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12 Feb 2008: b=t1AL-t2;. V=-M1GB1cos(t1) - M2G(B2cos(t2) AAcos(ALt1));. T=0.5((I1M1B12)t1d.2 I2t2d.2 M2(B2t2d.sin(b)).2 M2(AAt1d. B2t2d.cos(b)).2);. ... i[0 -B1 -2B1 -B1 0 -AAcos(AL)];. line2=[AAsin(AL)[0 B2sin(t2-t1)]]. i[-AAcos(AL)[0 -B2cos(t2-t1)]];.

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.

22 Jan 2008: Conversely, show that for any three positive numbers a, b, c less than πsatisfying the above conditions, we have cos (b c) < cosa < cos (b c), and that thereis a spherical ... The polar triangle A′B′C′ is defined by the unit vectorsin the

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].

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.

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)

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:.

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.

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). ).

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.

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

3 Jan 2008: Y = R s v which is constant. Transforming these to axes ( x, y ) fixed to the gear, Fig B-3, they become :-. ( ii ) x = r sin ( θ – φ ) = X cos φ – ... the rack :-. yC = Ro cos α ( Ro ψmin e ) sin α = R s – b e from which (viii ) ψmin = tan

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-.

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).

4 Dec 2008: Spacelike branching: gluon splitting on incoming line (a). p2b = EaEc(1 cos θ) 0. ... p2b m2q = EaEc(1 va cos θ) 0 ,. Hence Ec 0 soft divergence remains; collinear enhancement becomes a divergence as.

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.

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.

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).

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.