Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 3 11I d Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 3 11I c Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 3 11I b Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 3 11I a Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 3 10D b Created 2026-09-23 Updated 2026-09-24
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Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 9A Solution Created 2026-09-23 Updated 2026-09-25
By the shell theorem, only the mass inside the particle's radius contributes:The conserved Newtonian energy per unit particle mass isWriting and using conservation of enclosed mass, so that is constant, givesThis is the first Friedmann equation without a cosmological constant.
The derivation has two central defects. It uses instantaneous Newtonian gravity and omits the relativistic gravitational effect of pressure, so it is not valid for general cosmological matter. It also starts from a finite ball with a preferred centre and boundary, contrary to exact homogeneity and isotropy; the assumption only hides that defect locally.
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 8B Solution Created 2026-09-23 Updated 2026-09-25
For with Jacobian , the chain rule givesThis has Hamiltonian form with the same canonical matrix for every exactly whenTaking determinants gives ; the symplectic condition fixes the orientation, so . Hence the change-of-variables formula shows that the phase-space volume element is invariant. This is Liouville theorem in Hamiltonian mechanics.
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 7E Solution Created 2026-09-23 Updated 2026-09-25
Choose paths from to in the cut plane and deform them continuously with the endpoint, never crossing the specified cut. With , use the sheet reached from the positive real axis; when an endpoint is moved into the left half-plane, the path passes around the upper endpoint of the cut. The cut makes all such admissible paths homotopic with fixed endpoints, so the integral is single valued and its endpoint derivative is ; hence it is analytic.
For , continue the endpoint counterclockwise from the positive real axis through the upper half-plane. Inthe continued square root equals at . The logarithm is reached with argument , and therefore
The identitiesshow directly why continuation without an argument restriction is multivalued. The complete set of values of the multivalued inverse hyperbolic sine at is
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 6C c Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 6C b Created 2026-09-23 Updated 2026-09-24
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Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 5J c Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 5J b Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 5J a Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 4I iii Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 4I ii Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 4I i Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 41C b Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 2 41C a Created 2026-09-23 Updated 2026-09-24
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