Past exam of the mathematics course of the University of Cambridge 2006 ii Paper 1 30B Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 309 2 iv Solution Created 2026-10-03 Updated 2026-10-06
Rearranging the outgoing radial null geodesics of the Vaidya metric equation along givesFor the prescribed parametrization by , the chain rule yieldsSince , as . The asymptotic equivalence therefore impliesMore precisely and , using Big O notation. The positivity of justifies treating as a single-valued function of locally.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 2 32E a ii Solution Created 2026-09-24 Updated 2026-09-25
Termwise asymptotic equivalence fails for every , becauseFor example, as , the positive scale gives for . This illustrates the geometric-mean refinement of an asymptotic sequence.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 3 30E d Solution Created 2026-09-24 Updated 2026-09-25
For , the outer part of the action integral for a linear-to-square-root potential dominates:The WKB quantization condition therefore givesTaking the power yields the asymptotic equivalenceHence
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 332 2 vi Solution 2026-09-28
The approximation in part v requires the dimensionless stagnant depthto remain small. Since makes the numerator order one, this condition isUsing the solution from part v, the equivalent time range isor, at the level of asymptotic equivalence,The cooling becomes appreciable on the shorter scale , so these ranges overlap widely when .
The Rayleigh number based on the convecting depth isThe definition of implicit in part iii givesIt follows thatIf and , then . The thermal convection therefore remains strongly supercritical throughout the range in which the thin stagnant-layer approximation is valid.