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31 May 2024: y = r sin sin. z = r cos. In polar coordinates, the infinitesimal distance between two points can be simply de-. ... dt(r2̇) = r2 sin cos ̇2 ,. d. dt(r2 sin2 ̇) = 0 (1.12).

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 Jul 2009: B′ cos KX instead of A and B, with solution. N (0) ce(. ... A′ B′. 2sin 2T. B′. (2)2cos 2T. )cos KX. (5.10). Note that the oscillating part has to be retained in the present analysis up to the secondorder.

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).

26 Jul 2013: us(ξ,η)= u0s. 2(1 cos ξ)( cos ξ)(1 e2) (1 e2) cos(2ξ) s(ξ,η). ... Here hwis the minimal distance of the centroid to the wall without penetration, hw =(1 e2) (1 e2) cos(2θ)/2.

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)

7 Apr 2021: x(t) = X R sin(!B(t t0)) and y(t) = Y R cos(!B(t t0)). ... AN=. g. 4r. 1 cos. sin(6.25). The resulting magnetic field is.

18 Feb 2009: E = (E0, 0, 0) cos(kz ωt), B = (0,1. cE0, 0) cos(kz ωt), (9). ... for E0 real,. (Ephys)x = 2E0 sin kz sin ωt, and (Bphys)y = (2/c)E0 cos kz cos ωt.

26 Oct 2020: Substituting y,a,b and f by theirFourier series and using (3.3)-(3.5) we obtain an infinite dimensional system of linear equationsfor the Fourier coefficients ŷn:. ... For instance: choosinga(x) = f(x) = cos πx, b(x) = sin 2πx (which incidentally

5 Nov 2014: When small-amplitudewaves propagate along ex, the surface of the cylinder is de-scribed by r̂ = [h(s, t) b cos φ]ey b sin φez sex and the. ... p̃ ηωh̃i cos θAK1(kr)ũr h̃ωi cos θ[αArkK1(kr) BK2(kr) CK0(kr)]ũθ h̃ωi sin θ[BK2(kr)

26 Jul 2013: 1x, = U1ey , D4a. 1x,0 = b cos x tex a sinx t ey , D4b. ... UN =12 b. 2 2ab cos a2. D17. Allowing tangential motion does therefore reduce the swim-ming speed by the same amount, and the final results of thepapers, Eq.

18 Sep 2022: 45 –. by a sign, since this can be compensated by the cos function. ... Insisting that the two. cos functions agree, up to sign, gives us the condition.

24 Nov 2023: 26). 4. If a b = 0, and a, b 0 then cos θ = 0, necessarily, i.e. ... We have. a b = ab cos θ,(2, 1, 0) (3, 0, 1) =. 5. 10 cos θ,. cos θ = 6/. 50,. and thus θ = cos1(6/.

26 Oct 2015: When λ 0 then X(x) = a sin(λx) b cos(. λx) solves X′′ = λX. ... λx B cos. λx, but if we impose the boundary condition y(1) = y(1) then.

14 Jun 2017: cos. (ωλ ,1(ck,t). )sin. (ελ ,1(ck,t). )sin. (ωλ ,1(ck,t). ), (29). ... 0. + cos. (ωλ ,1(ck,t). )(t ck).  t ck(t ck)21.

28 Sep 2020: z = r cos θ. Then compute the Christoffel symbols by directly applying (1.8). ... l2(1 e cos φ) (4.16). where e is the eccentricity, a free parameter.

15 Oct 2020: S =. α β. β α. β. β α.  ,are given by λk = α 2 iβ cos kx , and wk = (i. ... λk (A) =λk (C ). λk (B )=. 1 12 iµ cos kx.

1 Dec 2023: GF′′=α2F, σG+ iαA[G cos X F′ sin X]F′′=G′′α2G, (4.1a,b). where F and G are functions of X alone. ... which have the leading-order solution. F =a(Y)[sin X cos ξY X π], G=a(Y) sin X cos ξY (4.5a,b).

28 Nov 2014: f ′′cot θf ′6f =A cos θ B. (2.14)Together with the particular solution, the full solution is then ψ = cr3(at bt3 1)for t =cos θ. ... The function H(ν) is plotted for cone angles in figure 1(b).

4 Nov 2014: as. Θ(0, t) = Θ0 A cos (2πt/tD) B cos (2πt/tY) (4.30). – ... We can solve these as. X(x)T(t) =[a sin(. λx) b cos(.