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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.

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.

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: 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].

7 Oct 2008: tn. g(t) dt. If b(t) = ε cos ϑ(t) then the integral in(2.4) is of the form. ... The first step is to rewriteb(t) in the form. b(t) = 12 [cos(ω̃1t) cos(ω̃2t)],whereω̃1 = ω1 ω2 and ω̃2 = ω1 ω2.

8 Apr 2008: In polar coordinates z = reiθ, Ar2r(B cos θC sin θ)D =0. ... Under the inversion 1/z ρeiφ = 1/reiθ this leads to Aρ(B cos φC sin φ)Dρ2 = 0which is also a circle.

30 Mar 2008: Here P, (cos e) is the ordinary Legendre polynomial, and V , is defined by equation. ... V , = sin e, V , = sin e cos 8, V , = t sin e (5 cos2 e - 1).

26 Feb 2008: Then. |R r|2 = (R r). (R r) = R2 r2 2Rr cos θ. ... u =GM. h2+ A cos(θ θ0). where A and θ0 are constants.

8 Apr 2008: 4x.).ResonanceConsider ÿ ω20y = sin(ω0t). We know that yh = A sin(ω0t) B cos(ω0t) then the. ... sin. 1 κ2τ. B cos. 1 κ2τ ). This is a damped oscillator with decay time (time that takes the am-plitude to decrease by 1/e of the original value)

5 Dec 2008: So general solution becomes. f = A(cos 2φ 1) B(sin 2φ 2φ). ... The boundary condition at φ = α gives. A(cos 2α 1) B(sin 2α 2α) = 12.

30 Sep 2008: General solution,. χ(x) = A sin(kx) B cos(kx). Boundary conditions from continuity of χ at x = 0 and x = a,. ... χ(x) = A cos(kx) B sin(kx). even parity condition,. χ(x) = χ(x) B = 0 χ(x) = A cos(kx).

3 Oct 2008: X′′ k2X = 0. T = A sin(ckt) B cos(ckt). X = C sin(kx), k =nπ. ... L, n = 1, 2, 3,. 46. Elementary solution:. u =. [. A sin. (. nπct. L. ). B cos. (. nπct.

3 Jul 2008: ζsatisfies. |f ′′(ζ)| = | sin(ζ|x|) sin(. ζ|y|)(|x|2 |y|2). 4ζζ. 2 cos(. ... 0. )(cos φ sin φsin φ cos φ. )(. cos φ sin φsin φ cos φ. )(0 D2. D1 0. ). =.

17 Jan 2008: a.b = ab cos θ and a b = ab sin θn̂ , (0.9). ... v= sin θ f. x+ cos θ. f. y. F. u= cos v.