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7 Dec 2023: r3; (12b). _c % 32r3ð1þB cos 2cÞ%a. rsin c; (12c). where we recall that B = (a2 % 1)/(a2 1) A (0,1) is the shapefactor of the spheroid. ... Fig. 3A). Otherwise, it will escape due tothe fact that :r p cos c.

1 Dec 2023: tan δ = (Ws 1) sin Φ(Ws 1) cos Φ Q, (2.18). ... Fx(δ, n, Bi) Bifx(|δ|) sgn(δ) Fz(δ, n, Bi) Bifz(|δ|) sgn(cos δ), (3.6a,b)for some functions fx and fz.

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-.

1 Dec 2023: u= eσ cos(2πy). I0(σ)1, v=0. (3.2a,b). Given this fixed, known velocity the concentration simply evolves via the advection–diffusion equation (2.7). ... u(x, y, t)= eσ cos(2πy)Rc′(x,y,t) 1. 0eσ cos(2πs)Rc. ′(x,s,t) ds1. (4.4).

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.

3 Jul 2024: x(t) = A cos(!t) B sin(!t) with! =. rk. m. Here A and B are two integration constants and! ... E = mgl cos 0. We can determine the period T of the pendulum using (2.6).

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.

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.

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.