16
The Great Explosion of 2023
Our previous chapter concluded on a note depicting a very alarming situation: (1) a 1990 world population of over five billion people and a population doubling time of around 17 years and (2) a projection to a year-2000 population of over eight billion and a doubling time of around 12 years. Clearly, it is urgent that something be done about this serious crisis of rampant growth of world population.
However, we need to look at the data more closely. Before we try to resolve this incredible crisis, perhaps it is necessary to examine how serious it really is. So we return to the equation that describes so-called coalition or hyperbolic growth:
where N is the population at time t and a is the growth coefficient; N0 is the population at time t = 0. We obtained the values a = 0.002675 per year and based on t = 0 in the year 1650.
Some populations are listed in table 16.1, computed from equation (16.1), for various years commencing with 1650 and ending with 2023. Magnitudes of observed populations are also shown and the differences between computed and observed populations are indicated. Several important facts can be seen in the table:
TABLE 16.1
For the years 1650 through 1980 there is surprisingly close agreement between computed and observed populations. This is apparent in the right-hand column of the table.
For 1990, however, there is a difference of +0.488. That is, the computed population is nearly 500 million more than the observed. This is a very interesting and important point. We shall come back to it later.
For the years 2000 and beyond, computed populations rise to ridiculous values: nearly 52 billion in 2020 and over 245 billion on, say, January 1,2023.
Okay, here it comes. Doomsday is established by setting the denominator of equation (16.1) equal to zero. This gives te = 1/a = 373.832, where te stands for explosion time. Since t = 0 corresponds to 1650, then explosion year is 2023.832.
So 2023 will be the year of the great explosion. Both Keyfitz (1968) and von Foerster et al. (1960) arrive at about the same conclusion. Specifically, on November 1, 2023 there will be an infinite number of people in the world and the doubling time will have shrunk to zero.
Interesting. Now back to reality. Of course, the world's population is going to continue to increase in the years to come. However, obviously the population will not become infinite. Inevitably and increasingly there will be forces that will slow down, terminate, or even reverse the growth rate.
These forces imposing limitations to population growth are comprised of agricultural, demographic, economic, environmental, scientific, sociological, and technological components. Collectively, they specify and emphasize the finite capacity of the world for maintenance of a human population.
Apparently, population crowding effects began to appear, on a worldwide basis, during the decade of the 1980s. We noted the very important fact that the computed population for 1990 was almost 500 million more than the observed population. It is remarkable that after a million years of human population increases, crowding effects and limitations-to-growth factors made their first appearances only quite recently.
Long ago it was found that an appropriate way to slow down exponential or Malthusian types of growth phenomena was to attach a crowding or finite-resources term to the growth equation. This turned a skyrocketing exponential growth into a stabilized logistic growth. It seems logical to modify our hyperbolic growth equation in a similar fashion.
To follow the approach of Austin and Brewer (1971), we arbitrarily attach a growth limitation term to the equation for unrestrained explosive hyperbolic growth. That is,
where the quantity in parentheses requires that the growth rate, dN/dt, become zero when N = N*. This quantity, N*, is the carrying capacity, that is, the maximum population the earth is able to sustain. The solution to equation (16.2) is
which we will call the modified coalition growth equation. This expression is not as formidable as it might appear; indeed, it is quite easy to handle. We note that if N* approaches infinity, that is, resources are sufficiently large to support an infinite population, equation (16.3) reduces to (16.1) as expected.
Unfortunately, (16.3) is in the form that says “t is a function of N” and not “N is a function of t” as we would prefer. This is not a problem; we still know how the world's population N increases with time t.
At this point we shift the time origin, t = 0, from the year 1650 to a more recent date: 1980. Also, with computations involving populations and growth rates through 1990, a value of N* is obtained. These calculations provide the following: a = 0.0303, , and with t = 0 corresponding to 1980.
Populations calculated from equation (16.3) are listed in table 16.2. We compute a 1995 world population of about 5.73 billion and a year 2000 population of around 6.22 billion. These values are in good agreement with those given by the United Nations (1993).
A plot of equation (16.3) is shown in figure 16.1, in which we note several things:
In the years to come, the world's population certainly continues to increase. However, because of the stabilizing effect of the crowding term in (16.2), the explosive increase of uncontrolled hyperbolic growth is avoided.
As shown in figure 16.1, the rate of population increase reaches a maximum value in 2004 when the population is 6.667 billion people. The corresponding annual increase is 101 million people. This maximum occurs when the growth curve passes through the so-called inflection point, as shown in the figure. After that, the rate begins to decline and eventually becomes zero.
TABLE 16.2
Population of the world, N, in billions Computed from modified coalition growth equation.
N Year
3.0 1954.8
4.0 1973.6
5.0 1086.0
6.0 1997.7
7.0 2007.7
8.0 2018.2
9.0 2032.2
9.5 2044.0
9.9 2068.9
FIG. 16.1
Population of the world with projections leading to (a) infinite population in 2023 and (b) stabilized population of /V* = 10.0 billion.
One thing needs to be emphasized. Projections of future populations are not made on the basis of simply fitting curves to observed data. Demographic analyses of trends of the world's population are done on a country-to-country basis and take into account such things as age distribution, life expectancy, migration, mortality, and fertility of a particular nation's population. World trends are established as the aggregation of country and regional trends. For those interested in these topics, a good place to start is the interesting article by Coale (1974). More advanced treatments of the subject are provided by Keyfitz (1968), by Pollard (1973), and by Song and Yu (1988).
Complementing precise demographic analyses are mathematical frameworks such as our modified coalition growth model. These models can answer a lot of questions.
PROBLEM 1. The growth rate, dN/dt, is a maximum when its derivative, d2N/dt2, is equal to zero. Utilizing equations (16.2) and (16.3), show that the coordinates and slope of the growth curve at the inflection point are
PROBLEM 2. In the previous chapter we established that during the period from 1,000,000 B.C. to 1950, the number of person-years, M, was about 1,867 billion. By exact, numerical, or graphical integration of equation (16.3), confirm that the value of M between 1950 and 2050 will be about 625 billion.
We have taken a fairly good look at the world's population. Here is a point on which to conclude. From equation (16.2) we determine that during 1995 the world's population increased by about 95 million people. If these 95 million were to comprise an entirely new country, where would it rank, in 1995 population, among the world's nations? The answer is number 11, after China, India, United States, Indonesia, Brazil, Russia, Pakistan, Japan, Bangladesh, and Nigeria. Just think: an enormous new country every year!
17
How to Make Fairly Nice Valentines
Have you noticed how dramatically the cost of greeting cards has risen duri
ng recent decades? Do you realize that each year we pay trillions of dollars for birthday cards and holiday cards? Many people find this difficult to believe.
Well, we can begin reducing our budgets for greeting cards by creating our own, at least some of them. The problem is that unless one has considerable artistic talent, it is impossible to trim expenses by making one's own fantastically beautiful Christmas cards, Labor Day cards, and Groundhog Day cards. Likewise, homemade Easter cards and Thanksgiving cards would probably look pathetic. By the same token, it is futile to try to do much about constructing your own birthday cards. What could really be done beyond slapping on some important anniversary numbers like 5, 18, 39, and 65?
However—here comes the good news—as mathematicians we can save substantial sums of money by fabricating our own greeting cards for those holidays with which simple mathematical shapes are associated. For example, for Halloween there is the symbol of the pumpkin, which is something like an oblate ellipsoid whose equation is . We could draw ellipses on stiff paper, paint them orange, and mail them out by the dozens.
FIG. 17.1
Definition sketch for the valentine.
Or what about Saint Patrick's Day and the symbol of the shamrock? This is a three-leaf clover and looks like a trifolium, whose equation is . We could compute the curve, plot it on many pieces of cardboard, color them green, and send out as many as we like.
Best of all, of course, would be Valentine's Day. This is the big one! We could bring about industrial strength savings of cash by manufacturing our own valentines.
As we shall see shortly, there are numerous mathematical equations we could employ to construct valentines—some easy and some not. Furthermore, unlike pumpkins and shamrocks, a self-made mathematically perfect valentine to your loved one could result in substantial rewards—on a par with those ensuing from gifts of red roses and chocolates.
An Easy Valentine: A Circle and a Parabola
Before we start the mathematical construction of valentines, it is necessary to decide the overall proportions and dimensions we want. A definition sketch is shown in figure 17.1.
Arbitrarily, we select the proportion B/H = 0.8. If you prefer a square valentine, then take B/H = 1.0. For a fat valentine, take B/H = 1.5, and for a skinny one use B/H = 0.5. We will not specify the “plunge” distance, h; this dimension will be an outcome of the mathematics. Finally, for our computations we shall use H = 10 cm and B = 8 cm. This will be a nice fit for a small envelope.
FIG. 17.2
Construction of a valentine with a circle and a parabola.
Our first, and mathematically easiest, valentine will be composed of half of a circle and half of a parabola. In figure 17.2, the valentine has been cut in half and laid on its side. The abscissa, x, coincides with the axis of the valentine and the ordinate, y, is perpendicular to x. The origin of this Cartesian or rectangular coordinate system is shown in the figure.
The Circle
We start with the circle. Since B = 8 cm then B/2 = 4 cm and so the radius of the circle must be r0 = 2 cm.
Next, we need to call on analytic geometry. The equation of a circle of radius r0 centered at x = x0 and y = y0 is
In our problem, x0 = 0, y0 = 2, and r0 = 2. Substituting these numbers into (17.1) and solving for y gives
We now substitute values of x into this equation and compute the corresponding values of y. For example, if x = +1 or –1, we get .
Computations like this provide the (x, y) coordinates of points of the circle. Through a number of such points we pass a smooth curve, and there is our circle. (Of course, for this curve you might prefer to use a compass.) The half of the circle we will use for our valentine is shown as the solid curve in figure 17.2; the unused other half is dashed.
The Parabola
Next we analyze the parabolic portion. The general equation of a parabola is
where k0, kv and k2 are constants. Now the height of our valentine is H = 10 cm and the top 2 cm consists of the semicircle; therefore the bottom parabolic portion must be 8 cm. So, with reference to figure 17.2, we need a parabola that passes through the points x = –8, y = 0 (and, by symmetry, x = +8, y = 0), and also x = 0, y = 4. Using equation (17.3), a bit of algebra yields k0 = 4, k1 = 0, and k2 = –1/16. So the equation of the parabola is
From this expression, we plot the parabola. For example, if , (17.4) gives y = 3 cm.
The portion of the parabola we need for our valentine is shown as the solid curve in figure 17.2; part of the unneeded portion is dashed. The plunge distance for this valentine is h = 2 cm.
Only one more step is needed to complete this first design. We simply connect a mirror image to our semivalentine and we are ready for the red paint.
How much red paint do we need? This is a lowbrow way of asking the question: what is the area of our semivalentine? Well, the area of the semicircle is So, with r0 = 2 cm, .
What is the area of the semiparabola? It is fairly easy to show that the area of a parabola is equal to two-thirds of the area of the circumscribing rectangle. Accordingly, we get Ap = (2/3)(8)(4) = 21.33 cm2. Consequently, the area of the semi-valentine is Acp = 27.61 cm2; the area of the entire valentine is double this amount, that is, A = 55.22 cm2. Knowing how many valentines you need to make—that is, the number of sweethearts you have—you can now easily compute how much red paint, stiff paper, and envelopes you must buy.
PROBLEM Now suppose you want to put some fancy tassel or ribbon along the edge of your valentine. To do this, you need to know the length of the perimeter of the valentine.
For the semivalentine, the length of the semicircular portion is easy to calculate; it is simply . However, the length of the semiparabola is a bit more difficult to determine. Using the methods of elementary calculus, it is possible to show that the length of the parabola can be computed from the equation
which in our problem becomes, using (17.1),
Consequently, the entire length of the perimeter of the valentine is S = 30.92 cm.
Perhaps this seems like a rather trivial problem. However, structural engineers face a similar computation involving the parabola when computing the length of the cables of large suspension bridges.
A More Difficult Valentine: A Logarithmic Spiral
In dealing with entirely different topics in other chapters, we ran across the famous mathematical curve called the logarithmic spiral. This curve, also known as the equiangular spiral, has been around for a long time. It was first studied in 1638 by the French mathematician René Descartes (1594–1650).
FIG. 17.3
Construction of a valentine with a logarithmic spiral.
The logarithmic spiral describes the shape of the beautiful Nautilus sea shell and the pattern formation in many kinds of plants and flowers. It is closely related to the remarkable golden number, , and to the famous Fibonacci sequence, 1, 1, 2, 3, 5, 8,…
The equation of the logarithmic spiral is
in which r0 and a are constants. This equation is plotted in figure 17.3.
In our first problem, involving the circle and the parabola, the analysis was carried out in a rectangular (x, y) coordinate system. We have now moved, with equation (17.7) to a polar coordinate system. Incidentally, we can easily shift from one system to the other because and
In these expressions, r is the distance from the origin, (0, 0), to any point on the curve and θ is the angle between the x-axis and the r-line; r0 is the value of r when θ = 0. A reminder: When using (17.7), be sure to express θ in radians, not degrees, (1 radian = 360°/2π = 57.296°.)
Half of our logarithmic spiral valentine is shown in figure 17.3. In order to make the overall dimensions of this valentine the same as the first one, that is, H = 10 cm and B = 8 cm, it is necessary to do some curve fitting. This operation is greatly simplified if we use some calculus.
PROBLEM 1. If the equation of a plane curve is expressed in polar coordinates, r = r(θ), then the slope of the curve at any point (r, θ) is given by
Using
(17.7) in (17.8), show that y is a maximum when tan θ = –(1/a) and that x is a maximum when tan θ = a. This information enables you to determine that the values r0 = 1.432 cm and a = 0.561 will assure that H = 10 cm and B = 8 cm. The plunge distance h = 0.22 cm.
The total area of the valentine is A = 60.25 cm2 and the entire perimeter is S = 28.26 cm.
PROBLEM 2. Confirm these numerical values for area and perimeter. You can calculate the area from the equation
and determine the arc length from the expression
A Much More Difficult Valentine:
A Cardioid and a Lemniscate
The analysis and design of this valentine are similar to our first one in which we connected a circle to a parabola. This time we are going to connect a cardioid to a lemniscate.
Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics (Princeton Paperbacks) Page 12