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30 Jan 2013: φ(Ai) = A′i. Trigonometry. If 4ABC has sides a, b, c and opposite angles α, β, γ, then. ... sin αa. =sin β. b=. sin γc. c2 = a2 b2 2ab cos γ.J.Rasmussen@dpmms.cam.ac.uk.

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.

16 Dec 2013: 1. Find the General Solution with f (t) = 1. Complementary function is y = A cos(3t) B sin(3t). ... Particular integral is y = 19General solution is y = 19 A cos(3t) B sin(3t).

6 Aug 2013: b) The canonical momentum conjugate to x is. px =L. ẋ= (m1 m2)ẋ m2lφ̇ cos φ. ... circular motion around the z axis with constantangular velocity. In terms of the z coordinate, the particle undergoes simpleharmonic motion, z(t) = a sin κt b cos κt

15 Aug 2013: b) xn = A cos(n0) B sin(n0) where A and B are uncorrelated random vari-ables with zero mean and identical variances. ... rXX[k] = 0.5A2 cos(k0). ]. 5. A stationary random process has autocorrelation function:.

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.

19 Oct 2013: expected. (ii) all A’s, B’s zero except A1:. φ = A1r cos θ = A1z , u = φ = A1ez. ... z. (iii) all A’s, B’s zero except B1:. φ =B1 cos θ.

31 Jan 2013: The polar triangle A′B′C′ is defined by theunit vectors in the directions B C, C A, and B A. ... Deduce that. sin α sin β cos c = cos γ cos α cos β.

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.

26 Jul 2013: uy=y1 = 1 ṠR cos y=y1 , 11a. vy=y1 = ṠR sin y=y1 , 11b. ... uy=y2 = 1 ṠR cos y=y2 U , 11c. vy=y2 = ṠR sin y=y2.

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).

21 May 2013: For M, N N, we use the abbreviatednotationB(M, N) = B(M, M; 0, N). ... A horizontal crossingof B = B(M1, M2; N1, N2)is an open path ofB linking some vertexvM1,n1 to some vertexvM2,n2; a verticalcrossinglinks somevm1,N1 to somevm2,N2.

16 Dec 2013: b) -1, 2, -1. (c) -1, 0, 2, 0, -1. (d) 1, 2, 2, 1. ... b) Explain how you would use the Hamming window. w (n) = 0.54 0.46 cos. (.

23 Oct 2013: 4. Let (fn) be a sequence of real-valued continuous functions on a closed, bounded interval [a,b], and. ... m=1 mam cos mx. (b) Show that f(x) =. m=1 am sin mx defines a continuous function on [π,π], but that the seriesm=1 mam cos mx need

26 Jul 2013: ̃1 =ir2. 4!cos 2" 1" ,! 16a". ũ1r = ir. 2sin 2" ,! 16b". ũ1" =ir. 2!1 cos 2""! 16c". B. Second-order solution. 1. ... 4-̃4!0" = 0. A4 sin 2" B4 cos 2" C4 sin 4"r2. ,!

15 Aug 2013: b) Explain how you would use the Hamming window. wn = 0.54 0.46 cos. ... 10. A digital filter has the transfer function (1bz1)/(1az1), where a > 0,b > 0, b < a.

6 Aug 2013: E m2gl cos φ. [9]. 2 A harmonic oscillator is weakly perturbed by a cubic potential λx3 so thatits Hamiltonian has the form. ... 19]. (b) Find the temperature T (x, t) of the system if s(x, t) = cos(px)δ(t t0)and T (x, t < t0) = 0, for the two cases

21 May 2013: n. i ii. P P x a x b=. =. ( ) ( )data;model data, ;modelb. ... and D from the two distributions. ( ) ( ) ( ) ( ) ( )[ ]exp cos sin cos 2 sin 2P A B C Dα α α α α.

29 Nov 2013: Dfm|x(h)).(b) Define f: R3 R2 by f(x,y,z) = (exyz, cos x2y). ... 10. Use the contraction mapping theorem to show that the equation cos x = x has a unique.

4 Oct 2013: B(cos(1 ξ)) A cos ξ = 1. Solving these two equations for A and B gives the Green’s function to be:. ... Then. I. λ=. 0. cos(λx)eαxdx = <(. 0. e(αλi)xdx. ),. = <[e(αλi)x.

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

26 Jul 2013: 3) leads to 2p = 0. A solutionof the harmonic equation for the pressure is, at the first order inkb, p = p1(r,θ) (A cos s B sin s) with s = ... xi = x0 b cos β sin (s φ) , (34)yi = y0 b sin β sin (s φ).

5 Feb 2013: The polar triangle A′B′C′ is defined by theunit vectors in the directions B C, C A, and B A. ... Deduce that. sin α sin β cos c = cos γ cos α cos β.

16 Dec 2013: 1. Find the General Solution with f (t) = 1. Complementary function is y = A cos(3t) B sin(3t). ... Particular integral is y = 19General solution is y = 19 A cos(3t) B sin(3t).

22 May 2013: i/2)n = (1/2)neinπ/2 = (1/2)n(. cos(nπ/2) i sin(nπ/2)). Thus for some B′ and C′,. ... p(n)11 = A (1/2). n(. B′ cos(nπ/2) C′ sin(nπ/2)). We then use facts that p(0)11 = 1, p.

26 Jul 2013: a) The no-shear regions are transverse to the flow direction. (b) Theno-shear regions are parallel to the flow direction. ... Thesolution to this problem was derived by conformal mapping in Philip (1972a) and isrecalled in Appendix B.

19 Oct 2013: be similar to that of the spherical harmonicP4(cos ), and so we adopt the estimate L 4:5. ... Then equations(1) and (2) imply T=T r2c 2i =gL and juj 2g T=2N2T , where we have takensin cos 21=2.

21 May 2013: n. i ii. P P x a x b=. =. ( ) ( )data;model data, ;modelb. ... and D from the two distributions. ( ) ( ) ( ) ( ) ( )[ ]exp cos sin cos 2 sin 2P A B C Dα α α α α.

15 Aug 2013: 1 x2t1+ 8 cos(1.2t) vt. yt = B(xt) wt. =(xt). 2. ... where. µt1 = Pt1(BTD1. yt1P1t1|tµt1|t), and Pt1 = (B. TD1. BPt1|t)1. •

4 Nov 2013: r1 = [2 cos(φ1)]cos(φ1 θ ). d. 2, (3). r2 = [2. ... 2 cos(φ2)]cos(φ2 θ ). d. 2. (4). In drainage, θ = θr and the drainage capillary entrypressure corresponds to the minimum radius of curvature r1:rdr = minφ1 r1(φ1; θr ).

26 Jul 2013: x=0, 59. with boundary conditions. y10, t = 0,y1x. 0, t = cos t,2y1x2. ... 1 cos 12 102 Re1010.66. When the swimmer does not have a body, R FU = 0, we obtainthe “elastic scallop,” the flexible version of the two-armswimmer discussed by Purcell

26 Jul 2013: The scale bar is 10 mm. (b)Speed-frequency characteristics of flexible nanowire motors. ... ment as. x(s, t) b(s)cos[t f(s)], (29). y(s, t) b(s)sin[t f(s)], (30).

5 Oct 2013: As is evident, the quantity cos (θn,n) is exponentially small in n. ... 100. 106. 1010. (a) (b) (c). Figure 1: (a): cos(θn,n) against n.

16 Jul 2013: The kernelsg. (ν)k,k′,′ can be written asg( ,L)k,Kkq,K2 = 〈g2 〉[1 cos (θk,q θkq,q )],g( ,T)k,Kkq,K2 = 〈g2 〉[1 cos (θk,q ... In thesecoordinates, ki = (Q/2)[cosh(ui ) cos(vi )] and | Q ki | =(Q/2)[cosh(ui ) cos(vi )], so that nonlinear

16 Dec 2013: an =1. π. ππ. cos(nt)f (t) dt = 0 n 6= 0. ... B : 4π n=1(1)n1. sin(12nω0t)n. C : 4π n=1(1)n1. cos(12nω0t)n. D : None of the above.

13 Aug 2013: σ1 = sin θ sin ψdφ cos ψdθ,σ2 = sin θ cos ψdφ sin ψdθ,σ3 = cos θdφ dψ. ... 1 a. 4. ρ4. ). 4(dψ cos θdφ)2+. ρ2. 4(dθ2 sin2 θdφ2. ).

10 Nov 2013: 2t cos ka)c†kσckσ. /a. (k). /a k-. B.Z. As expected, as k 0, spectrum becomes free electron-like:k 2t t(ka)2 (with m = 2/2a2t). ... k. cos k a†kak. Ĥ = JNS2 B.Z.k. ωka†kak O(S.

24 Jun 2013: FIG. 2 (color online). Steady-state circulation in highly concentrated B. subtilis droplets. ... 2(e)]. The averagevalue b4m is slightly smaller than the length ‘5mof B.

15 Aug 2013: 36. Example: standard nonlinear modelxt = A(xt1) vt= xt12 25 xt11 x2t1 8 cos(1:2t) vtyt = B(xt) wt. =

26 Jul 2013: 23). This equation can be integrated in t as. e(j )i (t, τ) = a(j )i (τ) cos(t) b(j )i (τ) sin(t). ... A(τ) cos(2t) B(τ) sin(2t). C(τ) cos(t) D(τ) sin(t) E(τ), (28).

11 Apr 2013: r1 = [2 cos(φ1)]cos(φ1 θ ). d. 2, (3). r2 = [2. ... 2 cos(φ2)]cos(φ2 θ ). d. 2. (4). In drainage, θ = θr and the drainage capillary entrypressure corresponds to the minimum radius of curvature r1:rdr = minφ1 r1(φ1; θr ).

19 Oct 2013: determining the velocity field everywhere given only the mass under each isentrope (B-surface),. ... the potential vorticity distribution on each isentrope, and the B distribution on the boundary.

5 Aug 2013: π/10. DwA. dB. Step E. dgeSubban. d Directio. n. D w cos π/. ... θ1 =π. 10+ α =. π. 10+ tan1. dA dB Dw sin π10Dw cos π10.

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.

26 Jul 2013: B]lab = B0[cos(ωt ), si n(ωt ), 0]T. (2). Experimentally this magnetic field provides an external torque to the head of the artificial flagellumbut no external force. ... ξ̄‖ =ξ‖. µ= 2π. ln(2 cos θ/γ ) 1/2,. ξ̄ =ξ.

27 Jan 2013: CYLINDRICAL SHELL AND ELASTIC CONTINUUM. 192. B.1 Volmir’s Shell Equations. 192. ... B.2 Coefficients for the Cylindrical Shell.193. B.3 Coefficients for the Elastic Continuum.

19 Oct 2013: determining the velocity field everywhere given only the mass under each isentrope (B-surface),. ... the potential vorticity distribution on each isentrope, and the B distribution on the boundary.

21 May 2013: n. i ii. P P x a x b=. =. ( ) ( )data;model data, ;modelb. ... and D from the two distributions. ( ) ( ) ( ) ( ) ( )[ ]exp cos sin cos 2 sin 2P A B C Dα α α α α.

16 Jul 2013: ν2 = 0.9770νmax cos(P 0.48), (4)which were taken from de Leeuw et al.;74. ... 3J = A cos2 θ B cos θ C, (6)where θ is a dihedral angle.

14 Mar 2013: 1 cos(5π/N ) (N 10) for the. high-frequency subbranch originating from 10L-MoS2. ... 1 cos(π/N ) (N. 2) at higher frequency, and ωC (N ) = 38.6.