By R. Sneider

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**Example text**

2 g 3 (c) With the ship at latitude 45◦ and firing the shell at close to 45◦ to the local horizontal, v and ω are almost parallel and the ω × v term can be set to zero. Further, with v in a northerly direction, (g × ω) · v = 0. Thus we are left with only the cubic term in τ . 85 10−4 m s−3 . e. some 10–15 m to the East of its intended target. 7 Parameterizing the hyperboloid x2 y2 z2 + − =1 a2 b2 c2 by x = a cos θ sec φ, y = b sin θ sec φ, z = c tan φ, show that an area element on its surface is dS = sec2 φ c2 sec2 φ b2 cos2 θ + a 2 sin2 θ + a 2 b2 tan2 φ 1/2 dθ dφ.

E. to a1 ξ˙ 2 + a2 η˙ 2 and b1 ξ 2 + b2 η2 . To find the new variables we will use the following result. If the reader is not familiar with it, a standard textbook should be consulted. e. Q1 = vT Cv and Q2 = vT Dv, with both C and D diagonal. Further points to note are: (i) that for the ui as determined above, (um )T Aun = 0 if m = n and similarly if A is replaced by B; (ii) that P is not in general an orthogonal matrix, even if the vectors un are normalized. (iii) In the special case that A is the identity matrix I: the above procedure is the same as diagonalizing B; P is an orthogonal matrix if normalized vectors are used; mutual orthogonality of the eigenvectors takes on its usual form.

It is therefore x(t) = A cos ωt + B sin ωt + qE . mω2 29 Vector calculus ˙ The initial condition x(0) = 0 implies that A = −qE/(mω2 ), whilst x(0) = 0 requires that B = 0. Thus, qE (1 − cos ωt), mω2 qE qE qB x=ω (1 − cos ωt). (1 − cos ωt) = z˙ = 2 m mω mω x= ⇒ Since z(0) = 0, straightforward integration gives qE sin ωt t− mω ω z= = qE (ωt − sin ωt). mω2 Thus, since qE/mω2 = mE/B 2 q, the path is of the given parametric form with ξ = ωt. It is a cycloid in the plane y = 0; the x-coordinate varies in the restricted range 0 ≤ x ≤ 2qE/(mω2 ), whilst the z-coordinate continually increases, though not at a uniform rate.

### A Guided Tour of Mathematical Physics by R. Sneider

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