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16 Oct 2021: Γ. θ. r. θ. =. Γ. r. θ. θ. =. 1. r. Γ. ϕ. ϕ. θ. =. -. sin. θ. cos. ... θ. ϕ. =. cos. θ. sin. θ. (1.11). There are some important lessons here.

8 Jul 2021: ψ. x. (x(1)s ,y(1)s ). = b cos(x t ), (18). ... ψ. x. (x(2)s ,y(2)s ). = b cos(x t ), (20).

18 Oct 2021: The wavefunction (2.12) can be viewed as a sum of sine and cos functions and describes a complex-valued wave of wavelength. ... 2. x. The most general solution has the form. x. (. t. ). =. A. cos. (. ω. (. t. -. t.

12 Jun 2021: This too can be. resolved as. T = T cos r̂ T sin ̂. ... ei!t sin A cos. rg. lt B sin. rg. lt.

7 Apr 2021: x(t) = X R sin(!B(t t0)) and y(t) = Y R cos(!B(t t0)). ... AN=. g. 4r. 1 cos. sin(6.25). The resulting magnetic field is.

8 Jul 2021: Using spherical coordinates, wefollow ref. [12], and prescribe the tangential surface veloc-ity, us, as us = (3U0/2)(sin θ β sin θ cos θ)eθ, where U0. ... For a given value ofthe dipole strength, the maximum growth rate is obtainedfor cos k = 1, i.e.

8 Jul 2021: φ̇n = ω0 βω0 (cos χn cos χn1 ) βτ ω20 (sin χn1 sin χn ), (16). ... θ1 sin θ2 ) sin α1 sin δ cos θ1 (sin θ1 sin θ2 )]}, (46).

8 Jul 2021: Γ ={(r,θ): r = R(θ) = b cos θ. c2 b2 sin2 θ. = ... 2. ξ2+ η2. )and h = a. cosh η cos ξ ,(4.33a,b).

5 Mar 2021: Γ ={(r,θ): r = R(θ) = b cos θ. c2 b2 sin2 θ. = ... 2. ξ2+ η2. )and h = a. cosh η cos ξ ,(4.33a,b).

12 Jun 2021: We then define the usual. polar coordinates. x = r cos , y = r sin. – ... l2(4.13). with A and 0 integration constants. (You might be tempted instead to write u =. A cos B sin k/l2 with A and B as integration constants.

18 Oct 2021: cos. (. ω. t. -. 𝐤. 𝐱. ). B. (. 𝐤. ). sin. (. ω. t. -. 𝐤. 𝐱. ). (3.143). ... This form of the solution is more useful simply because it’s quicker to write exponentials rather than cos and sin.

1 Dec 2021: 29/34. Toy example. a = cosh(x), b = sin(πx) 2, ρ = tanh(x) 2, F (x,t) = cos(20t) sin(πx),. ... ρ = 1, F (x,t) = cos(πt)(24 π2(1 x2)2), y(x, 0) = (1 x2)2, yt.

16 Oct 2021: X. and. Y. at each point, using. g. (. X. ,. Y. ). =. |. X. |. |. Y. |. cos. θ. ... θ. ϕ. =. cos. θ. sin. θ. (3.111). From these, it is straightforward to compute the components of the Riemann tensor.

27 May 2021: θ ),. ξx (t ) = ξ0 {[A(R) B(R) cos2 θ ]eiωt } and ξy (t ) = ξ0 {B(R) cos θ sin θ eiωt }. (17). B. Low-frequency limit of Lissajous figures. The ... 1 cos2θ )2 ξ2x. 2 tan θ cos φ. 1 cos2θ ξxξy 1.

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. =

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.

17 Jan 2021: On the other hand, along the beating direction,the minimum distance is dkðmin cos 23 cos 18Þ, where 23 degand 18 deg are the maximum tilt angles with respect to the ... erna. l flo. wu. b||. b||0. T. y=cos (θ). x. Telθ.

8 Jul 2021: φ̇2{(s 1) sin φ2, (s 1) cos φ2} 1 < s < 3,(2.17). ... MPθ 1 MPθ 2. (2.35). and. MPx1 = 2Δζ81ζ‖. (4φ1 5φ2) cos θ, (2.36).

23 Apr 2021: 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.

20 Apr 2021: Even braver people can choose. U(φ) = 1 cos φ. This has infinitely many minima. ... U(φ) = 1 cos φ. Again, we suppress coupling constants, but it is possible to add them back.The potential looks like.