Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 72 3 Solution 2026-10-14
Work in the fixed reference domain , and write . For the unconstrained hyperelastic material considered here, the material-first nominal stress is . The increment of the deformation gradient and the incremental elastic moduli areAll coefficients are evaluated at the base deformation. The Hessian symmetry is ; minor interchange symmetry is unnecessary. Linearizing the momentum equation givesThe reference mass density and reference outward unit normal vector are fixed, so the incremental boundary conditions areIn particular, no variation of a current surface normal belongs in this material-coordinate traction condition. If a material constraint were imposed, one would additionally linearize its constraint reaction in hyperelastic stress and its admissibility equation; the present moduli formula is the unconstrained one.
Past exam of the mathematics course of the University of Cambridge 2005 ib Paper 4 12A Solution Created 2026-09-24 Updated 2026-10-14
At , regularity makes linearly independent. The tangent space is , and an oriented unit normal vector isReversing its sign chooses the opposite orientation. The first fundamental form is the squared length of a tangent displacement:where , , . Its determinant is positive.
Use to distinguish the scalar coefficient of the second fundamental form from the normal vector. Since , its derivatives are perpendicular to it and hence tangent. Write and . Differentiating givesSubstitution of the tangent expansions provesThe shape operator is ; its matrix in the parametrized tangent basis is the negative transpose of the coefficient matrix shown. Its determinant, the product of the principal curvatures, is thereforethe Gaussian curvature. Transposition and the two minus signs do not change this determinant.
Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 66 5 ii Solution Created 2026-10-03 Updated 2026-10-14
Evaluate all quantities at and assume a regular surface point, . The tangent plane passes through and is spanned by . ThusChoose the unit normal . The first fundamental form has coefficientswhile the second fundamental form has coefficientsTo derive the curvature formula, differentiate . This gives , , and . Therefore the shape operator , expressed in the tangent basis, isIts eigenvalues are the principal curvatures. Their product, the Gaussian curvature, is its determinant:Regularity ensures . Reversing changes the signs of and both principal curvatures but leaves unchanged. At a singular parameter point this quotient and the cross-product normal vector are undefined; an alternative regular chart or separate geometric analysis is required.
Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 66 5 i Solution Created 2026-10-03 Updated 2026-10-14
The core parametric surface interrogation is evaluation of the position and its derivatives through second order:Position answers point queries; first derivatives give the differential, tangent directions and normal vector; second derivatives supply curvature. An evaluator should also report the parameter domain, patch adjacency, boundary curves and knot or crease locations, and whether these derivatives exist at the queried point. At a discontinuity boundary it should provide the relevant one-sided limits rather than silently average them.
For spatial searches, provide restriction or subdivision into equivalent smaller patches and certified bounding volumes, with error estimates when an approximate mesh is used. These allow intersection and closest-point algorithms to reject remote patches and refine the remaining candidates. A useful interface therefore combines differential evaluation with domain/topology and enclosure information. Inverse parameter lookup, ray intersection and nearest-point search can be built from these primitives using subdivision and local root or minimization solves; they need not be presumed exact basic evaluations of every representation.
Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 66 6 ii Solution Created 2026-10-03 Updated 2026-10-14
The corresponding subdivision surface interrogation needs mesh connectivity and patch adjacency, boundary and crease flags, local coordinate charts with transition maps, and subdivision into smaller patches with their original-chart mappings. Its evaluation enquiry returns a limit point, tangent derivatives and a normal vector; second derivatives are supplied only where the limit is sufficiently smooth. Certified bounding volumes and patch approximation errors allow a hierarchy for intersection and proximity searches. Boundary-curve enquiries complete the interface for clipping.
On regular parts of a subdivision surface, these evaluations may use its equivalent tensor-product or box spline patch. Near an extraordinary subdivision vertex, the local subdivision matrix determines the limit position through its constant left eigenvector, and its tangent modes determine the normal vector under the scheme's characteristic map of a subdivision surface regularity conditions. Refinement alone does not establish those conditions. A subdivision surface may have a well-defined tangent plane there without possessing second derivatives or Gaussian curvature. An interrogation interface must distinguish these cases and can work on surrounding regular charts when a higher derivative is unavailable at the vertex itself.
Together, subdivision, bounds, topology and limit evaluation support ray–subdivision surface intersection: bounds eliminate irrelevant patches; surviving patches are refined, and regular candidate intersections are corrected using position and first derivatives. The returned intersection is a point of the limit subdivision surface with an error certificate, rather than automatically a point on the last control mesh.
Past exam of the mathematics course of the University of Cambridge 2006 ia Paper 4 9C Solution Created 2026-09-24 Updated 2026-10-14
Let be the cylindrical unit vectors. The surface has an upward normal vector proportional to . If is the vertical component of the normal force, its radial component is . In the stated approximation, vertical acceleration and the vertical driving-force component are neglected, so vertical balance gives . The horizontal contact force is consequently . The projected velocity and its speed areThe driving force is parallel to this velocity, with components and . Use the acceleration in polar coordinates and Newton's second law to obtainThese equations apply where ; at rest the direction of an exactly velocity-aligned drive is not determined.
For a prescribed path, write , . The chain rule gives and . Subtract times the azimuthal equation from the radial equation. The driving force and both cancel:Thus the constraint for tangentially driven motion on an axisymmetric surface isThe denominator must give a nonnegative squared speed, and the calculation uses . At the polar origin a limiting Cartesian description is appropriate.
For the paraboloid and Archimedean spiral, putThen, for ,The branch tending to has positive angular velocity, henceSince , solve the azimuthal equation for the required force:This is positive for , consistent with a forward drive. Taking the limit gives
These are the large-time limits of the printed projected equations. Their mechanical approximation has a precise range limitation: on the exact surface, gives , which along this solution grows like . At fixed positive , it is not small as . Thus the requested limits are formal limits of that reduced model; neglect of vertical motion is not a uniform approximation to the full three-dimensional motion all the way to infinity.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 66 2 Solution Created 2026-10-03 Updated 2026-10-14
Work first in the parameter rectangle of the parametric surface. Set . With linearly independent, the chain rule givesThe intersection is transversal precisely when the implicit normal vector and the parametric normal vector are not parallel. Consequently , and the implicit function theorem makes a locally regular parameter curve. This is the setting for transversal surface intersection tracing by a predictor-corrector continuation method.
Find one seed on the required component. Subdivide parameter patches, enclosing their images by bounding volumes and testing whether a certified range of excludes zero. For a positive-weight NURBS, the active control-point convex hull supplies a convenient enclosure; interval arithmetic can tighten the range of the composed function. Search the surviving patches for a bracketed root or a local corrected seed. Merely checking signs at four corners is inadequate: a small closed intersection can lie wholly inside a patch with the same corner signs. An adaptive or certified patch search, or a supplied seed, avoids this failure. For a sphere centered at , use and ; for a torus, use its implicit polynomial and its differentiated polynomial in the same algorithm.
At a corrected seed , an unnormalized parameter tangent vector and a unit spatial tangent vector areChoose the sign consistently with the previous tangent vector. Normalize and predict , so is approximately a physical arc length step. Correct the predictor by Newton method applied toThe Jacobian has rows and . At the starting point its first row vanishes along , whereas its second row applied to is one. The two rows are therefore independent. The transverse-plane condition prevents the corrector from sliding arbitrarily along the intersection. Restrict correction to a neighborhood of the predictor, use damping, and reduce when correction fails or its displacement becomes excessive; this prevents jumps to another nearby branch.
To obtain a sufficiently dense sequence, impose a physical point-spacing bound, a bound on tangent turning, and a chord-error estimate. For local curvature , the small-step sagitta is approximately ; estimate curvature from successive tangent vectors and reduce until this and the desired spacing are acceptable. The correction residual should be small relative to the drawing tolerance. Uniform steps in or alone do not ensure spatial density because the surface metric can vary considerably.
For an open component, trace in both tangent directions from the seed. On reaching a parameter boundary, locate the endpoint by solving together with that boundary's parameter equation, rather than accepting an overshoot. For a closed component, stop only after nontrivial travel and return to the initial parameter location with compatible tangent orientation; near-coincidence in three-dimensional space alone can mistake a self-intersection or neighboring branch for closure. Cross periodic seams using the appropriate chart identification. Adaptive physical stepping, local correction, and component-aware termination give the required point sequence. A vanishing parameter gradient signals a tangency or singularity outside the transversal hypothesis and must be reported rather than passed through by this regular algorithm.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 76 2 Solution Created 2026-10-03 Updated 2026-10-14
Take any fixed material subvolume of the reference configuration, with outward normal vector , current particle velocity , and positive temperature . The first law of thermodynamics, including kinetic energy, isThe heat-flux sign is negative because is outward. The Second law of thermodynamics supplies the corresponding integral Lagrangian entropy inequality:The temperature in each supply or flux term is the local temperature, not a common temperature pulled outside the integrals.
Apply the divergence theorem and use the reference momentum conservation equation . In the total-energy equation, the kinetic derivative cancels the body-force work and the term. Since , the remaining local Lagrangian internal energy balance and Lagrangian entropy inequality areThey hold pointwise because was arbitrary. In particular,
The specific Helmholtz free energy is . Substitution yields its balanceand elimination of the heat supply yields the Clausius-Duhem inequalityFor , expand the derivative to obtainThe usual Coleman-Noll procedure admits independent reversible choices of and at a fixed state. Their coefficients must vanish, since either sign and arbitrary magnitude would otherwise violate the inequality. ThusInternal-variable evolution and heat conduction must satisfy this remaining inequality. Substituting the constitutive relations back into the energy balance givesThe last printed equation is missing a factor on its internal-variable term. The PDF defines with per unit mass, so the expression above is the dimensionally consistent consequence. Equivalently define the volumetric thermodynamic force conjugate to an internal variable ; the term is then with no extra density. The two conventions must not be mixed.
For thermoelasticity with a temperature-dependent constraint, admissible rates obeyThe reversible coefficient functional need now vanish only on this tangent hyperplane. Hence it can be a scalar multiple of the constraint normal:ThereforeThe reaction terms cancel on admissible rates, leaving the same internal-variable/conduction inequality and entropy heating equation. The multiplier is determined by equilibrium and boundary conditions rather than by the free energy alone.
For linear thermoelasticity with prescribed thermal volume, write , , and . Hold the internal variables fixed in this reversible linearization, and evaluate all derivatives below at and zero reaction. The stress-free reference state has . Since there, first-order nominal and Cauchy stress tensors agree. Their constitutive laws becomewhere equality of mixed free-energy derivatives givesThe linearized constraint is . Thus volume is thermally prescribed; strict temperature-independent incompressibility corresponds to . The coefficient is an entropy slope; the corresponding volumetric heat capacity at fixed deformation is .
Past exam of the mathematics course of the University of Cambridge 2007 ia Paper 3 3A ii Solution Created 2026-09-24 Updated 2026-10-14
Write the surface as the level set . The gradient is normal to a regular level surface because differentiating along any surface curve gives for its tangent vector. At the specified point,Thus a normal vector and a unit normal are and , respectively. The tangent-plane displacement is perpendicular to that normal, givingReversing the normal's sign gives the same plane.
Past exam of the mathematics course of the University of Cambridge 2007 ib Paper 4 7E Solution Created 2026-09-24 Updated 2026-10-14
For a moving circuit, Faraday's law takes the formHere is the instantaneous wire contour, its local material velocity, and any spanning surface. The contour orientation and surface normal vector obey the right-hand convention. The integral on the right is the magnetic flux, differentiated with both the field and circuit position allowed to vary. The electric field term is the electric force per unit electric charge around the wire, and supplies the motional electromotive force. Their sum is the electromotive force driving charge around the moving circuit. The condition makes the magnetic flux independent of the spanning surface.
Choose the initial loop normal vector to be , so rotation about gives . The spatially uniform magnetic field givesUsing Ohm's law for the resistive loop,Positive electric current is measured in the contour orientation associated with . Reversing that convention reverses the signed answer. When , the magnetic flux is constant and the induced electric current vanishes. This uses the usual thin resistive-circuit approximation with self-inductance neglected.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 77 1 ii Solution Created 2026-10-03 Updated 2026-10-14
Count scalar additions, subtractions, multiplications and divisions for the specific triangle-triangle intersection algorithm above; comparisons and index selection are separate. Computing one plane normal vector uses two vector differences, costing six subtractions, and one cross product, costing six multiplications and three subtractions. The three vertex-to-plane evaluations use nine further coordinate subtractions and three dot products, each with three multiplications and two additions. The first rejection test therefore costsFor randomly located triangles of characteristic size in a box of size , a second triangle straddles the first triangle's plane only when its position is in a slab of relative thickness . Consequently the first plane test rejects almost every pair: the expected arithmetic count for this implementation is , with the constant in the remainder determined by the subsequent branches. This is an average count, rather than the assertion that every input needs 39 operations.
For a generic surviving noncoplanar pair, the symmetric plane test costs another 39 operations, the cross product costs nine, and each of the four slice endpoints needs five scalar operations in the selected coordinate: one subtraction and division for the interpolation fraction, then one subtraction, multiplication and addition for the coordinate. Thus this branch usesA fully executed coplanar test can use six planar edge directions. Each needs two subtractions to form its perpendicular and six two-dimensional dot products, costing operations. Including both plane tests and the parallelism cross product gives an upper count before any early exit. Contact and degeneracy can shorten or change these counts. Precomputed normal vectors or a preliminary axis-aligned bounding box test also change the answer: the latter uses only comparisons after its coordinate extrema are available. The relevant small-triangle average is 39 operations for the specified plane-first implementation; a generic complete test uses 107.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 77 1 i Solution Created 2026-10-03 Updated 2026-10-14
Let the triangles have vertices and . Work initially with nonzero-area triangles. A triangle-triangle intersection algorithm begins with the unnormalized normal vector and the three signed plane evaluations . If all are strictly positive, or all strictly negative, the second triangle lies strictly on one side of the first one's plane, so the triangles are disjoint. Otherwise compute and and apply the symmetric rejection test. Strict inequalities are important: contact on a vertex or edge counts as intersection.
If the planes are not parallel, put . Each triangle meets the other one's plane in an empty set, a point or a line segment. Its endpoints are obtained from zero-valued vertices and the edges whose endpoint plane evaluations have opposite signs. For example, an edge contributesAll these points lie on the common line of the two planes. Choose a coordinate for which and form the coordinate intervals , of the two slices. This coordinate is one-to-one along the common line, so the exact final test isOnly that coordinate of each slice endpoint needs to be evaluated; no full three-dimensional line parameterization is necessary.
Parallel planes which have survived the strict plane tests are coplanar. Drop the coordinate corresponding to the largest component of and test the two projected planar triangles. For each edge of either projected triangle, project both vertex sets onto the perpendicular direction. A strictly disjoint pair of projection intervals certifies separation. If none of the six edge directions separates them, the planar convex sets overlap: a separating line for two disjoint convex polygons can be chosen parallel to an edge of one of them. This includes containment and boundary contact.
A zero normal vector signifies a degenerate triangle; reduce its convex hull to a point or its extreme line segment and use point containment or segment intersection instead. For floating-point input, reliable sign predicates and scale-aware error bounds are preferable to treating a tiny signed distance as an arbitrary exact zero. Plane rejection followed by interval overlap, with a planar test for coplanar input, determines all nondegenerate cases.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 77 2 Solution Created 2026-10-03 Updated 2026-10-14
At , orient the unit normal vector towards the eye. Define the unit directions from the reflecting point to the source and eye byThe incident propagation direction is . The law of specular reflection is , so is normal to the surface. Conversely, if the tangential components of vanish and , its unit-length summands have equal positive normal components and obey the reflection law. Therefore solve the two scalar equationsThe stated orientation condition gives . It removes the back-facing ambiguity, but not multiple images or caustics. Exclude and , where these directions are undefined.
At a root where the Jacobian matrix has rank two, the implicit function theorem makes the solution set a local curve in parameter space. Compute a unit tangent in its nullspace; equivalently take the normalized cross product of the two rows of . A continuation method predicts , where , then corrects with the Newton method applied to the three equationsThe last equation fixes a transverse section and allows tracing through a turning point of , where solving only for at successive fixed source parameters would fail. The corresponding space point is , with derivative along the traced parameter-space curve.
Find initial roots by subdividing the parameter cube, using interval arithmetic bounds on to discard impossible boxes and validated local solves in the remaining boxes. Trace in both directions, stop at the boundary or a recognized closed loop, and retain a record of traced boxes to avoid duplicates. Search the remaining boxes as well: starting from just one seed would miss disconnected closed reflection curves. Where loses rank, reduce the step and subdivide locally to find the outgoing branches instead of continuing a single guessed tangent. If physical visibility is required, discard roots for which either open segment or is occluded; test these segments against the surface or any supplied scene geometry.
Control the geometric approximation rather than just the parameter step. For any regular traced arc with a bound , the chord error bound gives an arc-to-chord Hausdorff distance at most . Budget, for example, half of for corrected-vertex error and half for chord error; accept a segment only whenBound the vertex error through the inverse transverse Jacobian matrix and a bound on , or use validated root enclosures. If analytic second-derivative bounds are unavailable, recursively subdivide and enclose each arc in a geometric tube of radius at most the remaining tolerance around its chord. A midpoint test alone is a useful heuristic but cannot certify the requested error for an arbitrary oscillatory curve.
The returned vertices and their ordered connecting segments give the reflection curve to the prescribed tolerance on every resolved branch. Finite certified termination requires regular evaluable patches, derivative or enclosure bounds, and finitely resolvable branches; eye-facing normals alone do not guarantee these properties. Singular components or an identically vanishing reflection constraint must be handled as such rather than silently reported as an ordinary curve.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 77 4 iii Solution Created 2026-10-03 Updated 2026-10-14
Treat the subdivision surface as an evaluable limit together with its refinement hierarchy. At a requested location , evaluate the limit position and two independent limit tangent vectors, then setOn regular regions, local spline patches supply position and derivative evaluations. Near extraordinary vertices, use the actual subdivision rule and its limit-position and tangent evaluation machinery, or a convergent local refinement with controlled position and normal vector errors. If a tangent pair degenerates, the regular offset evaluator is not defined there; a crease requires separate one-sided offsets and an explicitly chosen join.
For polygonal interrogation, subdivide the base mesh adaptively. At every new sample, evaluate the base limit and its limit normal vector, and displace that sample by . Connect these samples with the inherited connectivity, refining until flatness and normal-variation bounds meet the desired tolerance. If the position error is and unit-normal error is , the offset-position error is at most , in addition to the interpolation error between samples. A variable normal offset can be sampled at the same locations, with its approximation error included as well. The result retains the base subdivision hierarchy for locating points and refining patches.
Displacing the original control vertices along estimated vertex normals and applying the original subdivision mask just once does not, in general, give the exact surface offset. If is one linear refinement step, then , while displacing the refined base gives ; normalization and the normal calculation are nonlinear, so in general. Thus the exact offset is an interrogation wrapper around the base limit, and a conventional subdivision control mesh for it is a fitted approximation requiring refinement or error control. Refine and interrogate the base limit, then offset its evaluated points; do not assume offsetting commutes with subdivision.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 77 4 ii Solution Created 2026-10-03 Updated 2026-10-14
For a regular parametric surface , form , and define the offset evaluatorIt is already a parametric surface with the same parameter domain, even if its coordinates no longer belong to the original polynomial or rational patch family. Point interrogation requires evaluations of and one normalization. For tangent or normal vector interrogation, differentiate the evaluator. For example,and similarly for . The offset normal vector is the normalized cross product wherever it is nonzero. These derivatives also support intersection, projection and curvature calculations by the usual methods for parametric surfaces. For a variable normal offset, add and to the corresponding tangent derivatives.
With shape operator , the constant-distance offset differential is . Along a principal direction its factor is , where is a principal curvature. Consequently regular interrogation fails at ; on regular pieces the offset principal curvatures are with the continued normal orientation. Global self-intersections can occur even when both local factors are nonzero. Evaluate and differentiate the base surface and its normalized normal; an exact conversion back to the original patch type is unnecessary.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 77 4 i Solution Created 2026-10-03 Updated 2026-10-14
A parallel surface at signed distance from a design surface describes a constant-thickness skin, shell, coating or clearance envelope. A particularly important application is machining with a spherical cutter of radius : when the cutter touches without penetration, its center lies at , where is the outward unit normal vector. Thus a tool-center path can be computed on the surface offset rather than on the desired contact surface. Positive and negative surface offsets also model material addition and removal, and pairs of offsets describe the two faces of a thin manufactured wall. Offsets convert contact, thickness and clearance specifications into geometric surfaces which can be interrogated.
The construction is initially local. Where the offset develops self-intersections or reaches a curvature radius, not every point of the raw parallel surface is a usable physical boundary. Tool paths and solid envelopes may require trimming those pieces and checking global collisions.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 77 6 Solution Created 2026-10-03 Updated 2026-10-14
Use the limit subdivision surface and limit subdivision curve, rather than treating their initial control meshes as the reflecting geometry. Build adaptive refinement hierarchies for both. Each surface cell supplies a limit-position evaluator, local tangent vectors and a normal vector; each curve interval supplies a limit-position evaluator and tangent. Regular regions may be evaluated as local splines, while extraordinary neighborhoods require the evaluation machinery of the specified subdivision rules. A piecewise smooth surface is handled patch by patch, with explicit boundary or crease joins.
A useful initial polygonal construction refines the surface to small triangles and the source curve to small line segments. For a candidate mirror triangle in a plane , reflect the eye across that plane:For a source edge , a physical reflection point is the plane intersection of the line :Keep only , , points inside the mirror triangle, and eye/source points on the appropriate reflecting side. Unfolding the reflected path across the mirror proves the specular reflection condition. As moves along a straight edge, lies on the intersection of the mirror plane with the plane through , so the retained locus is a straight segment, a point or the empty set. Plane and triangle clipping computes its endpoints. If these defining points are collinear, or a denominator vanishes, use the corresponding limiting geometric case rather than a generic plane formula. A bounding volume hierarchy accelerates candidate search where geometric bounds exclude pairs; refined facet segments provide seeds and approximate connectivity.
To obtain the reflection curve of the actual limits, correct these seeds. In a local surface chart write , and on a source interval write . SetThe exact limit reflection locus is together with the front-side conditions. For a two-sided mirror, orient locally so that and require ; for a one-sided mirror use its prescribed orientation and discard the wrong side. At grazing incidence this orientation condition ceases to be strict and the case needs separate handling. Unlike the eye-facing assumption in the parametric question, no global orientation condition is supplied here.
Apply a continuation method: at a rank-two root of , predict in a nullspace tangent direction and correct by the Newton method on a transverse section. Map corrected parameters to , crossing patch and curve-interval boundaries through their hierarchy adjacency. Trace all branches, recognizing closed components and endpoints. Supplement facet seeds by recursive searches of the surface-cell/source-interval product, with interval arithmetic or other certified enclosures of ; otherwise a small loop absent from the initial polygonal approximation could be missed. Split singular neighborhoods where the Jacobian matrix loses rank, and test the open source-to-mirror and mirror-to-eye segments for occlusion when selecting actually visible components.
Allocate a geometric error budget to limit evaluation, root correction and polygonal chords. For example, require combined position-evaluation and corrected-root error at most and, on every regular arc, require the chord error bound . Use subdivision position and normal bounds for evaluations: a tiny position error alone is insufficient, since a wrong normal vector changes the reflected ray. Away from singularities an inverse transverse Jacobian matrix bound converts reflection residual into parameter error and a bound on converts this to space error. Refine until these bounds hold, or use validated enclosures of entire arcs and their distances to the chords.
Near a caustic the inverse bound can become large, so the hierarchy must refine more strongly or isolate the singular branch geometry. Mesh flatness alone does not certify a reflection tolerance. With regular convergent evaluators and finite resolvable components, the resulting ordered vertices and chords satisfyThe problem supplies no particular subdivision rules or smoothness guarantees, so this construction specifies what their evaluators and bounds must provide; an arbitrary nonsmooth or degenerate limit need not have an ordinary finite reflection curve. Refine for seeds, enforce reflection on the actual limit geometry, then refine the traced arcs to the required spatial tolerance.
Past exam of the mathematics course of the University of Cambridge 2009 iii Paper 58 1 Solution Created 2026-10-03 Updated 2026-10-14
Assume the level set is regular and non-null. The latter hypothesis is necessary: a null hypersurface has a degenerate induced metric and hence no ordinary intrinsic Levi-Civita connection determined by that metric. Normalize the normal vector and define the non-null hypersurface projection byHere denotes the induced metric restricted to tangent vectors, while its ambient extension annihilates . The intrinsic covariant derivative is for tangent vector fields. Projection of the ambient torsion-free connection and metric compatibility shows that is the Levi-Civita connection of . Define the extrinsic curvature by , or . It is symmetric: differentiating the defining function twice and projecting removes the derivative of its normalization, leaving a multiple of its symmetric Hessian. This sign convention makes the scalar second fundamental form .
Let be the tangent linear map characterized by . Differentiating and gives the Gauss formulaFix the Riemann curvature tensor conventionApplying the Gauss formula twice givesSubtracting the expression with interchanged, and then subtracting the bracket derivative, provesThus the Gauss–Codazzi equations for a non-null hypersurface give the intrinsic Riemann curvature tensor asAll free indices here are tangent to the hypersurface. The normal component of the same calculation isAfter antisymmetrizing and using , this yields the curved-ambient Codazzi equationThe extrinsic curvature measures how tangent planes turn in the ambient manifold; the Gauss equation says that this turning contributes to intrinsic curvature even when the ambient Riemann curvature tensor is zero.
For the quadric in Minkowski spacetime, write its position vector as . On the hypersurface, and differentiating this identity in a tangent direction gives . Hence is a unit normal and a spacelike vector, , and in flat Cartesian coordinates. Therefore . The ambient Riemann curvature tensor vanishes, soThis is unit-radius de Sitter spacetime. Reversing the normal changes to but does not change this curvature. An opposite overall convention for the Riemann curvature tensor would reverse both curvature signs below.
To retain a four-dimensional Lorentzian metric while reversing the curvature, take a flat ambient metric with two negative directions and the quadricNow the position vector is a unit normal with negative squared norm, so while remains true. The Gauss equation givesThis is unit-radius Anti-de Sitter spacetime. If a positive-definite induced metric is acceptable, an even simpler alternative is to keep the original one-time ambient Minkowski metric and change the quadric level to . Each sheet then gives four-dimensional hyperbolic space with the same negative curvature, but its induced signature differs from that of de Sitter spacetime.
Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 15 3 c Solution Created 2026-10-03 Updated 2026-10-14
The integral identity requires compactness of , compact support of , or suitable conditions eliminating flux at infinity. The printed question does not state such a hypothesis. We prove the Riemannian divergence theorem for a compact manifold, and more generally for a compactly supported smooth vector field on an oriented Riemannian manifold with boundary.
Write for the inclusion. Give the outward-normal-first boundary orientation, and let be the outward unit normal vector. The Riemannian volume form of the induced Riemannian metric isIndeed, if is a positively oriented orthonormal basis tangent to the boundary, is a positively oriented orthonormal basis in . Both sides therefore evaluate to one on the boundary basis.
At a boundary point decompose , where is tangent to . Evaluating on tangent vectors, the term with vanishes: all arguments of the alternating volume form then lie in the -dimensional boundary tangent space. Thus the interior product of a differential form satisfiesThe general Stokes theorem for an oriented manifold with boundary states for any smooth differential form of degree with compact support; compactness of makes the support condition automatic. Applying it to and using the defining equation for the divergence of a Riemannian vector field proves
For a concrete failure without the extra hypothesis, take with Riemannian metric and . The divergence of a Riemannian vector field is , sobecause vanishes at the only boundary point. Both integrals are finite. The missing is the limiting flux at infinity, so mere integrability of the divergence cannot repair the unrestricted assertion.
Past exam of the mathematics course of the University of Cambridge 2012 ia Paper 3 9C Solution Created 2026-09-24 Updated 2026-10-14
For an oriented smooth surface with piecewise smooth, consistently oriented boundary, Stokes theorem givesfor a continuously differentiable vector field defined on a neighborhood of . Choose the upward normal vector here. The surface is an annular band of an elliptic paraboloid, parametrized byUsing as the oriented vector area element givesThe sketch below shows the open band, not a capped solid. Its upper circle has radius one and its lower circle radius one third.
Annular paraboloid with upward normal and opposite induced orientations on the outer and inner boundary circles
. Details. Differentiating the given vector field yieldsThe surface integral is thereforeFor the boundary line integral, the upward normal vector induces counterclockwise traversal of the outer circle and clockwise traversal of the inner circle, as viewed from above. On a circle of radius , parametrized counterclockwise, is constant andSince each fourth power integrates to , the two-circle line integral isconfirming Stokes theorem and the Stokes flux through an annular paraboloid. Reversing the chosen orientation changes both integrals to .
