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6.4). Then, as one can verify, expressing s and h(s)in terms of the angle of displacement ϕ, we obtain s ϕdsR dψt==,√√2g(h−h(s))2gR(cosψ − cos ϕ0 )s0−ϕ009 Theparametrization of a curve by its own arc length is called its natural parametrization, and s iscalled the natural parameter.6.4 Some Applications of Integrationor389'1t=2Rgϕ−ϕ0dψ(sin2 ϕ20.− sin2(6.64)ψ2)Thus for a half-period 12 T of oscillation of the pendulum we obtain'11T=22Rgfrom which, after the substitutionϕ0−ϕ0sin(ψ/2)sin(ϕ0 /2)'RT =4gπ/20dψsin2 ϕ20− sin2,(6.65)ψ2= sin θ , we finddθ#,1 − k 2 sin2 θ(6.66)where k 2 = sin2 ϕ20 .We recall that the functionF (k, ϕ) =0ϕ#dθ1 − k 2 sin2 θis called an elliptic integral of first kind in the Legendre form.
For ϕ = π/2 it depends only on k 2 , is denoted K(k), and is called the complete elliptic integral of firstkind. Thus, the period of oscillation of the pendulum is'RK(k).(6.67)T =4gIf the initial displacement angle ϕ0 is small, we can set k = 0, and then we obtainthe approximate formula'R.(6.68)T ≈ 2πgNow that formula (6.66) has been obtained it is still necessary to examine thewhole chain of reasoning. When we do, we notice that the integrands in the integrals (6.63)–(6.65) are unbounded functions on the interval of integration. We encountered a similar difficulty in studying the length of a curve, so that we have anapproximate idea of how to interpret the integrals (6.63)–(6.65).However, given that this problem has arisen for the second time, we should studyit in a precise mathematical formulation, as will be done in the next section.Example 10 A body of mass m rises above the surface of the Earth along the trajectory t → (x(t), y(t), z(t)), where t is time, a ≤ t ≤ b, and x, y, z are the Cartesian3906Integrationcoordinates of the point in space.
It is required to compute the work of the bodyagainst the force of gravity during the time interval [a, b].The work A(α, β) is an additive function of the time interval [α, β] ⊂ [a, b].A constant force F acting on a body moving with constant velocity v performswork &F, vh' = &F, v'h in time h, and so the estimate+ ,+ ,inf F p(t) , v(t) (β − α) ≤ A(α, β) ≤ sup F p(t) , v(t) (β − α)t∈[α,β]t∈[α,β]seems natural, where v(t) is the velocity of the body at time t, p(t) is the point inspace where the body is located at time t, and F(p(t)) is the force acting on thebody at the point p = p(t).If the function &F(p(t)), v(t)' happens to be integrable, then by Proposition 1 wemust conclude that b+ ,A(a, b) =F p(t) , v(t) dt.aIn the present case v(t) = (ẋ(t), ẏ(t), ż(t)), and if r(t) = (x(t), y(t), z(t)), thenby the law of universal gravitation, we findF(p) = F(x, y, z) = GmMGmMr= 2(x, y, z),|r|3(x + y 2 + z2 )3/2where M is the mass of the Earth and its center is taken as the origin of the coordinate system.Then,&F, v'(t) = GmMand therefore bax(t)ẋ(t) + y(t)ẏ(t) + z(t)ż(t),(x 2 (t) + y 2 (t) + z2 (t))3/2(x 2 (t) + y 2 (t) + z2 (t))dt =2223/2a (x (t) + y (t) + z (t))bGmM bGmM=−.=− 2|r(t)| a(x (t) + y 2 (t) + z2 (t))1/2 a1&F, v'(t) dt = GmM2bThusA(a, b) =GmM GmM−.|r(b)||r(a)|We have discovered that the work we were seeking depends only on the magnitudes |r(a)| and |r(b)| of the distance of the body of mass m from the center of theEarth at the initial and final instants of time in the interval [a, b].SettingU (r) =GM,r6.4 Some Applications of Integration391Fig.
6.5we find that the work done against gravity in displacing the mass m from any pointof a sphere of radius r0 to any point of a sphere of radius r1 is computed by theformulaAr0 r1 = m U (r0 ) − U (r1 ) .The function U (r) is called a Newtonian potential. If we denote the radius of theEarth by R, then, since GM= g, we can rewrite U (r) asR2U (r) =gR 2.rTaking this into account, we can obtain the following expression for the workneeded to escape from the Earth’s gravitational field, more precisely, to move abody of mass m from the surface of the Earth to an infinite distance from the centerof the Earth.
It is natural to take that quantity to be the limit limr→+∞ ARr .Thus the escape work isA = AR∞ = lim ARrr→+∞ 2gR 2gR−= mgR.= lim mr→+∞Rr6.4.6 Problems and Exercises1. Figure 6.5 shows the graph of the dependence F = F (x) of a force acting alongthe x-axis on a test particle located at the point x on the axis.a) Sketch the potential for this force in the same coordinates.b) Describe the potential of the force −F (x).c) Investigate to determine which of these two cases is such that the position x0of the particle is a stable equilibrium position and what property of the potential isinvolved in stability.2. Based on the result of Example 10, compute the velocity a body must have inorder to escape from the gravitational field of the Earth (the escape velocity for theEarth).3926IntegrationFig. 6.63.
On the basis of Example 9a) derive the equation R ϕ̈ = −g sin ϕ for the oscillations of a mathematical pendulum;b) assuming the oscillations are small, obtain an approximate solution of thisequation;c) from the approximate solution, determine the period of oscillation of the pendulum and compare the result with formula (6.68).4. A wheel of radius r rolls without slipping over a horizontal plane at a uniformvelocity v. Suppose at time t = 0 the uppermost point A of the wheel has coordinates (0, 2r) in a Cartesian coordinate system whose x-axis lies in the plane and isdirected along the velocity vector.a) Write the law of motion t → (x(t), y(t)) of the point A.b) Find the velocity of A as a function of time.c) Describe graphically the trajectory of A. (This curve is called a cycloid.)d) Find the length of one arch of the cycloid (the length of one period of thisperiodic curve).e) The cycloid has a number of interesting properties, one of which, discoveredby Huygens10 is that the period of oscillation of a cycloidal pendulum (a ball rollingin a cycloidal well) is independent of the height to which it rises above the lowestpoint of the well.
Try to prove this, using Example 9. (See also Problem 6 of thenext section, which is devoted to improper integrals.)5. a) Starting from Fig. 6.6, explain why, if y = f (x) and x = g(y) are a pair ofmutually inverse continuous nonnegative functions equal to 0 at x = 0 and y = 0respectively, then the inequalityxxy ≤0f (t) dt +yg(t) dt0must hold.10 Ch.Huygens (1629–1695) – Dutch engineer, physicist, mathematician, and astronomer.6.5 Improper Integrals393b) Obtain Young’s inequalityxy ≤1 p 1 qx + ypqfrom a) for x, y ≥ 0, p, q > 0, p1 + q1 = 1.c) What geometric meaning does equality have in the inequalities of a) and b)?6. The Buffon needle problem.11 The number π can be computed in the followingrather surprising way.We take a large sheet of paper, ruled into parallel lines a distance h apart and wetoss a needle of length l < h at random onto it.
Suppose we have thrown the needleN times, and on n of those times the needle landed across one of the lines. If N2lis sufficiently large, then π ≈ ph, where p = Nn is the approximate probability thatthe needle will land across a line.Starting from geometric considerations connected with the computation of area,try to give a satisfactory explanation of this method of computing π .6.5 Improper IntegralsIn the preceding section we encountered the need for a somewhat broader concept ofthe Riemann integral.
There, in studying a particular problem, we formed an idea ofthe direction in which this should be done. The present section is devoted to carryingout those ideas.6.5.1 Definition, Examples, and Basic Properties of ImproperIntegralsDefinition 1 Suppose the function x → f (x) is defined on the interval [a, +∞[and integrable on every closed interval [a, b] contained in that interval.The quantitya+∞f (x) dx := limb→+∞ abf (x) dx,if this limit exists, is called the improper Riemann integral or the improper integralof the function f over +∞the interval [a, +∞[.The expression a f (x) dx itself is also called an improper integral, and in thatcase we say that the integral converges if the limit exists and diverges otherwise.Thus the question of the convergence of an improper integral is equivalent to thequestion whether the improper integral is defined or not.11 J.L.L.Buffon (1707–1788) – French experimental scientist.3946IntegrationExample 1 Let us investigate the values of the parameter α for which the improperintegral +∞dx(6.69)xα1converges, or what is the same, is defined.Since! b11−α |bdx11−α x=αln x|b11 xfor α = 1,for α = 1,the limitlimb→+∞ 1bdx1=αxα−1exists only for α > 1.Thus,∞1dx1,=αxα−1if α > 1,and for other values of the parameter α the integral (6.69) diverges, that is, is notdefined.Definition 2 Suppose the function x → f (x) is defined on the interval [a, B[ andintegrable on any closed interval [a, b] ⊂ [a, B[.
The quantityBf (x) dx := limbf (x) dx,b→B−0 aaif this limit exists, is called the improper integral of f over the interval [a, B[.The essence of this definition is that in any neighborhood of B the function fmay happen to be unbounded.Similarly, if a function x → f (x) is defined on the interval ]A, b] and integrableon every closed interval [a, b] ⊂ ]A, b], then by definition we setbf (x) dx := limAbf (x) dxa→A+0 aand also by definition we setb−∞f (x) dx := lima→−∞ abf (x) dx.6.5 Improper Integrals395Example 2 Let us investigate the values of the parameter α for which the integral10dxxα(6.70)converges.Since for a ∈ ]0, 1]1adx=xα!it follows that the limit11−α |1 ,a1−α xln x|1a ,1lima→+0 aif α = 1,if α = 1,dx1=αx1−αexists only for α < 1.Thus the integral (6.70) is defined only for α < 1.Example 30−∞ex dx = lima→−∞ a0ex dx = lima→−∞ x 0 e a = lim 1 − ea = 1.a→−∞Since the question of the convergence of an improper integral is answered in thesame way for both integrals over an infinite interval and functions unbounded nearone of the endpoints of a finite interval of integration, we shall study both of thesecases together from now on, introducing the following basic definition.Definition 3 Let [a, ω[ be a finite or infinite interval and x → f (x) a function defined on that interval and integrable over every closed interval [a, b] ⊂ [a, ω[.
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