Past exam of the mathematics course of the University of Cambridge 2022 ia Paper 3 4A Solution Created 2026-09-23 Updated 2026-09-25
Use as the parameter of the smooth curve:Its velocity vector and speed areThe unit tangent vector is therefore
The parameter-independent formula for the curvature of a space curve givesDifferentiating,Hence increases for and decreases for . Its global maximum occurs at and equalsThe corresponding point is
Past exam of the mathematics course of the University of Cambridge 2023 ib Paper 2 11F Solution Created 2026-09-23 Updated 2026-09-25
The tangent vectors ofareHence the first fundamental form has coefficientsPutThe upward unit normal is , so the second fundamental form has coefficientsThus the two forms are
For the final claim, fix a point of and make a rigid motion taking to the plane . The common tangent plane is horizontal, so locally is the graph . Along the projected curve of tangency,Differentiating the second identity givesBecause is a smooth curve, , so the Hessian is singular. Its determinant is zero, and the graph formula gives at every point of . This is tangency to a plane along a curve forces zero Gaussian curvature.
Past exam of the mathematics course of the University of Cambridge 2023 ib Paper 3 2E Solution Created 2026-09-23 Updated 2026-09-25
In the Poincare half-plane model, the Riemannian metricassigns to a smooth curve the lengthThe metric is conformal to the Euclidean metric, so hyperbolic and Euclidean angles agree. Its area element is
The two geodesics from and to infinity are the vertical lines and , while the third side is the unit semicircle. The triangle lies above that semicircle, so its area isIts interior angles are , , and zero at the ideal vertex, and henceTriangulating a geodesic polygon with sides into triangles givesthe area of a hyperbolic geodesic polygon and the polygonal Gauss-Bonnet theorem.
Past exam of the mathematics course of the University of Cambridge 2024 ib Paper 3 13F Solution Created 2026-09-23 Updated 2026-09-25
The argument principle says that if a positively oriented closed curve bounds a domain , and a meromorphic function has no zeros or poles on , thenwhere zeros and poles are counted with multiplicity.
Suppose and are holomorphic on a neighbourhood of andFor , has no boundary zero, because a zero would imply . Its argument-principle count is integer-valued and continuous in , hence constant. Thus Rouché's theorem states that
Now let . Since on the unit circle, Rouché's theorem shows that and have the same number of zeros in the unit disc. The latter has the zero , so has at least one zero there. For any with , the same boundary inequality shows that has the same positive number of zeros as . ThereforeThis is the unit-disc image from a boundary modulus lower bound.
Finally, take a sufficiently small positively oriented circle around zero. The residueis the winding number of around zero, so . The pole is simple, hence its residue is nonzero and . Forwe haveThe second term cancels the complete principal part at zero, so the integer residue of a logarithmic derivative proves that has a removable singularity there.
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 2 25F Solution Created 2026-09-24 Updated 2026-09-25
For , is the common kernel of ; is the transcendence degree of its function field, and is smooth when . A nonzero maximal Jacobian minor defines a nonempty open smooth locus (generic rank gives nonemptiness).
A line corresponds to . Lines through satisfy , a projective line. For a smooth curve , the dual morphism is .
For , eliminating from givesThe dual sextic is singular: its nine cusps are the tangent lines at the nine flexes of the cubic. Thus points of mapping to singularities are precisely inflection points, where the tangent has higher contact.
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 3 5K Solution Created 2026-09-24 Updated 2026-09-25
The residual-versus-fitted smooth curve bends downward, indicating nonlinearity, and the spread increases at high fitted values. The Q-Q plot is close centrally but departs in both tails, with observations 12, 17, and 67 notable. The scale-location trend confirms heteroscedasticity. The leverage plot identifies observations 1 and 4 as high leverage and 67 as a large residual, with potentially material Cook distance. The additive Gaussian linear model is therefore doubtful. Inspect those observations, transform sale price (often logarithmically), add nonlinear or interaction terms and omitted predictors such as lot size and age, and use weighted or robust regression if unequal variance remains.
Regular curve Created 2026-09-24 Updated 2026-09-24