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28 Nov 2008: B. B3 = B cos θ. B2 = B sin θ cos ψ. ... H = 2. 2m2 12 B σ cos θ. 12B. 2 sin2 θ 12.

5 Dec 2008: 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 9b to evaluate the integral.

13 Mar 2008: 4 A circular helix is given by. x = (a cos t, a sin t, ct). ... Its mass density is ρ0 cos θ. By evaluating a volume integral find the mass ofthe cone.

13 Mar 2008: F = (3x2yz2, 2x3yz, x3z2), G = (3x2y2z, 2x3yz, x3y2),. H =(. 3x2 tan z y2exy2. sin y, (cos y 2xy sin y)exy2. , ... u = ((x cos y cos y y sin y)ex, (x sin y sin y y cos y)ex).

5 Dec 2008: 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 ... 1 e cos φ,. where (ρ, φ) are plane polar coordinates and e is a

25 Feb 2008: Show that. θ̇2 =6gb. a2 3b2(cos θ cos α) (1). where b is the length OX, θ is the angle between OXand the vertical and α is the maximum value ... Combining these results,. 12 (. 13 a. 2 b2)θ̇2 gb cos θ = E/M.

26 Feb 2008: Tx = T sin θ cos φ. = T (r/l)(x/r). = T x/l,. ... ζ = eiωt sin λ(A cos(. g/l t) B sin(. g/l t). ).

26 Feb 2008: e1 B. (cos(βt) i sin(βt). )e2}. = eαt(a cos(βt) b sin(βt). ... J =. (0 1. (g/l) cos θ 0. ). The phase diagram has this form:.

26 Feb 2008: constant k:. mẍ = kx. so x = A sin(. k/m t) B cos(. ... mẍ = kx cẋ F cos t. The general solution consists of two parts: (a) the particular integral; and (b) the com-.

30 Oct 2008: k=0. (1)k[2ω(n m)]2k2 [f. (2k1)(b) cos 2ω(n m)b. f(2k1)(a) cos 2ω(n m)a]}. ... m. ). {. k=0. (1)k[ω(2n 2m 1)]2k1. [f(2k)(b) cos ω(2n 2m 1)b f(2k)(a) cos ω(2n 2m 1)a].