Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

(c) The differential equation y 11 + k 2 y = 0 has auxiliary equation r 2 + k 2 = 0. (This is the r of Section 17.1,
CHAPTER 17 REVIEW D 357
not the r measuring distance from the earth's center.) The roots of the auxiliary equation are ± ik, so by ( II) in
Section 17.1' the general solution of our differential equation for t is y( t) := Cl cos kt + C2 sin kt. It follows that
y' (t) = -c1k sin kt + c2k cos kt. Now y (0) = Randy' (0) = 0, so c1 = Rand c2k = 0. Thus y(t) = Rcos kt and
y'(t) = - kRsinkt. This is simple harmonic motion (see Section 17.3) with amplitude R, frequency~. and phase angle 0.
The period is T = 21r j k. R Rj 3960 mi = 3960 · 5280 ft and g = 32 ft/s 2 , so k = v'i(R Rj 1.24 x 10- 3 s- 1 and
T = 27r/k Rj 5079 s Rj 85 min.
(d) y(t) = O # coskt = O # kt = !+1rnfor someintegern => y'(t) = - kRsin(f+7rn)=±kR.Thusthe
particle passes through the center of the earth with ~peed kR Rj 4.899 mi/s Rj 17,600 mi/ h.
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