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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.

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/.

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.

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).

7 Nov 2017: u u u1, 1, 0, 1,3. 2sin 1 cos , 1r q q q q b q= = = f q( ) ( ) ( ) ( ) ( ). where θ is the azimuthal angle from the squirmer orientation e and ... From equation (1) we have that the tangential velocity on the squirmer is given byu 3 2 sin 1 cos ,

2 Jul 2024: b = R sin. 2. 2. = R cos. 2. If b > R, clearly there is no scattering. ... incident(r) = eikz = eikr cos. is also a solution to the Schrödinger equation.

19 Jul 2019: x(t) = X R sin(!B(t t0)) and y(t) = Y R cos(!B(t t0)). ... 4r. 1 cos. sin(1.27). This again gives rise to the magnetic field (1.26).

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).

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.

10 Sep 2021: e1 = cos χ [cos θ (cos φex sin φey ) sin θ ez ] sin χ [ sin φex cos φey], (1). ... the helix is R = sin(ψ )/(π N ) and the pitch is p = 2 cos(ψ )/N.

25 Jan 2007: pair abmeans add or subtract the same expression with a and b interchanged. ... cos θ =gabU. aV b. |U a||V b|. (103). If the two vectors are timelike, and both are future directed, the rapidity θbetween them is given by.

11 Mar 2010: β2π. π. 0. sin θ dθ eβµB cos θ (10.5.3). =I. ... for the 2π, and the integration over θ is doneby setting x = cos θ.

10 Sep 2021: helical turn and b = cos ψ. ... ez Fp = ζ T1 LT / cos ψ. 0[U (βT 1)b(R sin ψ U b)] d s. =

26 Jul 2013: 1(a). Our new result is illustrated in Fig. 1(c), where we showthe dynamics of swimmers undergoing reciprocal motionwith velocity Uswim UðtÞe and UðtÞ U cos!t withU 5 ... a) No swimming, (b) steady swimming at speed U 5 m=s, (c) reciprocalswimming at

9 May 2023: taking the opportunity to remind ourselves of this. In Electromagnetism, the magnetic field obeys r B = 0 which means that it can be. ... magnetic field B in terms of the current density. B(x) =µ04.

7 Nov 2017: b β(b b‖) , (43). respectively, where. P (θ ) = (1 U0)b b‖ cos θ (1 U0)(b b‖) cos2 θ, (44). ... uγ̇ th =. sin2 θ 2 b‖b. (cos θ q)2, (45).

3 Dec 2022: EI. ET. B. B. BR. I. T. Figure 66: Incident, reflected and transmitted waves with normal polarisation. ... E(x, t) = E(!) ek2z cos(k1z! t). B(x, t) =|k|! (ẑ E(!)) ek2z cos(k1z!

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. ].

26 Jul 2013: mal a or b at given Pe). Starting from a random initial condi-. ... For t 1, the swimming stroke,. b cos t b1 þ sin t bk; (44).

13 May 2020: 220. (|A|2 3. 4x20. ),. εv =1. 220(Br sin 0t Bi cos 0t)2. ... 1. 220(Br cos 0t Bi sin 0t). 2. =1. 220|B|2,. as required.

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.

26 Jul 2013: evaluated at =. In Fourier notation and given that t= cos t, they become. ... 0. = 0. 2. Pressure and stress field. We have. ̃20 =. 2De1 De21 iDe1. B. 2 cos 2 sin 22, 43. as well as zero pressure p2 = 0, so that the average

24 Jul 2023: Ω sed = mg16πa2η. ln(a. b. )(κ̃x sin θ κ̃y cos θ. ... sin θ, (3.40). Usedx = mg. 8πaηln(a. b. )cos θ sin θ, (3.41).

1 Dec 2023: sin α) 2π(2 cos φ, sin φ)/ log Bi1 (see (3.2)). ... Ṽs =U cos(φΦ), ω̃=URH. sin(φΦ). (4.5a,b). Dow. nloa. ded. from. htt.

19 Oct 2013: exponential counterpart, cos(µ) = 12 (e. ... Then φ̂ = φ̂(x, z) exp(kz) cos(kx),equivalent to a single pair of progressive deep-water waves travelling in the x directions.

25 Jul 2018: 708 J. B. KIRKEGAARD AND R. E. GOLDSTEIN. Fig. 5. Drift efficiency 〈cos θ 〉 with finite tumbling time. ... 1 I1(k)I0(k). cos μ. )(〈cos θ 〉 〈q cos θ 〉) , (B.2).

3 Oct 2023: 13. Calculate the following sums:. (a)5. n=1 sin(nθ). (b)N. n=1 cos(nθ). ... a) (x,y) = (a cos θ,b sin θ),. (b) (x,y) = (a cosh θ,b sinh θ),.

27 May 2019: 1, 2), B = B(r){cos(ot),sin(ot),0}is the background magnetic field, ẑ is the normal to the air–water interface, B(r) is the strength of the ... 4h2sintyðtÞ; (18). where n̂ = {sin[c(t)],cos[c(t)],0}, b 3mB0ðC

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.

1 Nov 2017: 3ρ̄r. (b) Consider linear perturbations about the homogeneous solution. Show that the density. ... t = B (1 cos θ). where you should determine A, B and θ.

25 Jan 2007: with no dependence on and ) is = a br: (116)Next we have solutions, like the dipole potential, / cos , = Er cos cr2 cos : (117)What E does the rst term give? ... Hence, from (186)B = 0I4 Z 11 sdz0(s2 z02)3=2 e= 0I4s Z =2=2 cos d e; (z0 = stan)= 0I2se:

20 Jan 2009: Repeat the calculation in spherical geometry, i.e. for concentric spheres of radiir = a, b, b > a. ... For (iii), from a decent diagram of spherical polars, find k = er cos θ eθ sin θ. ].

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).

28 Nov 2013: X = 0 is given byF′ G = c0, (cos X σ0) G sin XF = 0, (4.4a,b). ... Θ = Ae(1κ2)τ cos κX Z, W = Ae(1κ2)τ cos κX, (5.1a,b)for any amplitude A.

19 Jul 2019: 161 –. 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.

2 Jul 2024: is given by. H(t) = eE0 x cos(!t). Our goal is to find the eigenstates of the time-dependent Hamiltonian. ... The diagonal matrixelements are particularly simple. h i|H| ii = eE0 h i|x| ii cos(!t) = 0.

11 Nov 2006: Then. z1z2 = r1r2 (cos θ1. cos θ2 sin θ1. sin θ2i (sin θ1. ... 1.30). Proof. From (1.25). cos nθ i sin nθ = ei (nθ).

2 Nov 2018: 12 e2iωγerf. (iω(γ 1). (iω)12. )(1 γ). ]. 27. Let. Ca,b(α) :=. ba. cos(αt)eiωt2dt =. π12 exp. ( iα24ω. )4(iω). 12. [erf. (a(iω). ... 2(iω)12. )],. C1,a,b(α) :=. bat cos(αt)eiωt. 2dt =. i. 2ω. [cos(aα) exp(iωa2) cos(bα) exp(iωb2). ]

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.

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:.

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.