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15 Feb 2017: cos (λ,1(ck, t)) rλ,1(ck, t)sλ,1(ck, t). ... r′λ(t) = (1 q(t)λ) sin(θλ(t)) cos(θλ(t))rλ(t). and yields the seminal result:.

9 Nov 2015: 87 –. while the jump condition gives. B( cos(1 ξ)) A cos ξ = 1. ... The generalsolution of this homogeneous equation c1 cos x c2 sin x and the inhomogeneous boundary.

15 Aug 2005: Denoting by θn. a(t, yn), it is clear that. F (x) = exA(tn,yn) =. (cos xθn θ. 1n sin xθn. θn sin xθn cos xθn. ),. ... 1](x) = ϕ1(x). (cos 2θnx. θ2n sin 2θnx. 2θ3nx x. θ2n+ sin 2θnx.

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.

6 Jan 2021: Phys. Comm. 181 (2010) 232, arXiv:0903.1805. 4. B. C. Allanach, C. ... Sridhar, ‘Explaining Tevatron leptons, photons, missing ET events with. supersymmetry”, Proceedings of the IX International Symposium on Particles, Strings and Cos-.

30 Nov 2009: f (2k1)(1) f (2k1)(1)],. 1. 1f (x) sin(bx) dx cos b. ... where. F (a, b) = cos(a b) exp. (. a2. 4z2. ) [.

22 Jan 2018: b "arctan ðcos2 W þ q sin2 WÞR. 2ð1 " qÞsin W cos Wh! : ... Q2 1 þ 2hð1 " qÞsin W cos WRðcos2 W þ q sin2 WÞ!

26 Jul 2013: xm(x, t) =x 'a cos(x Re ω(t)t φ)and ym(x, t) = 'b sin(x Re ω(t)t),. ... xy|y=0 b sinx. 2ψ1. y2|y=0, (3.15). ψ2,x |(x,0) = a cos(x φ)2ψ1.

26 Jul 2013: 4 m1=1. m2=1. 1. m12 m221 m12. 2m12 m2. 2cos 2 b m1x cos 2 b m2y , 18from which we note that large contributions to Eqs. ... cos m2w W. m1=1. 1m Cm 12m1 12 ln S sdwds , 20b.

6 Feb 2019: This implies a = b,so using the formula cos θ 1 = 2 cos2(θ/2), we obtain,. ... 4.11). The antiderivative of cos(2 arctan(2x)) is arctan(2x) x, so there exist (new) realconstants a, b and c such that.

22 Nov 2017: rA A O ‘( cos u1, sin u1) ð4:1Þ. and. rB B O ‘( cos u1 þ cos u2, sin u1 þ sin u2), ð4:2Þ. ... 0 its magnitude andt̂ ( cos u2, sin u2) the unit tangent vector joining A and B.Under the assumption of creeping flow, the drag forces areFA 2 zvA

21 Oct 2022: x′ = x cos ωt y sin ωt. y′ = y cos ωtx sin ωtz′ = z (2.18). ... 12ml2θ̇2 mgl cos θ (2.39). From which we may derive Lagrange’s equations using the coordinate θ directly.

5 Oct 2020: φ = U(r a2/r) cos θ,. with associated streamfunction. ψ = U(r a2/r) sin θ,. ... As before. u. t= ν. 2u. y2in y > 0. u = U cos ωt on y = 0.

8 Jul 2021: Using spherical coordinates, wefollow ref. [12], and prescribe the tangential surface veloc-ity, us, as us = (3U0/2)(sin θ β sin θ cos θ)eθ, where U0. ... For a given value ofthe dipole strength, the maximum growth rate is obtainedfor cos k = 1, i.e.

1 Oct 2016: A = 12 ln[cosh (ŷ ŷ0) cos (x̂ x̂0)], (15). B = 12 ln[cosh (ŷ ŷ0) cos (x̂ x̂0)]. ... We obtain. F H = ξH aω(1 ρ) sin cos 4.92 pN μm (54).

6 Nov 2016: x y. 2. )cos. (xy. 2. ), (0.8d). sin(x) sin(y) = 2 sin. ... 1. (cos θ i sin θ)p. =1. cos pθ i sin pθ. =

22 Jan 2019: ζ,fibLfib (sin2 α ρfib cos2 α) 6π μah, (26)B = [ζ,t Lt (1 ρt )ty tz hζ,fibLfib (1 ρfib ) sin α cos α]Rf l, (27). ... where. A = {ζ,t Lt (cos α[1 (1 ρt )t 2z ] h sin α(1 ρt )ty tz ) μ̃Lfib cos α 6π μah cos α}, (53)B = Lfib sin α,

13 Jan 2015: repr. oduc. ed w. ithou. t per. mis. sion. In the last step, we used spherical polars with z = r cos θ. ... is. φ = a b. r. (90). Next we have solutions, like the dipole potential, cos θ,.

3 Oct 2023: a b = ab cos θ. (23). a. bθ. Figure 10: Angle used in the definition of the scalar product a b. ... 26). 4. If a b = 0, and a, b 0 then cos θ = 0, necessarily, i.e.

10 Apr 2011: B s =(. η2. 4 ζ 2. )1/2 [Bx sin ω By cos ω. ] ... s = η2. [sin θ sin ωx̂ sin θ cos ωŷ cos θ ẑ. ].