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26 Jan 2017: Conversely, show that for any three positive numbers a,b,c less than πsatisfying the above conditions, we have cos(b c) < cos a < cos(b c), and that there isa ... Deduce the formula. sin α sin β cos c = cos γ cos α cos β.

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.

13 Dec 2017: 5. Use the contraction mapping theorem to show that the equation x = cos x has a uniquereal solution. ... sin y1y′1 cos y2y. ′2 = e. y′1. cos y2y′1 sin y1y.

13 Dec 2017: real a=2.,b=1.;. border Gamma1(t=0,2pi) { x = a cos(t); y = bsin(t); }. border Gamma2(t=0,2pi) { x = 0.5a cos(t); y = 0.5bsin(t); }. // ... border Gamma(t=0,2pi) { x = cos(t); y = sin(t); }. mesh Th=buildmesh(Gamma(20));.

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:.

10 Oct 2017: 1 q cos p. ). 6. Show that the following transformation is canonical, for any constant, λ :. q1 = Q1 cos λ P2 sin λ , q2 = Q2 cos λ P1 sin λ ,. p1 = ... Q2 sin λ P1 cos λ , p2 = Q1 sin λ P2 cos λ.

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.

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 ,

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

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).