,y) specifies a position in an abstract plane.
A vector Q−P specifies the directed change from P to Q. Its components are coordinate differences.
Bézier constructions repeatedly combine points and vectors in well-defined ways. A control point influences shape; it is not automatically a location visited by the final curve.
Linear interpolation is the simplest starting point
Given points P₀ and P₁, define L(t)=(1−t)P₀+tP₁, with 0≤t≤1.
At t=0 the result is P₀. At t=1 it is P₁. At t=1/2, it is their midpoint.
Because the weights sum to one and are nonnegative, L(t) stays on the straight segment between the endpoints.
Worked example 1: a linear blend
Let P₀=(0,0) and P₁=(6,4).
Then L(t)=(6t,4t). At t=1/4, the point is (1.5,1).
At t=1/2 the point is (3,2). The formula specifies the entire straight segment, not just those two sample points.
The quadratic curve adds a third point
A quadratic Bézier curve uses P₀, P₁ and P₂.
P₀ and P₂ are its endpoints. P₁ is the interior control point that influences bending.
The curve is generally not the broken line P₀→P₁→P₂, nor a parabola through all three control points. It is a weighted combination of the three positions.
The quadratic blending weights
For 0≤t≤1, the weights are:
w₀=(1−t)², w₁=2t(1−t), w₂=t².
They are nonnegative and add to one because (1−t)²+2t(1−t)+t²=[(1−t)+t]²=1.
This is the basis of the curve’s convex-hull property.
Worked example 2: weights at the midpoint
Set t=1/2.
Then w₀=1/4, w₁=1/2 and w₂=1/4. Therefore
B(1/2)=P₀/4+P₁/2+P₂/4.
The control point gets half the weight at this parameter, but the curve still need not pass through it.
Calculate the sample curve
Choose P₀=(0,0), P₁=(2,4), P₂=(6,0).
The x-coordinate is x(t)=4t(1−t)+6t²=4t+2t².
The y-coordinate is y(t)=8t(1−t)=8t−8t².
Both are quadratic polynomials in t. The curve itself is described parametrically.
Worked example 3: the middle point
At t=1/2,
x=4(1/2)+2(1/4)=2+0.5=2.5.
y=8(1/2)(1/2)=2. Thus B(1/2)=(2.5,2).
This is not the control point (2,4), and it is not necessarily the midpoint (3,0) of the two endpoints.
The parameter is not automatically horizontal distance
In B(t)=(x(t),y(t)), t indexes progress through the polynomial construction.
It does not generally equal x, horizontal distance, travelled distance or elapsed time. In the sample curve x=4t+2t², which is not equal to t.
A pupil should not assume that t=1/2 means the curve has been traversed halfway by physical arc length.
Worked example 4: quarter parameter
Set t=1/4.
x=4(1/4)+2(1/16)=1+1/8=1.125.
y=8(1/4)(3/4)=1.5.
Thus B(1/4)=(1.125,1.5). The calculation provides an exact point without freehand drawing.
Endpoint conditions are automatic
At t=0, the quadratic weights are (1,0,0), so B(0)=P₀.
At t=1, they are (0,0,1), so B(1)=P₂.
These identities hold for every placement of the three control points. They are structural properties of the formula, not special features of our numerical example.
Worked example 5: verify endpoints
With the chosen points, at t=0 we have x=0 and y=0.
At t=1, x=4+2=6, while y=8−8=0.
The curve therefore begins at (0,0) and ends at (6,0). Its middle control point influences the path between them.
A useful geometric construction: de Casteljau
The de Casteljau algorithm produces the Bézier point through repeated linear interpolation.
First interpolate between P₀ and P₁ to obtain Q₀(t). Interpolate between P₁ and P₂ to obtain Q₁(t).
Then interpolate between Q₀(t) and Q₁(t). The final result is B(t). This construction is easy to check visually and extends to higher degrees.
First-stage interpolation formulas
Define Q₀(t)=(1−t)P₀+tP₁.
Define Q₁(t)=(1−t)P₁+tP₂.
Then B(t)=(1−t)Q₀(t)+tQ₁(t).
Expanding the last line yields the quadratic Bernstein weights. Thus algebra and the geometric subdivision method are equivalent.
Worked example 6: de Casteljau at one-half
For P₀=(0,0), P₁=(2,4), P₂=(6,0), the midpoint of P₀P₁ is Q₀=(1,2).
The midpoint of P₁P₂ is Q₁=(4,2).
The midpoint of Q₀Q₁ is (2.5,2). It matches the polynomial evaluation at t=1/2.
Worked example 7: de Casteljau at one-quarter
At t=1/4, Q₀=(0.5,1).
Q₁=P₁+(1/4)(P₂−P₁)=(2,4)+(1,−1)=(3,3).
Blend those points with t=1/4: B=(3/4)(0.5,1)+(1/4)(3,3)=(1.125,1.5). This verifies the earlier direct calculation.
Why the control point attracts the curve
At interior parameter values, w₁=2t(1−t)>0.
Thus P₁ contributes positively to the weighted result. Moving P₁ changes the curve’s interior, especially near the parameter range where w₁ is large.
But its influence is blended with the endpoint positions. The curve generally does not travel through P₁ as if it were a compulsory waypoint.
The convex-hull property
For 0≤t≤1, the weights are nonnegative and add to one.
Therefore B(t) is a convex combination of P₀,P₁,P₂, placing it inside or on the triangle formed by those points.
This is a general geometric containment result for quadratic Bézier curves. It is useful for rough shape checks, although a containing triangle is not an exact description of the curve.
Worked example 8: an impossible curve position
Our control points all have y-coordinates between zero and four.
The quadratic curve cannot produce y=10 for any parameter 0≤t≤1, since its points lie in the control triangle.
Indeed its explicit y(t)=8t(1−t) has maximum two. A claimed curve point with y=10 violates both the convex-hull bound and the formula.
The convex hull is a region, not the path
The triangle P₀P₁P₂ encloses a two-dimensional region.
The Bézier curve is a one-dimensional parametric trace inside that region. Most points in the triangle are not on the curve.
Confusing containment with equality would imply that a curve fills its control triangle, which is false in this ordinary polynomial construction.
Differentiate the quadratic curve
The derivative is taken component by component:
B′(t)=2(1−t)(P₁−P₀)+2t(P₂−P₁).
This tangent-direction vector tells how the point changes as the parameter increases.
Its magnitude gives speed per unit parameter, not necessarily physical speed unless t has been defined as a time variable with appropriate units.
Worked example 9: starting tangent direction
At t=0, B′(0)=2(P₁−P₀).
Our P₁−P₀=(2,4), so B′(0)=(4,8).
The tangent points in the same direction as segment P₀P₁. This is the endpoint tangent property of a quadratic Bézier curve when the derivative is nonzero.
Worked example 10: ending tangent direction
At t=1, B′(1)=2(P₂−P₁).
Our P₂−P₁=(4,−4), so B′(1)=(8,−8).
The tangent points along the final control-polygon segment, not necessarily the straight segment P₀P₂.
Tangent direction and tangent slope
For a parametric curve B(t)=(x(t),y(t)), with x′(t)≠0, the slope is
dy/dx=y′(t)/x′(t).
If x′(t)=0 while the derivative vector is nonzero, the tangent may be vertical rather than absent.
The full vector B′(t) is often safer than forcing every tangent into one numerical slope.
Worked example 11: initial and final slopes
Our start derivative is (4,8), so initial tangent slope is 8/4=2.
At t=1 the derivative is (8,−8), so final slope is −1.
The curve begins rising steeply and ends descending toward the final endpoint. Its tangents need not have equal magnitudes or opposite slopes.
The midpoint tangent is horizontal
For the sample curve, x′(t)=4+4t and y′(t)=8−16t.
At t=1/2, derivative is (6,0).
Therefore the tangent slope is zero at the parameter midpoint. The curve is locally horizontal there, consistent with its maximum y-coordinate two.
Worked example 12: find the highest point
The vertical coordinate is y(t)=8t−8t².
Differentiate: y′(t)=8−16t. Set this to zero to obtain t=1/2.
Since y″(t)=−16<0, the point is a maximum in the vertical coordinate. Substitution gives B(1/2)=(2.5,2).
A second derivative describes changing tangent
Differentiate B′ again:
B″(t)=2(P₂−2P₁+P₀).
For a quadratic Bézier, this vector is constant across t. It expresses the second derivative with respect to the chosen parameter, not automatically the geometric curvature itself.
A parameter acceleration vector can be nonzero even when the curve’s tangent speed behaves in a complicated way.
Worked example 13: calculate the second derivative
With P₀=(0,0), P₁=(2,4), P₂=(6,0):
P₂−2P₁+P₀=(6,0)−(4,8)=(2,−8).
Double this to obtain B″=(4,−16). It matches x″=4 and y″=−16 from direct differentiation of the coordinate polynomials.
Geometric curvature involves both derivatives
For a smooth regular planar parametric curve, curvature magnitude is
κ=|x′y″−y′x″|/(x′²+y′²)^(3/2).
It measures how rapidly the curve’s tangent direction changes per unit arc length.
This is an advanced calculus formula. It is not equal to |B″| in general, because parameter speed also matters.
Worked example 14: curvature at the top
At t=1/2, x′=6, y′=0, x″=4 and y″=−16.
Then |x′y″−y′x″|=|6(−16)−0|=96. The denominator is (6²)^(3/2)=216.
Thus κ=96/216=4/9 in inverse coordinate units. The calculation assumes a regular Euclidean parametric curve.
Cubic Bézier curves use four control points
A cubic Bézier curve has points P₀,P₁,P₂,P₃ and formula
B(t)=(1−t)³P₀+3t(1−t)²P₁+3t²(1−t)P₂+t³P₃.
Again t ranges from zero through one. The curve begins at P₀ and ends at P₃.
Two interior control points allow greater flexibility in shaping endpoint tangents and interior bending.
The cubic weights also add to one
The coefficients are the four terms of the binomial expansion
[(1−t)+t]³=1.
For 0≤t≤1 each weight is nonnegative, so the cubic curve lies in the convex hull of its four control points.
This is the same central principle as the quadratic case, extended from three to four control points.
Worked example 15: a symmetric cubic
Choose P₀=(0,0), P₁=(1,3), P₂=(4,3), P₃=(5,0).
At t=1/2 the cubic weights are 1/8,3/8,3/8,1/8.
The x-coordinate is 0+3/8+12/8+5/8=2.5. The y-coordinate is 0+9/8+9/8+0=2.25.
Endpoint tangent vectors of a cubic
The cubic derivative at the first endpoint is
B′(0)=3(P₁−P₀).
At the final endpoint it is
B′(1)=3(P₃−P₂).
These identities let a designer choose P₁ and P₂ to influence entry and exit directions while keeping the endpoints fixed.
Worked example 16: cubic starting tangent
For the symmetric cubic, P₁−P₀=(1,3).
Therefore B′(0)=(3,9) and the initial tangent slope is three.
At the other end P₃−P₂=(1,−3), so B′(1)=(3,−9), with slope −3. The endpoint slopes reflect the chosen symmetric control geometry.
The parameter midpoint need not bisect arc length
For the symmetric cubic, t=1/2 happens to produce a geometrically central point because the control geometry is symmetric.
But for arbitrary control points, equal parameter intervals usually do not correspond to equal distances along the curve.
A drawing tool seeking uniform physical spacing along the curve must handle arc length, not merely evaluate t at regular increments.
Arc length uses an integral
For a sufficiently smooth curve B(t)=(x(t),y(t)), the arc length from t=a to t=b is
S=∫ from a to b √[x′(t)²+y′(t)²]dt.
The integrand is the magnitude of the parametric derivative.
This formula differs from the straight endpoint distance and the total control-polygon length, although those quantities can provide useful bounds or comparisons.
Worked example 17: compare two easy lengths
For our quadratic, endpoints are (0,0) and (6,0). Their straight distance is 6.
The control polygon travels from (0,0) to (2,4), then to (6,0). Its length is √20+√32=2√5+4√2≈10.129.
The actual Bézier arc length lies between these values for this nondegenerate curve, but is not equal to either simply by definition.
Why the curve length cannot be read from the control polygon
The broken control polygon and the smooth Bézier trace connect the endpoints differently.
The convex-hull property and triangle inequality constrain where the curve can lie, but do not make its length equal the two control segments.
An exact arc-length calculation needs the derivative integral or an appropriately controlled numerical approximation.
Subdivision gives two smaller Bézier curves
De Casteljau’s intermediate points at a chosen t value define new left and right control polygons.
Each polygon represents one portion of the original Bézier curve under an appropriate reparameterisation.
The subcurves join at exactly the computed B(t). This is useful for drawing and numerical calculations that refine local portions without changing the underlying curve.
Worked example 18: split at one-half
For our quadratic, de Casteljau gives Q₀=(1,2), Q₁=(4,2), and B(1/2)=(2.5,2).
The left subcurve has control points (0,0), (1,2), (2.5,2).
The right subcurve has control points (2.5,2), (4,2), (6,0). Together they reproduce the original quadratic curve, with remapped parameters.
Control polygons help estimate where a curve lies
A convex hull encloses the whole Bézier trace over t∈[0,1].
Subdividing the curve makes smaller control polygons whose hulls generally give tighter local containment regions.
This geometric structure is valuable for plotting and intersection algorithms. But containment does not, by itself, determine exact arc length or physical accessibility.
A cubic can be subdivided the same way
The de Casteljau algorithm interpolates neighbouring control points repeatedly until one final point remains.
For four cubic controls, the first interpolation row has three points, the second has two, and the final row has one.
Recording the outer edges of this interpolation triangle constructs the left and right cubic control polygons. The method works for any Bézier degree.
Worked example 19: first row of cubic midpoint subdivision
For P₀=(0,0), P₁=(1,3), P₂=(4,3), P₃=(5,0), at t=1/2:
First interpolated points are (0.5,1.5), (2.5,3) and (4.5,1.5).
The next row gives (1.5,2.25) and (3.5,2.25). Their midpoint is (2.5,2.25), matching the earlier cubic computation.
Bézier control points are not always interpolation nodes
In polynomial interpolation, one often constructs a curve passing through specified data nodes.
An ordinary Bézier curve passes through its endpoint controls but generally not its interior control points.
This distinction is central. The Bukit Timah Lagrange interpolation lesson explains exact passage through observed nodes, whereas Bézier control points shape a parametric curve.
A Bézier curve can be fitted to data
One can choose control points to approximate observed positions under an optimisation criterion.
That becomes a fitting problem rather than a direct interpolation through all given observations.
It may require choosing a parameter value for each observation, a loss function and any endpoint or smoothness constraints. The simple control-point formula alone does not establish the optimal fit.
The same curve can have different parameter speeds
Reparameterising a geometric curve can alter how quickly its coordinates change with the parameter, even when the traced geometric set stays the same.
Therefore B′(t) is not universally a physical velocity. To interpret it as velocity, t must correspond to meaningful time units.
A tutor should avoid labelling all parametric derivatives “metres per second” when t is just a number from zero to one.
Connecting two cubic segments smoothly
Suppose one cubic ends at point Q and another begins at the same Q.
Position continuity, often called C⁰ continuity, requires their endpoint positions to match.
For tangent-direction or derivative continuity, additional conditions on adjacent control points and parameter scales are needed. Merely touching at one point does not guarantee a visually smooth join.
Worked example 20: an obvious corner
Let one segment end at Q=(0,0) with final tangent direction (1,0).
Let a second segment begin at Q with initial tangent direction (0,1).
The pieces meet at the same position, but the tangent directions are perpendicular. The join has a visible corner rather than tangent continuity.
Tangent continuity and derivative continuity differ
A geometrically smooth tangent join can occur when adjacent tangent vectors point in the same direction but have different positive magnitudes.
Equal derivative vectors under compatible parameterisation are a stronger condition, often called C¹ continuity.
The distinction matters in advanced curve design: matching direction alone need not mean matching parameter speed.
Worked example 21: choose a tangent-aligned join
Imagine first cubic ending at Q=(0,0) with previous control point (−1,0).
Its final derivative is three times (1,0), pointing right.
Choose the next cubic’s first interior control point (2,0). Its initial derivative is three times (2,0), also pointing right. The tangent directions align, although the derivative magnitudes differ.
A quadratic has a polynomial coordinate rule
The example x(t)=4t+2t² and y(t)=8t−8t² gives two polynomial coordinates.
But eliminating t may not yield a simple globally single-valued graph y=f(x) for every arbitrary Bézier construction.
A parametric curve can represent shapes that double back horizontally or have vertical tangents. This is one reason parametric equations are more flexible than a single y-as-function-of-x formula.
The quadratic midpoint is not the chord midpoint
Our endpoints are (0,0) and (6,0), whose midpoint is (3,0).
The curve’s parameter midpoint is (2.5,2), above and left of that chord midpoint.
The difference reflects the influence of the control point and the nonlinear blending weights.
Bernstein polynomials explain the general degree
A degree-n Bézier curve can be written
B(t)=Σ from i=0 to n [C(n,i)(1−t)ⁿ⁻ⁱtⁱPᵢ].
The binomial coefficients C(n,i) provide the Bernstein blending basis.
For t between zero and one, the weights are nonnegative and sum to one by the binomial theorem. This generalises the convex-hull and endpoint principles.
Worked example 22: identify cubic Bernstein weights
For degree three, C(3,0),C(3,1),C(3,2),C(3,3) are 1,3,3,1.
Thus the weights are (1−t)³, 3t(1−t)², 3t²(1−t), and t³.
Their sum is one. This is why the cubic formula uses those precise coefficients rather than arbitrarily selected multipliers.
Affine transformations preserve the construction
For an affine map T(P)=MP+q, where M is a linear transformation and q is a translation, applying T to every Bézier control point gives the transformed curve T(B(t)).
The reason is that the blending weights sum to one.
This makes Bézier representations useful for translations, rotations and other affine geometric operations. The map need not preserve Euclidean lengths unless it has the appropriate properties.
Worked example 23: translate the sample quadratic
Translate all controls by vector (10,−2).
The original curve point B(1/2)=(2.5,2) becomes (12.5,0).
Recomputing with translated controls yields the same result because the weights sum to one. The geometric shape has moved but not changed internally.
Uniform scaling changes sizes predictably
If every control coordinate is multiplied by a positive scale factor k about the origin, every Bézier point is multiplied by k.
All Euclidean lengths scale by k. Areas enclosed by suitable scaled closed constructions scale by k².
This is a property of the affine blending rule, not proof that a physical model can be scaled without engineering constraints.
Worked example 24: scale by two
Original quadratic controls are (0,0),(2,4),(6,0), and midpoint point is (2.5,2).
Scale all controls by two to (0,0),(4,8),(12,0).
The midpoint of the scaled curve is (5,4). The tangent directions are preserved and derivative vectors are scaled by two.
Bézier curves are not always parabolic arches in real life
A quadratic Bézier is a polynomial parametric curve related to a parabolic conic in nondegenerate situations.
But calling a real arch “quadratic Bézier” requires a precise representation or measured model.
The Bukit Timah parabola focus-directrix guide describes the geometric parabola locus. This article’s primary focus is control-point construction and parameterised design.
A control point can create a degenerate curve
If all three quadratic control points are the same, B(t) is a constant point.
If all three are collinear, the image of the curve lies on that line, though its parameterisation can vary along the line.
These cases remind us that “quadratic formula in t” does not guarantee a visibly curved geometric trace for every control configuration.
Worked example 25: all controls collinear
Let P₀=(0,0), P₁=(2,0), P₂=(6,0).
Every weighted result has y-coordinate zero, so the curve lies on the x-axis.
Its x-coordinate is still 4t+2t². It is parameterised nonlinearly even though the traced geometric path is straight.
Numerical sampling can miss sharp features
To draw a Bézier curve on a screen, software may evaluate points at a finite set of t-values.
A coarse sample can miss important local curvature or give an inaccurate polygonal approximation to arc length.
Adaptive subdivision can refine sections where the control geometry shows more deviation from the chord. The plotted polyline is an approximation of the underlying mathematical curve.
Comparing a chord and control-polygon length
The straight distance between endpoints is a lower bound on the length of any rectifiable curve connecting them.
The control polygon provides an upper bound for a Bézier curve’s arc length in the standard construction, under the relevant geometric assumptions.
These bounds help numerical checking, but a bound is not the exact length. The correct arc-length calculation uses the derivative or a controlled approximation.
A spline is a larger family of connected pieces
Multiple Bézier segments can be joined to form a piecewise polynomial curve.
A spline may impose smoothness across those joins. Other spline representations, including B-splines, use control points and basis functions with different locality and knot conventions.
The terms Bézier and spline are related but should not be treated as interchangeable definitions for every curve.
Engineering meaning requires constraints
An invented curve can be perfectly smooth on graph paper yet unsuitable as a physical path because of grade, curvature, clearance, width or access constraints.
This guide deliberately specifies none of those local engineering requirements.
Mathematics supplies geometric models, not automatic safety verification. Applying a curve to real infrastructure requires the relevant professional checks and measurements.
Common error 1: assume the curve passes through every control point
The quadratic generally passes through P₀ and P₂ but not P₁.
For our example, B(1/2)=(2.5,2), not (2,4).
The repair is to calculate the blending weights and recognise P₁ as a shaping point rather than an obligatory visited point.
Common error 2: treat t as x
The parameter t belongs to [0,1]. The x-coordinate can take an entirely different range and is given by x(t).
In the example, x(1/2)=2.5, not 0.5.
Check the coordinate formulas before assigning physical units or labels to the parameter.
Common error 3: forget the binomial coefficient two
A pupil writes B(t)=(1−t)²P₀+t(1−t)P₁+t²P₂.
The weights then sum to 1−t(1−t), not one. The curve may lose its intended affine properties.
The correct middle coefficient is 2t(1−t), from expanding [(1−t)+t]².
Common error 4: mistake a control polygon for arc length
The control polygon contains straight segments joining control points.
The Bézier trace bends within their convex hull and generally has a different length.
A learner should distinguish endpoint chord, control polygon, and derivative-based curve arc length.
Common error 5: assume equal parameter increments give equal distances
The parameter t does not generally measure arc length.
Even spacing t values at 0,0.25,0.5,0.75,1 may produce unequal distances between curve points.
If equal physical spacing is the objective, arc-length parameterisation or an appropriate numerical method is needed.
Common error 6: use a zero denominator in tangent slope
For parametric derivatives, dy/dx=y′/x′ requires x′≠0.
If x′=0 and y′≠0, a vertical tangent may exist.
Use the derivative vector rather than claiming the entire tangent is undefined merely because its finite slope cannot be computed.
Common error 7: treat a geometry construction as a real path
A fictional control polygon and its resulting curve do not specify entrance locations, path closures or safe travel constraints.
A mathematical boundary cannot be converted into verified local directions by adding a neighbourhood name.
Keep geometric teaching claims separate from actual Bukit Timah mapping and infrastructure information.
Primary preparation: fractions and line segments
Younger pupils can find midpoints and quarter points between two positions.
They can then construct a point between two moving midpoints, seeing the de Casteljau idea before writing polynomial formulas.
The foundational skill is recognising a weighted average of positions, not memorising calculus.
Lower Secondary: coordinates and quadratic functions
Students ready for algebra can evaluate x(t),y(t), expand the weights and check endpoints.
They can compare straight line segments with curves generated by a middle control point.
Later derivative work gives a deeper explanation of tangent directions, smooth joining and changing curvature.
Advanced Mathematics and course boundaries
Bézier curves and Bernstein bases are useful in mathematical modelling, computer graphics and computational geometry.
The SEAB 2027 G3 Mathematics directory provides Secondary prerequisites; the 2027 A-Level syllabus directory provides more advanced pathways.
This article is enrichment and does not claim that general Bézier curvature or spline theory is required in every school Mathematics course.
A twenty-minute family diagnostic
Minutes 1–5: Interpolate between (0,0) and (6,4) at t=1/2, obtaining (3,2).
Minutes 6–10: Write the quadratic formula and confirm its weights add to one. Use controls (0,0),(2,4),(6,0).
Minutes 11–15: Calculate B(1/2)=(2.5,2) by direct weights and de Casteljau’s midpoints.
Minutes 16–20: Find endpoint tangent directions (4,8) and (8,−8). Explain why the control point is not generally on the curve.
Seven days of independent practice
Day one studies linear interpolation. Day two constructs quadratic blending weights. Day three uses the de Casteljau algorithm.
Day four computes tangent vectors and slopes. Day five introduces cubic controls. Day six splits a quadratic into two subcurves.
Day seven distinguishes curve containment, control polygon length and arc length, then tests an unfamiliar curve with an independent numerical check.
Practice questions with checked answers
A. Formula for quadratic Bézier weights? Answer: (1−t)²,2t(1−t),t².
B. Control points (0,0),(2,4),(6,0); B(1/2)? Answer: (2.5,2).
C. Same curve B(1/4)? Answer: (1.125,1.5).
D. Initial tangent vector? Answer: (4,8).
E. Final tangent vector? Answer: (8,−8).
F. Highest y-coordinate and its parameter? Answer: 2 at t=1/2.
G. Cubic midpoint for (0,0),(1,3),(4,3),(5,0)? Answer: (2.5,2.25).
H. Is t always proportional to arc length? Answer: no.
Frequently asked questions
What is a Bézier curve?
A polynomial parametric curve formed by weighted combinations of control points, with Bernstein polynomial blending weights.
Does it pass through all control points?
Generally no. It passes through its endpoints; interior controls influence the shape.
What does the parameter t represent?
Progress through the mathematical parameterisation, not automatically horizontal coordinate, time or arc length.
What is de Casteljau’s algorithm?
A method that repeatedly interpolates neighbouring control points to calculate an exact Bézier point and enable subdivision.
Why does the curve stay inside the control hull?
For 0≤t≤1 the weights are nonnegative and sum to one, so every curve point is a convex combination of its controls.
How do we calculate tangent direction?
Differentiate the coordinate polynomials, or use the endpoint identities involving the adjacent control-point differences.
Is a quadratic Bézier always a physical parabola?
It is a parametric polynomial construction. Physical shape identification requires a suitable geometric model or measurement.
How is arc length calculated?
Through an integral of the magnitude of the parametric derivative, or through a numerically controlled approximation.
Are these real Bukit Timah path designs?
No. The curves and controls are wholly fictional mathematical constructions.
How can parents recognise understanding?
The learner can explain weighted positions, compute a point by two methods, check endpoints and derive a tangent without confusing controls with actual curve points.
Official references and related Mathematics in Bukit Timah
Consult the SEAB 2027 G3 Mathematics syllabus directory for relevant school prerequisites. All graphics, values and coordinate assumptions are fictional enrichment examples.
Continue with Parabola Focus and Directrix, Lagrange Polynomial Interpolation, Matrix Transformations, Differentiation and Tangents and Fractals and Recursive Geometry.
The immutable Clementi Secondary 1 Mathematics small-group teaching reference anchors concept-first explanation, correct working, independent verification and clearly diagnosed errors.
The lasting habit is to start with the control geometry, use correct Bernstein weights, distinguish the curve from its polygon, verify endpoint tangents and keep any physical interpretation within the model’s stated limits.
Companion Mathematics in Bukit Timah articles in this new four-part batch: Chinese Remainder Theorem, Congruences and Repeating Cycles · Graph Isomorphism, Degree Sequences and Network Symmetry · Separable Differential Equations, Growth, Decay and Equilibrium. These new lessons extend parametric curve design, simultaneous modular constraints, network structural equivalence and calculus models, with original worked examples and explicit assumptions.