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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT 256 9 "cohen.mws" }{TEXT -1 87 " A Maple worksheet to e
xamine the results presented in the paper Cohen, J. E. ,1995, " }
{TEXT 257 57 "Population growth and the earth's human carrying capacit
y" }{TEXT -1 11 ", Science, " }{TEXT 258 3 "269" }{TEXT -1 145 ": 341-
348. Pay close attention to the sections enclosed between stars ***li
ke this*** ; these are problems that you will be expected to discuss!
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 594 "Cohe
n says on p. 341 that the annual rate of increase of the earth's popul
ation was 0.04% per year from AD 1 to 1650, rose to a peak of 2.1% p
er year around 1965-1970, and is now 1.6%. (You might want to compare
these assertions with your own analyses of the data from problem 5 on
ws1!) What are the consequences of these differences in rates? Using
the data from Fig 1. we can predict the earth's population if the gro
wth rate was 0.04%, 1.6%, or 2.1% over the period from AD 1 to the pre
sent. We use an exponential growth model with r the Malthusian parame
ter and P the population size." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "r
estart: with (DEtools): with(plots):" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 94 "Popgrowth:=diff(P(t), t)=r * P(t); # This is our equa
tion in symbolic form (r is a parameter)." }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 106 "We will run the model to
the year 2000. The initial population in the (fake) year AD 0 was 0.
252 billion." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "vars:=[ P(t) ]; do
main:= 0 .. 2000; init:=[ P(0)= 0.252 ];" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 300 "\nThe substitute command is very handy for making change
s in the values of a parameter. Here t is the independent variable, P
is the dependent variable, and r is treated as constant for any give
n run of the model. But we are free to adjust r and to examine the ef
fect of such a change on the model." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0
77 "diffeq:= subs(r = 0.0004, Popgrowth); # This is our equation with \+
r = 0.0004." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 116 "DEplot( dif
feq, vars, domain, [init], stepsize=20, title=`World Population in Bil
lions: r=0.0004` , arrows = LINE );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 48 "In this case we can also get a formula solution." }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "formula:= dsolve( \{diffeq, P(0) = \+
0.252\}, P(t) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "evalf( \+
subs( t = 1650, formula) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
33 "evalf(subs( t = 2000, formula) );" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 269 "Was a growth rate of 0.0
004 (0.04%) fast enough to reach the current (year 2000 for convenien
ce) population size of 2.6 billion? What if we used the current growt
h rate of 1.6% instead? Go back to where we substituted r = .0004 int
o the DE, and try r = .016 instead." }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 176 "# Use the mouse to skip to the next visible > if yo
u like. To open the optional section, just hit Return now, or click on
the little box; to close it, click on the little box." }}}{SECT 1
{PARA 258 "" 0 "" {TEXT -1 71 "Optional section for those comfortable \+
with piecewise defined functions" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 36 "restart: with(plots): with(DEtools):" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 426 "\nSInce none of the listed rates held for all of recen
t (AD) recorded history, let's try to see if we can use the figures Co
hen gave us in a more realistic way. Recall that r = .04% up to 1650
, then increases to r = 2.1% around 1965, and decreases to r = 1.6% \+
in the present. Let's plot r as a function of t. Can you see how we c
an get a formula for r at the times in between as well? (HInt: what if
you use style = LINE?)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117
"plot( [ [0, .0004], [1650, 0.0004], [1965, .021], [1995, .016] ], sty
le = POINT, symbolsize = 18, title = `r vs. t`);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 "Ok, did you figur
e it out? If not, take a look at the next line, and see if you can mak
e sense of it." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 134 "r:= piec
ewise( t < 1650, 0.0004, t < 1965, (.021-.0004)/(1965-1650) * (t - 165
0) + .0004,(.016-.021)/(1995-1965) * (t - 1965) + .021);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "formula:= dsolve( \{diff(P(t), t) =
r * P(t), P(0) = 0.252\}, P(t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 455 "For reasons that I do not underst
and, sometimes the above command works, and sometimes it doesn't. If \+
it works, it gives some spurious \"undefined\" answers, but for whatev
er it's worth, let's see what it gives for the population in 1995 and \+
2000. The following is obtained by cutting and pasting the relevant pa
rt of the above formula for P(t); I was unable to get Maple to do this
. Recall that population figures are given in billions ( 10^9 in the \+
US)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 150 "pop_1995:= evalf( \+
subs( \{_C1 = 0.2520, t = 1995\}, _C1*exp(3485000001/10000000000*t - 2
87202600103071/800000000000 - 1666666667/20000000000000*t^2)) );" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 151 "pop_2000:= evalf( subs( \{_
C1 = 0.2520, t = 2000\}, _C1*exp(3485000001/10000000000*t - 2872026001
03071/800000000000 - 1666666667/20000000000000*t^2)) );;" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "Of course
we can also get a graph by using DEplot (or by plotting our formula s
olution)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "Popgrowth:= di
ff(P(t), t) = r * P(t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 138
"DEplot(Popgrowth, [P(t)], 0 .. 2000, \{[0, 0.252]\}, stepsize=20, arr
ows = none, title=`World population in billions: r defined piecewise`)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 329 "How do these result
s compare to Cohen's Fig. 1, or with his reference in the text to the \+
current, 1995, world population as 5.7 billion? The axis scale is in \+
billions, as in Cohen's graph. Note, however, that his graph is on a
LOG scale and your graphs are on a linear scale. Lets take advantage
of our formula answer here. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 111 "logplot( rhs(formula), t = 0 .. 2000, axes=FRAME, title=`Wo
rld population in billions: r defined piecewise`);" }}}{EXCHG {PARA 0
"" 0 "" {TEXT -1 310 "\nIs it possible that separate growth rates, ove
r successive time periods, can give population predictions consistent \+
with the data of Cohen's Fig. 1? If we have not succeeded in doing t
his what do you think has gone wrong? In any event is this a plausibl
e model? a useful one for predicting future growth? " }}}{PARA 0 ""
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "restart: with(plots): with(
DEtools):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 116 "Cohen mentions in the Fig. 1 legend that the population \+
growth was \"faster than exponential\" between 1400 and 1970. " }}
{PARA 0 "" 0 "" {TEXT -1 30 "1. What does he mean by this?" }}{PARA
0 "" 0 "" {TEXT -1 33 "2. What evidence does he have? " }}{PARA 0 "
" 0 "" {TEXT -1 115 "3. How could you rewrite the equation P' = rP t
o describe a population that has faster than exponential growth? " }}
{PARA 0 "" 0 "" {TEXT -1 980 "\nCohen proposes \"idealized mathematica
l models\" for the race betwen human population growth and human carry
ing capacity. Suppose that it is possible to define human carrying ca
pacity K(t) as a numerical quantity measured in numbers of individuals
. Suppose also that P(t) is the total number of individuals in the po
pulation at time t. The logistic model then gives us the rate equatio
n P' = rP(K - P). (NOTE: this is not the standard version of the logi
stic equation, which would have the right hand side divided by K; qual
itatively, however, the results are the same.) In the simplest case K
(t) is taken to be constant, reflecting the stability of the natural e
nvironment over a time span that includes many generations: let's say,
to begin with, K1= 0.252789 billion. Again the constant r is the Ma
lthusian parameter, in this simulation roughly 1.4%. Cohen's simulati
ons run for 2500 years with a step size of 20 years, and again an ini
tial population of 0.252 billion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 115 "Bounded_growth:= diff(P(t),t) = r \+
* P(t) *(K1 - P(t));\nvars:= [ P(t) ]; domain:= 0.. 2500; init:= [ P
(0)= 0.252];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "DEplot(sub
s( \{r = 0.0014829, K1 = 0.252789\}, Bounded_growth), vars, domain, [i
nit], arrows = LINE, stepsize=20); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 175 "\nHow does this pattern of population growth compare to what y
ou modeled above? What effect does the substitution of other constant
values of r and K1 have on the solutions? " }}{PARA 0 "" 0 "" {TEXT
-1 564 "\nCohen's intuition is that, at least for human populations, t
he population itself can affect the environment in such a way that the
carrying capacity will change. Thus he goes on to investigate a more
complicated model in which the carrying capacity K(t) is allowed to v
ary, since \"every human being represents hands to work, and not just \+
another mouth to feed\" (George Bush, pe`re). He assumes the rate of \+
change of carrying capacity is proportional to the rate of change of p
opulation size, with the Condorcet parameter c as the constant of prop
ortionality. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 176 "We define Eq4 and Eq5 to accomodate Cohen's time-va
rying carrying capcity K(t) rather than the fixed carrying capacity K1
that we used before. Do you see where Eq5 comes from?" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 93 "Eq4:= diff(P(t), t) = r*P(t)*(K(t)- P(t));\nEq
5:= diff(K(t), t) = c* ( r*P(t)*(K(t) - P(t)) );" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 79 "vars:= [ P(t), K(t) ]; domain:= 0 .. 2500 ; i
nit2:=[P(0)=0.252, K(0)=0.252789];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 184 "\nFirst try modelling the populaton with the Condorcet paramet
er c < 1, say, c = 0.5. This means that each additional person increa
ses the carrying capacity by less than s/he consumes." }}{PARA 0 "> "
0 "" {MPLTEXT 1 0 63 "rate_eqns:= subs( \{ c = 0.5, r = 0.0014829 \},
[ Eq4, Eq5 ] ); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 110 "Kplot
:= DEplot(rate_eqns, vars, domain, [init2], stepsize=20, scene=[t, K],
linecolor = black, arrows=NONE): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0
107 "Pplot:= DEplot(rate_eqns, vars, domain, [init2], stepsize=20, sce
ne=[t, P], linecolor = cyan, arrows=NONE):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 69 "display( [Kplot, Pplot ], title = `Condorcet growth
model (c=0.5)`);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 502 "\n***Can you identify which curve
is which, even without the colorization? How do they compare to Cohe
n's Fig. 4? We can now look at the case where each person increases r
esources by MORE than s/he consumes (c > 1): try using c = 2.5 in the
lines above. Is the result consistent with Cohen's discussion at the
end of page 343? Try running the model with c = 1.0 (each person add
s to the carrying capacity just as much as s/he consumes). Does the p
opulation grow as Cohen describes in each case?***" }}}{EXCHG {PARA 0
"" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 211 "For the mathema
tically intrepid, footnote 35 gives Cohen's reasoning. Can you put th
is in more intuitive terms: for instance, what does he mean by \"virtu
al carrying capacity\" and \"virtual Malthusian parameter\"?" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 117 "\n***It is also interesting to co
nsider the case of a negative Condorcet parameter. What would happen \+
to the world's " }{TEXT 259 7 "present" }{TEXT -1 15 " population in \+
" }{TEXT 260 2 "50" }{TEXT -1 226 " years, assuming a present carrying
capacity of 18 billion, a 1.6% intrinsic growth rate, and a Condorcet
value c = -0.5? (Go back and change all the relevant parameter value
s, including initial conditions and the domain.)***" }}{PARA 0 "" 0 "
" {TEXT -1 114 "\nWe define Eq6 to accomodate Cohen's model of diluti
on, but not depletion, of resources; L is the Mill parameter." }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "Eq4:= Eq4; Eq6:= diff(K(t),t) = (L
/P(t))*(r*P(t)*(K(t) - P(t)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 68 "rate_equations:= subs( \{ L = 3.7 , r = 0.0014829 \} , [ Eq4, \+
Eq6 ] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 206 "Kplot:= DEplot
(rate_equations, vars, domain, [init2], stepsize=20, scene=[t, K], arr
ows=NONE, linecolor=BLUE):\nPplot:= DEplot(rate_equations, vars, domai
n, [init2], stepsize=20, scene=[t, P], arrows=NONE ):" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "display(\{Kplot, Pplot\}, title = `
Cohen Fig. 4`); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 676 "\n***Cohen claims that the long t
erm behavior of K(t) and P(t) is independent of the value of r. Is th
is true, and if so, how do you account for it? What effect does the Ma
lthus parameter have on the solutions? (Recall you looked at similar q
uestions in the context of the logistic model.) Read footnote 39. Wh
y did Cohen chose 3.7 for the parameter L? What is he assuming about \+
the shape of the relation between K(t) and time that led him to use th
is value? Do you believe he made a reasonable choice? Do you think C
ohen's curve-fitting approach is likely to give a better estimate of t
he upper bound on human populations than the estimates in Fig 3? Why \+
or why not?***" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
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