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Thus for an atomic reactorEq. (5.147) is slightly altered:βN (t) + n.(5.148)N (t) = α −rThis equation can be solved by the same device as Eq. (5.147), sinceβN (t)1(α−β/r)N (t)+n is the derivative of the function α−β/r ln[(α − r )N (t) + n] ifα−βr= 0. Consequently the solution of Eq. (5.148) has the form!N(t) =N0 e(α−β/r)t −n(α−β/r)t ]α−β/r [1 − eN0 + ntif α −if α −βrβf= 0,= 0.It can be seen from this solution that if α − βr > 0 (supercritical mass), an explosion occurs. If the mass is pre-critical, however, that is, α − βr < 0, we shall verysoon haven.N (t) ≈ βr −αThus, if the mass of radioactive substance is maintained in a pre-critical state butclose to critical, then independently of the power of the additional neutron source,that is, independently of n, one can obtain higher values of N (t) and consequentlygreater power from the reactor.

Keeping the process in the pre-critical zone is adelicate matter and is achieved by a rather complicated automatic control system.5.6.4 Falling Bodies in the AtmosphereWe are now interested in the velocity v(t) of a body falling to Earth under theinfluence of gravity.2945Differential CalculusIf there were no air resistance, the relationv̇(t) = g,(5.149)would hold for fall from relatively low altitudes. This law follows from Newton’ssecond law ma = F and the law of universal gravitation, by virtue of which forh $ R (where R is the radius of the Earth)F (t) = GMmMm≈ G 2 = gm.(R + h(t))2RA body moving in the atmosphere experiences a resistance that depends on thevelocity of the motion, and as a result, the velocity of free fall for a heavy body doesnot grow indefinitely, but stabilizes at a certain level.

For instance, during skydivingor parachuting, if the parachute of a skydiver does not open for a long time, the speedof a midsized skydiver in the lower strata of the atmosphere stabilizes at 50–60 m/s.We shall assume that under these conditions (that is, for this body and a speedrange from 0 to 60), the resistance force is proportional to the velocity of the bodymoving in the air. By equating the acting forces on the body, we arrive at the following equation, which must satisfy the velocity of the free-falling body in the atmosphere:mv̇(t) = mg − αv.Dividing this equation by m and denotingαm(5.150)by β, we finally obtainv̇(t) = −βv + g.(5.148 )We have now arrived at an equation that differs from Eq. (5.148) only in notation.We remark that if we set −βv(t) + g = f (t), then, since f (t) = −βv (t), one canobtain from (5.148 ) the equivalent equationf (t) = −βf (t),which is the same as Eq. (5.143) or (5.145) except for notation.

Thus we have onceagain arrived at an equation whose solution is the exponential functionf (t) = f (0)e−βt .It follows from this that the solution of Eq. (5.148 ) has the form11 −βte ,v(t) = g + v0 −ββand the solution of the basic equation (5.150) has the formmmv(t) = g + v0 − g e−(α/m)t ,ααwhere v0 = v(0) is the initial vertical velocity of the body.(5.151)5.6 Examples of Differential Calculus in Natural Science295It can be seen from (5.151) that for α > 0 a body falling in the atmosphere reachesa steady state at which v(t) ≈ mα g. Thus, in contrast to fall in airless space, thevelocity of descent in the atmosphere depends not only on the shape of the body, butalso on its mass. As α → 0, the right-hand side of (5.151) tends to v0 + gt, that is,to the solution of Eq. (5.149) obtained from (5.150) when α = 0.Using formula (5.151), one can get an idea of how quickly the limiting velocityof fall in the atmosphere is reached.For example, if a parachute is designed to that a person of average size will fallwith a velocity of the order 10 meters per second when the parachute is open, then,if the parachute opens after a free fall during which a velocity of approximately50 meters per second has been attained, the person will have a velocity of about12 meters per second three seconds after the parachute opens.mIndeed, from the data just given and relation (5.151) we find mα g ≈ 10, α ≈ 1,v0 = 50 m/s, so that relation (5.151) assumes the formv(t) = 10 + 40e−t .Since e3 ≈ 20, for t = 3, we obtain v ≈ 12 m/s.Note that at the opening of the parachute, which is significantly bigger on thefront surface compared with the body of the skydiver, the force of resistance isalready proportional to the square of the speed.

This means that after the opening ofthe parachute, the slowdown will be quicker than that obtained from the calculationwith formula (5.151), corresponding to Eq. (5.150).It is useful to rewrite and solve the equation of type (5.150) with this new assumption of the quadratic dependence of the resistance force on the speed of movementand to make the necessary correction to the result obtained in the first calculation.325.6.5 The Number e and the Function exp x RevisitedBy examples we have verified (see also Problems 3 and 4 at the end of this section)that a number of natural phenomena can be described from the mathematical pointof view by the same differential equation, namelyf (x) = αf (x),(5.152)32 It should be noted that here, like in many others studied examples of problems of natural science,we illustrate the calculation method.

In the calculation are included some original data (empiricalconstants, various assumptions) on which depends the level of concordance of the calculations withthe reality. This data are not given by mathematics but by a corresponding science. For example,in order to obtain data on the aerodynamic properties of the bodies, in particular the nature of thechange in the resistance force occurring during their movement in the atmosphere, it is necessaryoften to perform a long series of experiments in wind tunnels. Mathematics helps to calculate oneor another mathematical model of the phenomenon, it often provides a ready calculation procedure,but the correctness of the initial data and the adequacy of the model is the work of another relevantscience (biology, chemistry, physics, .

. . ).2965Differential Calculuswhose solution f (x) is uniquely determined when the “initial condition” f (0) isspecified. Thenf (x) = f (0)eαx .We introduced the number e and the function ex = exp x in a rather formal wayearlier, assuring the reader that e really was an important number and exp x reallywas an important function. It is now clear that even if we had not introduced thisfunction earlier, it would certainly have been necessary to introduce it as the solutionof the important, though very simple equation (5.152). More precisely, it would havesufficed to introduce the function that is the solution of Eq. (5.152) for some specificvalue of α, for example, α = 1; for the general equation (5.152) can be reduced tothis case by changing to a new variable t connected with x by the relation x = αt(α = 0).Indeed, we then have t= F (t),f (x) = fαdf (x)=dxdF (t)dtdxdt= αF (t)and instead of the equation f (x) = αf (x) we then have αF (t) = αF (t), or F (t) =F (t).Thus, let us consider the equationf (x) = f (x)(5.153)and denote the solution of this equation satisfying f (0) = 1 by exp x.Let us check to see whether this definition agrees with our previous definition ofexp x.Let us try to calculate the value of f (x) starting from the relation f (0) = 1 andthe assumption that f satisfies (5.153).

Since f is differentiable, it is continuous.But then Eq. (5.153) implies that f (x) is also continuous. Moreover, it followsfrom (5.153) that f also has a second derivative f (x) = f (x), and in general thatf is infinitely differentiable. Since the rate of variation f (x) of the function f (x)is continuous, the function f changes very little over a small interval h of variationof its argument. Therefore f (x0 + h) = f (x0 ) + f (ξ )h ≈ f (x0 ) + f (x0 )h.

Let ususe this approximate formula and traverse the interval from 0 to x in small steps ofsize h = xn , where n ∈ N. If x0 = 0 and xk+1 = xk + h, we should havef (xk+1 ) ≈ f (xk ) + f (xk )h.Taking account of (5.153) and the condition f (0) = 1, we havef (x) = f (xn ) ≈ f (xn−1 ) + f (xn−1 )h == f (xn−1 )(1 + h) ≈ f (xn−2 ) + f (xn−2 )h (1 + h) =5.6 Examples of Differential Calculus in Natural Science297Fig.

5.34= f (xn−2 )(1 + h)2 ≈ · · · ≈ f (x0 )(1 + h)n =x nn.= f (0)(1 + h) = 1 +nIt seems natural (and this can be proved) that the smaller the step h = xn , the closerthe approximation in the formula f (x) ≈ (1 + xn )n .Thus we arrive at the conclusion thatx nf (x) = lim 1 +.n→∞nIn particular, if we denote the quantity f (1) = limn→∞ (1 + n1 )n by e and showthat e = 1, we shall have obtained)*xx nf (x) = lim 1 += lim (1 + t)x/t = lim (1 + t)1/t = ex ,(5.154)n→∞t→0t→0nsince we know that uα → v α if u → v.This method of solving Eq. (5.153) numerically, which enabled us to obtain formula (5.154), was proposed by Euler long ago, and is called Euler’s polygonalmethod.

This name is connected with the fact that the computations carried outin it have a geometric interpretation as the replacement of the solution f (x) of theequation (or rather its graph) by an approximating graph consisting of a broken linewhose links on the corresponding closed intervals [xk , xk+1 ] (k = 0, .

. . , n − 1) aregiven by the equations y = f (xk ) + f (xk )(x − xk ) (see Fig. 5.34).We have alsothe definition of the function exp x as the sum of the encountered1 npower series ∞x.Thisdefinition can also be reached from Eq. (5.153) byn=0 n!using the following frequently-applied device, called the method of undeterminedcoefficients. We seek a solution of Eq. (5.153) as the sum of a power seriesf (x) = c0 + c1 x + · · · + cn x n + · · · ,(5.155)whose coefficients are to be determined.(n)As we have seen (Theorem 1 of Sect. 5.5) Eq. (5.155) implies that cn = f n!(0)But, by (5.153), f (0) = f (0) = · · · = f (n) (0) = · · · , and since f (0) = 1, we have2985cn =1n! ,Differential Calculusthat is, if the solution has the form (5.155) and f (0) = 1, then necessarilyf (x) = 1 +111x + x2 + · · · + xn + · · · .1!2!n!We could have verified independently that the function defined by this series isindeed differentiable (and not only at x = 0) and that it satisfies Eq.

(5.153) and theinitial condition f (0) = 1. However, we shall not linger over this point, since ourpurpose was only to find out whether the introduction of the exponential functionas the solution of Eq. (5.153) with the initial condition f (0) = 1 was in agreementwith what we had previously meant by the function exp x.We remark that Eq. (5.153) could have been studied in the complex plane, thatis, we could have regarded x as an arbitrary complex number. When this is done,the reasoning we have carried out remains valid, although some of the geometricintuitiveness of Euler’s method may be lost.Thus it is natural to expect that the functionez = 1 +111z + z2 + · · · + zn + · · ·1!2!n!is the unique solution of the equationf (z) = f (z)satisfying the condition f (0) = 1.5.6.6 OscillationsIf a body suspended from a spring is displaced from its equilibrium position, forexample by lifting it and then dropping it, it will oscillate about its equilibriumposition.

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