Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 59 3 Solution Created 2026-10-03 Updated 2026-10-14
A solid body transform is a rigid transformation: it preserves distances within the object. In three-dimensional coordinates it has the form , with translation and an orthogonal matrix . For a proper physical rotation, the coefficient constraints areEquivalently the columns of are an orthonormal basis with positive orientation. A reflection also preserves distance but has and is excluded when “solid-body motion” means a rotation and translation. There are three rotational and three translational degrees of freedom, not twelve independent matrix coefficients.
Computationally use the homogeneous coordinates representationThe last row is ; matrix multiplication composes transforms. Alternatively store a unit quaternion together with , maintaining its unit norm. A transpose gives the rotational inverse: .
Write the cubic Bézier curve as , where . The binomial theorem gives . Transforming the controls producesThis proves pointwise equivariance, including the original parameter values; it is stronger than a statement that the two curves merely have the same shape.
Orthogonality was not used in the calculation. The identical proof applies to every affine map , including scales, shears and even singular affine maps. It applies to every representation with scalar, control-independent partition of unity basis functions: Bézier curves of every degree, B-splines, surfaces built on a tensor-product surface basis, triangular Bézier patches, and more general linear control-point representations. Positivity is not needed for affine equivariance of a geometric basis. For linear stationary subdivision matrices with row sums one, each refinement step commutes with the affine transformation; taking a convergent limit proves the corresponding assertion for subdivision curves and subdivision surfaces. Rules with geometry-dependent nonlinear weights require separate equivariance tests and are not covered automatically.
The affine class is the largest class that preserves every such Euclidean-control construction with unchanged basis weights, under the usual continuity assumption. In fact the degree-one case would forcefor all and . This segment identity, followed by continuity, forces to be affine. Nonlinear transforms such as perspective division generally fail it.
There is a further extension when the representation itself is enlarged: apply a projective transformation linearly to homogeneous controls , then dehomogenize the weighted sum. The transformed objects are rational Bézier curves or non-uniform rational B-splines, with transformed weights as well as transformed Euclidean controls. This works wherever the new denominator is nonzero. It does not assert projective invariance of ordinary polynomial Bézier curves with their old weights.
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 59 5 Solution Created 2026-10-03 Updated 2026-10-14
A tensor-product surface basis is formed from two independent univariate basis families and on separate parameter intervals. Given a rectangular control net , its parametric surface isThe parameters and indices vary independently over a product domain. The representation can be evaluated first in either parameter, giving a convenient computational advantage. Assume a finite basis, or locally finite spline bases, so differentiation and summation below are justified locally.
For positivity, if each factor basis is nonnegative, then . If both factors are strictly positive at a point, their product is strictly positive there. Nonnegative weights are the usual geometric meaning of positivity; zero boundary weights are allowed. With partition of unity, this gives the local convex hull enclosure of the active control points.
For continuity, suppose the first family is and the second , including matching derivatives across their respective spline knots. On every polynomial piece, and by matching at the knot lines,Every factor on the right is continuous in its own variable, so its product is jointly continuous. Thus all these mixed derivatives are continuous, in particular the surface is jointly . The separate coordinate smoothness can be better: orders in and in . For piecewise polynomials of degrees , an interior knot of multiplicity in the first factor normally gives matching in , inherited along that knot line; similarly for . The statement is an inherited lower bound. Special control points can cancel derivative jumps, so a particular surface may be smoother than a generic member of its basis.
For summation to unity, independent summation factors exactly:This also explains why translations of all control points translate the surface by the same vector, as in affine equivariance of a geometric basis. These three arguments establish tensor-product inheritance of geometric basis properties directly from the factor identities.
A triangular Bézier patch is a non-tensor-product example. For barycentric parameters with , a quadratic patch isThe six basis functions have total degree two and sum to . Their index constraint and triangular parameter domain do not arise from two independently indexed univariate factors on a rectangle. An individual triangular polynomial could be embedded in another larger representation, but this triangular basis and control net are not a tensor-product surface definition.
Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 70 4 iii Solution Created 2026-10-03 Updated 2026-10-14
A triangular Bézier patch is a non-tensor-product example. On barycentric parameters with , defineThe indices satisfy a total-degree constraint, and couples the two independent parameters. The control net is triangular with points, rather than a rectangular product index set. The multinomial theorem gives nonnegative basis functions summing to one, but the entire basis is not a product of two independent univariate bases.