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23 Feb 2012: Sketch the imageof S in R3. (a) U = (0, 2π) R, S(θ, z) = (cos θ, sin θ, z).(b) U = (0, 2π) (1, 1), S(θ, z) = (cos θ. ... 1 z2, sin θ. 1 z2, z). (c) U = (0, 2π)(0, 2π), S(θ, φ) = ((a b cos φ) cos θ, (a b cos φ) sin θ, b sin φ)

23 Feb 2012: Sketch the imageof S in R3. (a) U = (0, 2π) R, S(θ, z) = (cos θ, sin θ, z).(b) U = (0, 2π) (1, 1), S(θ, z) = (cos θ. ... 1 z2, sin θ. 1 z2, z). (c) U = (0, 2π)(0, 2π), S(θ, φ) = ((a b cos φ) cos θ, (a b cos φ) sin θ, b sin φ)

26 Mar 2012: Letp : (0, ) [0, 2π) B() be the map : p(r, θ) = r cos θ e1 r sin θ e2, and let T :(0, ) [0, 2π) U be the composition T = S ... Define p : (0, ) (0, 2π) B() by p(r, θ) = r cos θ e1 r sin θ e2, and letp(g1) = E dr2 2F drdθ G dθ2.

17 Apr 2012: vr = rν cos φ[c(ν 1)Pν1 b(ν 1) cos θ Pν. ], ... 2.12a). vθ = rν cos φ[cP′ν1 2a csc θ Pν b.

26 Oct 2012: 2(ceR)2(cµR). 2 2(ceR)2(cµL)2 2(ceL)2(cµR)2 2(ceL)2(cµL)2) cos θ}= A(1 cos2 θ) B cos θ,. ... 1. A(1 cos2 θ) B cos θd(cos θ). = 2πA. [. x x3. 3. ]1. 1. =16. 3πA. =16. 3π. g4Z256π2Γ 2Z. [.

5 Mar 2012: βH′[φ<] = Z0> βH0[φ<] d2x g〈cos[λ(φ<(x) φ>(x))]〉> O(g2). (b) Working to first order in g, show that, under the RG transformation, theparameters obey the ... βH′[φ′] =. d2q. (2π)2K(b). 2q′2|φ(q)|2. d2x′ g(b) cos [λ(b)φ′(x′)].

3 Sep 2012: vr = rν cos φ[c(ν 1)Pν1 b(ν 1) cos θ Pν. ], ... 2.12a). vθ = rν cos φ[cP′ν1 2a csc θ Pν b.

15 Oct 2012: 3 given by:φ(s,v) = α(s) r(n(s) cos v b(s) sin v). ... 9. Consider a surface of revolution parametrized by φ : (0, 2π) (a,b) R3, whereφ(u,v) = (f(v) cos u,f(v) sin u,g(v)).

4 Jan 2012: Edited by Reinholdus Klotz. Lipsiae : B. G. Teubner, 1861. Bound with 27, 28. ... 79. Paterculus. M. Vellei Paterculi ex historiae romanae ad M. Vincium cos libris duobus quae supersunt.

18 Oct 2012: y =. . . . . . 0 , x < 0 ,. 1 cos x , 0 < x < ǫ ,. cos(x ǫ) cos x , x > ǫ. ... 1 cos x. x. )2. dx = π. (b) Use Parseval’s theorem and the result of question 7b to evaluate the integral.

31 Oct 2012: x = ax′ , y = by′ , z = cz′ ,. x′ = r′ sin θ′ cos φ′ , y′ = r′ sin θ′ sin φ′ , z′ = r′ cos θ′ ,. where a, b and c are positive constants. Calculate ... If an extra transverse force F sin(πx/L) cos(πct/L) per unit length

8 Feb 2012: Prove that a spherical circle of radius ρ has circumference 2π sin ρ and area2π(1 cos ρ). ... Deduce that. sin α sin β cos c = cos γ cos α cos β.

20 Apr 2012: ex2. sin 2x). ,. (ii)d12. dx12(x cos x) using (a) the Leibniz rule and (b) repeated application of the product rule,. ... i) (x2 a)3/2,. (ii) ln(cos x),. iii) exp. {. 1(x a)2. }. ,.

28 Nov 2012: φ(u,v) = ((a r cos u) cos v, (a r cos u) sin v,r sin u). ... φu φv| =EG F 2. For our example EG F 2 = r(r cos u a).

7 Feb 2012: a magnetic moment m aligned along its axis. Let there be a magnetic field B, so that. ... each molecule has a potential energy mB cos θ (θ being the angle between the axis.

10 Nov 2012: cosh a = cosh b cosh c sinh b sinh c cos α. ... Let G be the set of Möbius transformationsz 7 azbczd with a, b, c, d Z[i] and ad bc = 1.

29 Oct 2012: cos mx. (b) Show that f (x) =. m=1am sin mx defines a continuous function on [π, π], but that the series. ... m=1mam cos mx need not converge. 11. Show that, for any x X = R {1, 2, 3,. },

28 Nov 2012: φ(u,v) = ((a r cos u) cos v, (a r cos u) sin v,r sin u). ... φu φv| =EG F 2. For our example EG F 2 = r(r cos u a).

2 Sep 2012: sider the circular trajectory [see Fig. 1(b)]. R ðÞ bðcos; sin; 0Þ: (10)For this trajectory, we have jR0ðÞj b and tðÞ ð sin; cos; 0Þ. ... Whenthey are not parallel, the phase dependence is proportionalto RðÞ. For example, for perpendicular

8 Feb 2012: Prove that a spherical circle of radius ρ has circumference 2π sin ρ and area2π(1 cos ρ). ... Deduce that. sin α sin β cos c = cos γ cos α cos β.