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Bezier properties; point at u=0.2

Short answer

For the four control points, cubic Bezier at u=0.2 with weights 0.512, 0.384, 0.096, 0.008 gives P(0.2)=(2.912, 2.544). Key properties are endpoint interpolation, end tangents along first/last edges, convex hull containment, affine invariance, global control, and symmetry.

Explain properties of Bezier curve. Find the coordinate at u = 0.2 with respect to the control points (1,1), (4,6), (8,-3) and (12,2) using Bezier function.

Properties of Bezier curve

  1. Passes through the first and last control points.
  2. Degree is one less than the number of control points.
  3. End tangents are along the first and last edges of the control polygon.
  4. The curve lies inside the convex hull of the control points.
  5. Blending functions are positive and sum to 1.
  6. Affine invariance: transformations can be applied to control points only.
  7. Global control: moving any control point changes the whole curve.
  8. Symmetric: reversing the order of points gives the same curve.

Point at u = 0.2

Control points P0(1,1), P1(4,6), P2(8,−3), P3(12,2): cubic (n = 3).

P(u) = (1−u)³P0 + 3u(1−u)²P1 + 3u²(1−u)P2 + u³P3

FunctionFormula at u = 0.2Value
B0(0.8)³0.512
B13(0.2)(0.8)²0.384
B23(0.2)²(0.8)0.096
B3(0.2)³0.008

(Sum = 1.000)

x = 0.512·1 + 0.384·4 + 0.096·8 + 0.008·12 = 0.512 + 1.536 + 0.768 + 0.096 = 2.912

y = 0.512·1 + 0.384·6 + 0.096·(−3) + 0.008·2 = 0.512 + 2.304 − 0.288 + 0.016 = 2.544

Answer: P(0.2) = (2.912, 2.544)

✓ Answer checked by Claude Opus 5.5 (high effort)

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