Now, if we rotate this profile about the y-axis, we generate a solid of revolution that looks something like a trumpet, a champagne glass, or a fancy ice cream cone. If we put two of these cones end to end, we get a solid shape called a pseudosphere. Its volume, V = 4πa3/3, and surface area, A = 4πa2, are the same as those of a sphere of the same radius, a. In addition, a mathematician would point out that just as a sphere possesses constant positive curvature (and hence serves as a model for non-Euclidean elliptic geometry), the pseudosphere has constant negative curvature (and so serves as the basis for non-Euclidean hyperbolic geometry).
Our two turtles are probably much more interested in the race. We shall bet on turtle B.
14
More Great Turtle Races: Logarithmic Spirals
More turtle racing: This time we shall use several or many turtles and instruct all of them to move at the same speed. So it is not really racing; as we shall see, it is more a matter of symmetrical pursuit.
We start with a configuration involving two turtles. Turtle A is located at the west end of a straight line of length a, facing east. Turtle B is at the east end facing west. When the starting gun is fired, each turtle heads directly toward the other. Obviously, the two turtles will collide at the midpoint of the line and each will have covered a distance a/2.
Since that was not too difficult we move on to the three-turtle problem. This is not as easy. As shown in figure 14.1, initially the turtles are at the corners of an equilateral triangle with side length a. Turtle A is facing turtle B who is facing turtle C who is facing turtle A. Again, as the race commences, each turtle always heads directly toward his target.
The salient features of our analysis are that the three turtles will always be separated by an angle α = 120° = 2π/3 radians and will always be equidistant from the origin, r. We utilize the positions and paths of A and B for our analysis; by symmetry, the behavior of C will be the same.
FIG. 14.1
Definition sketch for the three-turtle problem.
Expressing the coordinates (x, y) of point P and (m, n) of point Q in terms of the angles θA and θB, we establish the relationships
The criterion to “…always head directly toward…” is
Substituting (14.1) into (14.2) yields the differential equation
This equation is easily solved with the substitution y = wx. Carrying out some algebra and simple integrations, we obtain
where k is an integration constant. This expression can be written in polar coordinate form by substituting and θ = arctan(y/x):
To determine the value of the constant K, we employ the initial condition for either A, B, or C. Selecting the last, we have when θ = 90° = π/2 radians. This gives the answer:
or in final form
As we know, this is the equation of the logarithmic or equiangular spiral. In chapter 9, we learned that this beautiful curve makes its appearance in many places in nature. In hydrodynamics, for example, the logarithmic spiral is produced when a vortex flow is combined with a source or sink flow—like the spiral you get when you drain the bathtub. Numerous properties of this interesting curve are discussed by Lockwood (1961), by Steinhaus (1969), and by Thompson (1961).
Back to the three turtles. Their paths are shown in figure 14.2. Here's a good question: How far does each turtle go before it collides with the others at the center of the triangle? To answer this question we write the expression
where ds is an incremental length along the turtles' path and dr and dθ are incremental changes in r and θ. From this relationship we obtain the expression
FIG. 14.2
Paths of the turtles in the three-turtle problem.
Using (14.6) in (14.8), we find the length of the travel path, S = (2/3)a. Thus, each turtle travels a distance equal to two-thirds of the length of the triangle's side. If a = 1.0 m = 100 cm and υ = 1.0 cm/s, the turtles will meet at the center of the triangle in T = S/υ = 66.7 s.
The obvious next problem concerns four turtles and a square. In this case, it is not difficult to show that the turtle paths are described by the equation
FIG. 14.3
Paths of the turtles in the four-turtle problem.
and the length of the path is S = a. Collision time is 100 seconds. This case is shown in figure 14.3.
Let us move on to five turtles and a pentagon, six turtles and a hexagon, eight turtles and an octagon, and n turtles and an n-gon. The general case solution is
in which α = 2π/n and n is the number of sides of the n-gon; θ0 is a constant whose value is determined by the geometry of the initial condition.
The length of the path for the general case is
where in each instance a is the length of a side of the n-gon.
As the number of sides increases, the central angle, α = 2π/n, decreases. Utilizing the approximations sin α = α and cos α = 1 – α2/2, equation (14.10) becomes
and (14.11) reduces to
These approximations are valid for values of n larger than about 10.
As we increase the number of sides, our n-gon begins to look like a circle. So for very large n the quantity na is the circumference of a circle of radius r0. That is, C = na = 2πr0. In this case, setting θ0 = 0, since now it does not matter where we start on the circle, equation (14.12) becomes
and from (14.13) the total length of a path is
Are you ready for the great experiment? We applied for and received a government grant to carry out an experiment featuring 360 small turtles spaced at 1° intervals around the circumference of a circle of radius r0 = 10 m. Each turtle has instructions to head directly toward the turtle immediately ahead and to move at a speed υ = 1.0 cm/s.
Backing up briefly, we can establish that one of the properties of the logarithmic spiral is that it always intersects the radius vector r at a constant angle β. In our turtle problem we determine that this intersect angle β = α/2, where α = 2π/n is the central angle. For example, if n = 3, then α = 2π/3 and β = π/3 radians = 60°. If n = 4, then α = π/2 and β = π/4 radians = 45°.
In our great experiment, n = 360, and so α = π/180 and β = π/360 radians = 0.5°. Consequently, in our 360-gon, which is almost but not quite a circle, each turtle heads 0.5° inboard from the tangent to the equivalent circle.
Since we have a large value of n, equations (14.14) and (14.15) can be utilized. From (14.14) we see that r = r0 = 10 m when θ = 0. If θ = 2π radians = 360° (i.e., one complete revolution of marching turtles around the circle), we determine that r = 9.466 m. Thus, the turtle circle has shrunk by about 53.4 centimeters after one revolution. In this first circuit they traveled approximately 61.6 m; since their speed is υ = 1.0 cm/s, it took them 6,120 seconds or around 1 hour and 42 minutes for the first lap.
The second lap takes a bit less time since the circle is slightly smaller. The third lap is shorter still, and so on. From equation (14.15) we determine that the total distance traveled by each turtle to get to the center is S = 1,146 m and it takes 31 hours and 50 minutes to get there. We prefer to ignore what happens when 360 tired turtles all arrive at r = 0 at the same instant—a pretty nasty traffic pileup.
There is one feature in common in all of the preceding problems: the turtles are pursuing one another on plane surfaces. An interesting generalization of our turtle problem is given by Aravind (1994), in which the plane surface is replaced by a spherical surface (constant positive curvature) and by a hyperbolic (or pseudosphere) surface (constant negative curvature).
A final word regarding a recommended reference. Virtually all of the topics covered in our book are examined in a clear and concise way in the comprehensive volume by Gellert et al. (1977). If you have a serious interest in the study of mathematics, this excellent reference book should be in your personal library.
15
How Many People Have Ever Lived?
In 1990 the population of the world was approximately 5.32 billion people. This is an increase of 844 million over the 1980 population, which was an increase of
755 million over the 1970 population, which was an increase of 671 million over the 1960 population,…, and so on.
Interesting, but perhaps we are going the wrong way in time. Who cares about the population of the past—the demography of yesterday? We want to know about the population of the future—the shape of things to come. Two comments: First, the best way to make forecasts, at least for the near future, is to base projections on past and present information. Second, in the next chapter we shall indeed examine the topic of the world's population in the years to come.
So let us start by taking a look at the past. We begin with the following simple relationship:
in which N is the magnitude of a particular growing quantity at time t and a is a growth coefficient or an interest rate. This equation indicates that the rate at which a quantity is growing in magnitude is assumed to be directly proportional to the magnitude at that instant.
The solution to equation (15.1) is
where e, of course, is the base of natural logarithms and N0 is the magnitude of N at time t = 0. This is the equation for so-called exponential growth.
Without getting too complicated with our mathematics, (15.1) can be written in a more generalized form:
which is the same as (15.1) except that we have replaced a with the quantity a(N). This symbol says simply that the growth coefficient a is no longer constant but instead depends on the value of N. Let us be more specific about the form of a(N). We shall say that the growth coefficient or interest rate is directly proportional to N; that is, a(N) = aN/N0. This gives
The quantity N0 is brought into the analysis at this point in order to keep the dimensions of the equation as simple as possible.
We note in (15.4) that the rate at which the particular quantity (e.g., population) is growing, dN/dt, is proportional not to the first power of N as in exponential growth, but to the second power, that is, the growth is like N squared. This kind of growth has been termed coalition growth by von Foerster et al. (1960).
As we shall see, coalition growth is truly “explosive” growth. It involves an increase to an infinite value of N in a finite time and it features a continuously shrinking “doubling time.” It incorporates, for example, the unbelievably good deal you got at your local financial house: your probably insane banker agrees to make your savings interest rate directly proportional to the amount you have on deposit (e.g., if you have $1000 in your account you get 4% interest; if you have $2000 you get 8%; $3000 gives 12%).
TABLE 15.1
Population of the world, 1650 to 1990
The solution to equation (15.4) is the amazingly simple expression
where N0 is the value of N when t = 0. This is the equation for coalition growth. Because of its mathematical form it is also called hyperbolic growth.
We go back to the population of the world. Information concerning the population is given in table 15.1 for the period from 1650 to 1990. We select 1650 as the year for which t = 0. The data shown in the table are plotted in figure 15.1. For the moment, disregard the solid curve shown in the figure.
It is observed that for a long time the world's population increased very slowly. Indeed, not until the beginning of the twentieth century did the population begin to rise sharply, and only after about 1950 did the population show really alarming increase.
FIG. 15.1
Population of the world, 1650 to 1990.
We suspect that this growth of population may have been more rapid than exponential growth. Hence, for starters, we assume it may be hyperbolic (or coalition) growth. To confirm this we need to see how the data “fit” the mathematical model. It is easy to rewrite equation (15.5) in the form
In the language of analytic geometry, this equation has the linear form where k0 and k1 are constants. Accordingly, if 1/N is plotted against t we should get a straight line if our assumption of hyperbolic growth is correct. The constants k0 and k1 provide the values of and a.
FIG. 15.2
Population of the world. Plot to determine numerical values of a and N0.
Such a plot is shown in figure 15.2. The fit of the data is remarkably good; the correlation coefficient is 0.9990. From a least squares analysis we obtain a = 0.002675 per year and Substitution of these numbers into equations (15.5) and (15.6) produces the solid lines shown in figures 15.1 and 15.2.
Some questions: How many people were there in the world when Columbus discovered America? Taking t = 1492 – 1650 = –158 and substituting into (15.5) (with a = 0.002675 and N0 = 0.525 × 109) gives N = 0.369 × 109 or 369 million. How about that very eventful year 1066? Answer: 205 million. What was the world's population in the year 4000 B.C.? From equation (15.5) we get 33 million.
It is risky business to make population projections very far into the future based on some kind of formula or equation. It is equally risky to estimate or try to calculate past populations. Even so, our almost pitifully simple equation (15.5) gives answers about previous populations that agree amazingly well with values obtained by anthropologists and demographers using entirely different methodologies.
For example, Deevey (1960) estimates that the world's population was 125,000 a million years ago; equation (15.5) gives 196,000. Westing (1981) cites a population of 3,000,000 in the year 40,000 B.C.; we get 4,670,000. Austin and Brewer (1971) indicate a population of 100 million for the year 0 A.D.; we compute 97 million; and so on.
Our main question is: how many people have ever lived on earth? In 1990 the world's population was 5.32 billion. Is this a large or small percentage of the total number who have ever lived?
This question is answered by returning to equation (15.5). We integrate this equation to determine the total number of person-years, M, of all who have ever lived. That is,
where we take t0 = –1,000,000 as the date from which we start counting the number of people who ever lived.
For example, how many people lived during the period t0 = –1,000,000 and t = –500,000? Substituting numbers into (15.7) gives M = 135.6 × 109 person-years. Assuming a life span of duration yr gives persons. In other words, about 5.4 billion people lived and died during the 500,000 years between 1,000,000 B.C. and 500,000 B.C.
Some anthropologists and historians indicate that the year 4000 B.C. marks the “dawn of civilization.” If so, how many people lived in the predawn period, 1,000,000 B.C. to 4000 B.C.? From (15.7) we determine M = 1,003 × 109 person-years for that period and, with yr as the average life span, obtain P = 40.1 billion people.
A listing of the cumulative number of people who have ever lived, commencing with the year 1,000,000 B.C., is shown in table 15.2. To follow the scheme of other studies of the subject, the average life span is taken as yr.
As they say, the bottom line is the answer. The last entry of table 15.2 indicates that through the year 1990, about 80 billion people have lived on earth. Some observations:
The world's population in 1990 was 5.32 billion. This is about 6.5% of the number of people who have ever lived. Said another way, about one person of fifteen who have ever lived is currently living.
TABLE 15.2
About half of the 80 billion people lived between 1,000,000 B.C. and 4000 B.C.; the other half since then.
During the nearly 2,000-year period from the year 0 A.D. to 1990, about 32 billion people have lived. This is 40% of the total who have ever lived.
Our final answer, P = 80 billion, agrees fairly well with results obtained by others; they range from 50 billion to 110 billion. Suggested references are Deevey (1960), Goldberg (1983), Keyfitz (1966), and Westing (1981).
We conclude our chapter with a word about the doubling time. By definition, this is simply the time required for a growing quantity to double in magnitude. For example, a quantity may be following an exponential growth relationship. If so, from equation (15.2) it is easy to establish that the doubling time, t2 is
For example, if the population growth rate of a certain country is a = 3.5% per year, then the doubling time is 20 years. This type of exponent
ial population growth is sometimes called Malthusian growth. We note that one of the features of such growth is a constant doubling time.
In contrast, suppose that a quantity is growing according to a hyperbolic growth relationship. From equation (15.5) we obtain the following expression for the doubling time:
In this case, the doubling time, t2, is not constant, as in exponential growth, but instead changes with time. In the year 1000 A.D., for example, t = 1000 – 1650 = –650, and so, from equation (15.9), the doubling time t2 = 512 yr. In 1650, t2 = 187 yr; in 1950, t2 = 37 yr; in 1990, t2 = 17 yr.
This is getting to be rather scary. Does this mean that in the year 1990 + 17 = 2007, there will be 2 × 5.52 = 10.64 billion people in the world? And then it doubles again nine years after that? We shall examine these alarming questions in our next chapter, “The Great Explosion of 2023.”
Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics (Princeton Paperbacks) Page 11