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11 Feb 2014: α =. C. A dx. α =. C. A dx =. B dS. Aϕ =g. 4πr. 1 cos θsin θ. B =g. ... 1 cos θsin θ. B =g. 4πr2̂r. References. [1] Gibbons and Rychenkova, 9608085.

21 Jan 2014: 7. Suppose γ : [a,b] R2 is a smooth curve parametrized by arc length. ... r cos z,r sin z,z).(c) U = (0, 2π) (0, 2π), σ(θ,φ) = ((a b cos φ) cos θ, (a b cos φ) sin θ,b sin φ).

29 Aug 2014: cosh ξ cos η ,. z =c sinh ξ. cosh ξ cos η , (B.1). ... UndVndη. , (B.17). ψ̂. ξ= 3. 2sinh ξ. (cosh ξ cos η)5/2.

21 Jan 2014: 7. Suppose γ : [a,b] R2 is a smooth curve parametrized by arc length. ... r cos z,r sin z,z).(c) U = (0, 2π) (0, 2π), σ(θ,φ) = ((a b cos φ) cos θ, (a b cos φ) sin θ,b sin φ).

18 Jan 2014: L =1. 2m(. ρ̇2 ρ2θ̇2). 12kρ2 q. (. Eρ cos θ B. ... A =B. 2. (. ρ sin θ, ρ cos θ, 0).

4 Feb 2014: φ(s,v) = α(s) r(n(s) cos v b(s) sin v)where n = n(s) and b = b(s) are the normal and binormal vectors of α. ... 9. Consider a surface of revolution parametrized by φ : (0, 2π) (a,b) R3, whereφ(u,v) = (f(v) cos u,f(v) sin u,g(v)).

6 Feb 2014: 1. Find the General Solution with f (t) = 1 (t 0).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).

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(.

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)

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).

23 Oct 2014: ui(0) = sin( πxi. b. ), u′i(0) =. π. b. gd(xi) cos. ( ... b. gd(xN ) cos. ( πxNb. ))T,. (39). where U (t) denotes the N -dimensional vector with entries ui(t) u(xi, t),.

20 Oct 2014: 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

22 Dec 2014: eπ2. sinh(π) j = 0,. j cos( jt)sin( jt)π2. j21. j Z+. ... a = 0, b = π, (t [0,π],q(t) = 20 cos(2t)), α1 6= 0, β1 6= 0, α2 = β2 = 0,.

9 Nov 2014: 2. V (t) =µπ2c2. 4L. n=1. n2[An cos. (nπct. L. ) Bn sin. ... m=1. i=1. [Cmi cos mθ Dmi sin mθ] cos(kmict) Jm(kmir). (5.32).

14 Oct 2014: obtain:. Φ(ϕ) = A sin(λϕ) B cos(λϕ),. with positve constants A, B. ... Aλrλ Bλr. λ) sin(λϕ) (Cλrλ Dλr. λ) cos(λϕ). E0 F0 ln(r).

13 Oct 2014: b) Now consider an almost ideal FJC with N monomers of type a. ... Show that these can be written as. W(n)(x) = A cos(k(n)x) B sin(k(n)x) D cosh(k(n)x) E sinh(k(n)x) ,. and

3 Feb 2014: From simple trigonometry, this leads tothe following expressions. jGxx j sin2 a; jGyy j cos2 a; jGxy j sin a cos a (1). ... a diameterwith end-points ðjSxx;jSxyÞ ðj sin2 a;j sin a cos aÞ and ðjSyy; jSxyÞ ð0; j sin a cos aÞ; see Fig.

25 Mar 2014: xz. -y. -xz zz. (a) (b). angle = πJt. 2Î1yÎ2z2πJ12tÎ1zÎ2z cos (πJ12t) 2Î1yÎ2z sin (πJ12t) Î1x. ... cos (τ) sin (τ)Îy sin (τ) sin (τ)Îx Îx Spin echoes 20.

6 Feb 2014: S = b cos θ tan γffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 cos2θ tan2γ. p ; [2]. ... cos γA1cos γB1. =cos γA2cos γB2. [21]. must be satisfied to enable stacking.

29 Aug 2014: û% |r =1 = K cos(nθ ). . 0. cos φ cos θ sin φ. ... Thus, the previous equations canbe put in the form. s û% = K cos φk.