By K. F. Riley
This pupil answer guide offers entire suggestions to all of the odd-numbered difficulties in crucial Mathematical equipment for the actual Sciences. It takes scholars via each one challenge step by step, to allow them to truly see how the answer is reached, and comprehend any errors of their personal operating. scholars will study by means of instance the way to choose a suitable technique, enhancing their problem-solving abilities.
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Additional resources for Essential Mathematical Methods for the Physical Sciences: Student Solution Manual
699 η2 ). 47 Three particles each of mass m are attached to a light horizontal string having fixed ends, the string being thus divided into four equal portions, each of length a and under a tension T . Show that for small transverse vibrations the amplitudes xi of the normal modes satisfy Bx = (maω2 /T )x, where B is the matrix ⎛ ⎞ 2 −1 0 ⎝−1 2 −1⎠ . 0 −1 2 Estimate the lowest and highest eigenfrequencies using trial vectors (3 4 3)T and (3 − 4 T T √ √ Use also the exact vectors 1 2 1 and 1 − 2 1 and compare the results.
As expected, ∂B/∂φ = 0. The azimuthal component of the equation gives − ∂B = μ0 C ∂ρ for ρ≤a ⇒ B(ρ) = B(0) − μ0 Cρ. Since B has to be differentiable at the origin of ρ and have no φ-dependence, B(0) must be zero. This, together with B = B0 for ρ > a requires that C = −B0 /(aμ0 ) and B(ρ) = B0 ρ/a for 0 ≤ ρ ≤ a. (b) With B = ∇ × A, consider A of the form A = (0, A(ρ), 0). Then ∂ ∂ (ρA), 0, (ρA) ∂z ∂ρ 1 ∂ = 0, 0, (ρA) . ρ ∂ρ (0, 0, B(ρ)) = 1 ρ We now equate the only non-vanishing component on each side of the above equation, treating inside and outside the cylinder separately.
E. that which lies above the plane z = 0 with 0 ≤ θ ≤ π/2. It would not do so for the part with z < 0 for which x 2 + y 2 has to be greater than a 2 ; this is not catered for by the given parameterization. 51 Line, surface and volume integrals Having carried out this check, we calculate expressions for dS and hence r · dS in terms of θ and φ as follows: r = a sin θ cos φ i + a sin θ sin φ j + a 2 cos2 θ k, and the tangent vectors at the point (θ, φ) on the surface are given by ∂r = a cos θ cos φ i + a cos θ sin φ j − 2a 2 cos θ sin θ k, ∂θ ∂r = −a sin θ sin φ i + a sin θ cos φ j.