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### #199 - Tools Of Our Trade -- Part 2: More Ways To Understand Growth, 18-Sep-1990

In RHWN #197, we introduced some tools that allow anyone to understand
things that grow exponentially, which is to say, things that grow by a
constant percentage of the whole in a constant time period, such as a
bank account that grows at the rate of 6% each year. Most things that
affect the environment are growing exponentially. Tools for
understanding exponential growth are very useful because they allow us
to see why we're in the situation we're in today.

RULE 5: If you know some original amount, called N_sub_o, and some
amount that it has grown to, N_sub_t, at some later time, t, you can
calculate the value of k (the annual rate of change, expressed as a
decimal fraction) by the following equation:

k = (ln(N_sub_t/N_sub_o))/t [Rule 5]

where:

calculator will give you the natural log of any number instantly);

N_sub_o is some original amount;

N_sub_t is the amount that it has grown to at some later time, t;

k = the annual percentage increase expressed as a decimal fraction (in
other words, it's the value we've been calling p, divided by 100) [for
example, if something is growing at 6% per year, p = 6 and k = 0.06];

t = time (in any units you care to choose).

In other words, take the amount N_sub_t at time t, divide it by N_sub_o
(the original amount), take the natural log of the result, then divide
that natural log by t. That will give you k; if you then multiply k by
100, you have p. Then you can use k and p with Rules 1 through 4 (given
in RHWN #197) to learn important characteristics of the thing that's
growing.

For example, the consumer price index grew from 133.1 in 1973 to 246.8
in 1980. What was the annual percentage rate of increase, p? To find p,
first we calculate k. The annual increase rate (expressed as a decimal
fraction), k, was:

(ln(246.8/133.1))/t

Since t = 1980-1973 = 7 years, and 246.8/133.1 = 1.85 and the natural
log of 1.85 = 0.62, the value of k = 0.62/7, or 0.088; thus p = 0.088 x
100 or 8.8% per year increase. With this number in hand, we can use
rules 1 through 4 to learn its other growth characteristics.

RULE 6: If a quantity is growing exponentially and you want to know how
rapidly it is increasing by a factor of x (i.e., how fast it's
increasing by a factor of, say, 5, or, as we might say, how long it
takes to increase 5-fold), use this equation:

x folding time = ln(x)/k [Rule 6]

Example: If a bank account is growing at 6% per year, how long will it
take for the amount in the bank to increase five-fold? The natural log
of 5 = 1.61; 1.61/0.06 = 26.8. Therefore, a bank account growing at 6%
per year will increase 5-fold in a period of 26.8 years.

RULE 7: If an exponentially-growing quantity has increased by a factor
of x during some period of time, t, then the fractional increase, k,
per unit time can be found by the equation

k = ln(x)/x folding time [Rule 7]

Example: The annual production of a very useful chemical increased 6-
fold from 1949 to 1958. What was the annual percentage increase?

My scientific calculator tells me that the natural log of 6 = 1.79. The
x folding time = 1958-1949 = 9. Thus k = 1.79/9 = 0.198 or 19.8% per
year, an impressive annual rate of increase, which, by Rule 1, would
lead to a doubling of that chemical's production every 70/19.8 = 3.5
years, and, by Rule 4, would lead to an increase by a factor of a
million during one human lifetime. Impressive growth, indeed. (See
Table 1 in RHWN #197.)

RULE 8: If you want to know the total amount of an exponentially-
growing quantity produced during time-period t, use this equation:

C = (N_sub_o/k)*(e**((k*t)-1)) [Rule 8]

where:

** means "raise to the power of"

* means "multiply by"

C = total amount produced during time-period t;

N_sub_o = the original amount;

k = the increase (expressed as a decimal fraction) per unit of time
(for example, per year);

e = 2.718, the base of natural logarithms;

t = time (in any units).

For example, if 40,000 automobiles were produced in 1920 and car
production increased 10% per year for the next 15 years, how many cars
were produced, total, during the 15 years? N_sub_o = 40,000; k = 0.10;
t = 15. Therefore, C = (40,000/0.10) (e**((0.10*15)-1)) = 1,392,676.

In writing the equation above, we've used an asterisk to indicate
multiplication; in other words, 2*2 = 4, and we've used ** to indicate
exponentiation, meaning "raised to the power of;" i.e., 2**3 = 8.

RULE 9: During one doubling time, the growth in an exponentially-
growing quantity equals all the growth that has occurred in all
previous time. Thus if production of chemical XYZ is growing at 10% per
year (doubling in 7 years), during the next 7 years we will produce an
amount of chemical XYZ equal to all of the chemical XYZ that has been
produced up to today.

These 9 rules can help us all understand why environmental problems
have sneaked up on us, and why the growth of some things must be
curbed.

Let's take an example. Let's say we learn from a magazine article that
the synthetic organic chemical industry produced 38 billion pounds of
chemicals in 1945 and since then has grown steadily at 6.5% per year.
From that small amount of information, we can learn a lot. For example,
from Rule 3 we can learn that annual synthetic organic chemical
production in 1990 was about 708 billion pounds. By Rule 1 we can learn
that total production is doubling every 10.8 years (we'll round it off
and call it 11 years). By Rule 9, we can learn that during the most
recent doubling-time (from 1979 through 1990), which happens to be the
time since Love Canal was discovered, the chemical industry produced an
amount of chemicals equal to the amount it produced during all time
prior to the discovery of Love Canal. By Rule 8, we can learn that
since 1945, the chemical industry has produced a total of 10.3 million
million (or 1.03 x 10**13) pounds of chemicals, all of which has gone
somewhere, and Rule 9 tells us that during the next 11 years they'll
produce another 10.3 million million pounds. From Rule 4, we can learn
that if the chemical industry keeps growing at 6.5% per year, during
one human lifetime it will increase its size 90-fold; in other words,
if the chemical industry keeps growing at its current rate, for every
chemical factory in existence today, there will be 90 chemical
factories just 70 years from now.

These facts help us understand why more and more people are complaining
about chemicals affecting their lives: it isn't because of growing
"chemophobia," it is because people are being exposed to more and more
chemicals that are being released into the air and water that we all
breathe and drink. A massive change--the "chemicalization of the
environment"--has occurred during the past 50 years, and the problem is
doubling in size every 11 years.

These facts can lead us to one more conclusion: since the chemical
industry in all its trade publications constantly stresses the need for
continuous growth at rates that match or exceed the historical rate of
growth, we can conclude that executives in the chemical industry do not
understand the implications of exponential growth. No one who has seen
the heavily industrialized sections of chemical-producing states like
New Jersey, Ohio, Louisiana, Texas and California can believe that we
could survive 90 chemical factories where each chemical factory stands
today. Growth of this industry has got to be curbed. By what means? By
whatever non-violent means are necessary. Whatever it takes.

Further reading: Ralph Lapp, THE LOGARITHMIC CENTURY (Englewood Cliffs,
NJ: Prentice-Hall, 1973). And: chapters 1 and 2 of Donella H. Meadows
and others, THE LIMITS TO GROWTH (NY: Universe Books, 1972). And:
Albert A. Bartlett, "The Exponential Function," THE PHYSICS TEACHER
(October, 1976), pgs. 393-401.

--Peter Montague

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Descriptor terms: mathematics; chemical industry;