CHAPTER 3 SUPPLEMENTARY EXERCISES
4.5 MODELS OF EXPONENTIAL GROWTH AND DECAY
Populations grow and decline. Intuitively, population changes are generally proportional to population size. While populations of people, animals, insects, bacteria, and such quickly come to mind, radioactive decay, consumer demand, bodily drug absorption, attenuation of light beams, electronic signals, and dollar interest on principal are examples of the principle.
⧫A mathematical “model” is a symbolic representation of a reality as in the examples cited previously.
A basic mathematical model for growth (or decay) assumes that at a time,t, for a pop- ulation size, y, the rate of change of population with time,dy/dt, is proportional to the population at that instant.
In symbols,
dy dt =𝛼y where𝛼is a coefficient of proportionality.
↑ The previous equation, containing a derivative, is called adifferential equation. Differ- ential equations are of basic importance in the modeling of phenomena involving change.
We seek the population as a function of time, y(t). The solution to this elementary differential equation is
y(t) =y0e𝛼t
wherey0is the value ofy(t)whent=0. You can check this solution by substitution into the differential equation. This relatively simple model has amazingly wide application. When 𝛼is positive,y(t)increases with increasing time,t; a model for exponential increases and growth. When𝛼is negative,y(t)decreases with increasingt; a model for exponential decline and decay.
Basic exponential models typically assume that reproduction is continuous, that organ- isms are identical, and that environmental conditions are constant. There are many applica- tions in the biological sciences such as the prediction of fish populations and carbon dating.
The exponential model is fairly robust, meaning that it provides reasonable estimates even when assumptions are approximations.
Example 4.5.1 Population Growth
An exponentially increasing population numbered 50,000 in 1980 and 75,000 in 2000.
Forecast the population, y(t), in 2020.
Solution:
Set t=0 for the year 1980. Then, t=20 for the year 2000. Use these to solve for𝛼in the exponential equation.
75,000=50,000e20𝛼 or 3∕2=e20𝛼 Using logarithms,
ln(3∕2) =20𝛼 and 𝛼=0.020273 Therefore, y(t) =50,000e0.02073t
In 2020, when t=40, y(40) =50,000e0.020273(40)≈112,500.
⧫Thomas Robert Malthus (1766–1834) prophesied mankind’s eventual doom. He con- tended that human populations increase geometrically, while areas under cultivation (agriculture) can only grow arithmetically. The widely known “Malthusian Doctrine”
has, so far, been a “victim” of the agricultural revolution. The coefficient,𝛼, has been called the “Malthusian Parameter.”
⧫Consider several “circles of friends.” To contact a friend of a friend of a friend, there are a potentially large number of choices. A variant of this idea is the basis for the Kevin Bacon board game.
In the June 4, 1998 issue ofNature, amathematical model described the “six degrees of separation” (or “small world thesis”) devised by mathematicians Steven Strogatz and Duncan Watts. Their shortcuts help to explain why strangers often find they share com- mon acquaintances and why epidemics can spread quickly.
MODELS OF EXPONENTIAL GROWTH AND DECAY 125 Radioactive Decay
Many natural elements are radioactive; their intensity measured by a Geiger Counter as they decay. A popular measure of decay is an element’shalf-life. The half-life is the time required for an elements’ radioactive intensity to fall to one-half of its initial value. Physicists use exponential models for radioactive decay.
Negative exponential growth, exponential decay, occurs when the proportionality factor is negative valued. The model becomesy′(t) = −𝛼y(t) (𝛼>0)whose solution is
y(t) =y0e−𝛼t 𝛼 >0
↓ It is easy to verify thaty(t) =y0e−𝛼tsatisfiesy′(t) = −𝛼y(t). Simply differentiatey(t)to obtainy′(t).
As before,y(t)is the radioactive intensity at a time,t;y0, the initial intensity; and𝛼, the decay constant (varies among radioactive elements). For the half-life, sety(t) = (1∕2)y0to yield
(1∕2)y0=y0e−𝛼t Using logarithms,
ln(y0e−𝛼t) =ln(1∕2y0) lny0+ln e−𝛼t=ln y0+ln 1∕2
ln e−𝛼t=ln 1∕2
−𝛼t=ln 1∕2
⧫Half-lives can vary from millionths of a second to billions of years. For instance, the half-life of Thorium-232 is 14 billion years. Tritium, a radioactive isotope of hydrogen, composed of one proton and two neutrons, has a half-life of about 12.5 years. Iodine-131 has a half-life of about 8 days. Recently discovered elements have half-lives of a fraction of a second.
Example 4.5.2 Radioactive Decay
Dairy cows that eat hay with higher levels of radioactive Iodine-131 produce milk unfit for human consumption. Suppose that a dairy farmer has purchased hay that has about 20 times the allowable limit of Iodine-131 present. The half-life for this isotope is 8.04 days. How long should the dairy farmer store the hay before use?
Solution:
Firstly, calculate the decay constant𝛼.
Using the model y(t) =y0e−𝛼t, 𝛼 >0
y0e−8.04𝛼= (0.50)y0 e−8.04𝛼=0.50
−8.04𝛼=ln(0.50) 𝛼≈0.0862
∴y(t) =y0e−0.0862t
The dairy farmer seeks the Iodine-131 to be 1/20 of its present level so:
(1∕20)y0 =y0e−0.0862t (1∕20) =e−0.0862t ln(0.05) = −0.0862t
t≈35 days
∴the hay should be stored for 35 days before use.
Some amazing recent scientific findings are attributed to carbon dating. Carbon-14 is a radioactive isotope of the stable Carbon-12 atom with a half-life of 5730 years. Knowing this, the antiquity of objects is estimated from their current levels of Carbon-14.
Example 4.5.3 Carbon-14 Dating
Develop a model for the remaining Carbon-14 in an object after t years. The half-life of Carbon-14 is 5730 years.
Solution:
Using the exponential decay model, y(t) =y0e−𝛼t, and the half-life for Carbon-14, estimate
𝛼as:
y0e−5730𝛼= (0.50)y0 e−5730𝛼=0.50
−5730𝛼=ln(0.50) 𝛼≈0.00012097 Therefore, remaining Carbon-14 is y(t) =y0e−0.00012097t.
MODELS OF EXPONENTIAL GROWTH AND DECAY 127
⧫Chemical compounds are composed of varied elements, each having a distinct atomic weight. Many elements have unstable isotopes that differ in properties of atomic weight and decay rates. Archeologists use the isotope Carbon-14 to estimate the age of artifacts.
Ages of artifacts older than 50,000 years have been estimated from Carbon-14 dating.
Elements that have longer half-lives, such as some isotopes of uranium or potassium, have been used to date geological events millions or billions of years old.
⧫The prehistoric art of the Lascaux Caves in France is one of the famous revelations attributed to Carbon-14 dating. Discovered in the 1940s, these caves, and others in the region, have led to conjectures about early people. For this and other fascinating stories related to dating of early objects such as the Shroud of Turin and Stonehenge, begin a web search with Lascaux Caves.
Not all exponential models for growth and decay have the natural basee, although it is most common and most important. Many empirical models have bases other thane.
Consider an exponential model, baseb, for a population,y(t), y(t) =y0bat b>0
wheretis time; base,b, a positive real number; a, a constant; andy0=y(0).
Example 4.5.4 Bacteria In a Culture
A culture of 250 bacteria doubles hourly. Without depletions, what is the population after 5 hours?
Solution:
As the population doubles hourly, it grows as 2t, t in hours. The model is y(t) =250(2)t. After 5 hours, 250(2)5=8000. There are 8000 bacteria after 5 hours.
Example 4.5.5 Growth of Escherichia coli Bacteria
If E. coli bacteria double their number every 20 min, how many are there in a day’s growth?
Solution:
There are 72 twenty-minute periods every 24 hours. Therefore, after 24 hours, a single bacterium becomes
y(72) =1(2)72 ≈ 4.7×1021. An astronomic number!
⧫“The mathematics of uncontrolled growth are frightening: A single cell of the bac- teriumE. coliwould, under ideal circumstances, divide every twenty minutes. That is not particularly disturbing until you think about it, but the fact that bacteria multiply geometrically: one becomes two, two becomes four, four becomes eight and so on. In this way in a single day one cell ofE. colicould produce a super colony equal in size and weight to the entire planet Earth.”
Michael Crichton (1969) The Andromeda Strain, Dell, N. Y. p 247.
EXERCISES 4.5
1. A city experiences exponential growth according toy(t) =50,000e0.03t. Using the year 2010 as a base(t=0), estimate the city population for 2015.
2. The population of a town in 2010 was 300 and is expected to increase to 400 by 2020.
If exponential growth applies, forecast the population for 2030.
3. The size of a bacteria culture doubles hourly. If their initial number is 500, forecast the number after 7 hours.
4. A bacteria culture increases daily by 25%. Initially 1000 are present. How many bac- teria are present after 2 weeks?
5. An initial population of 250 bacteria grows at a daily rate of 6%. When will the popu- lation have doubled?
6. An isotope of sodium (sodium-24) has a half-life of approximately 15 hours. If 1000 g of the isotope is present initially, how many grams will there be 5 days hence?
7. How many grams of Stontium-90 remain from 500 g after 200 years? Strontium-90 has a half-life of 28 years.
8. Plutonium is a heavy radioactive man-made metallic element. How long before 300 mg of Plutonium-239 decays to 100 mg? The half-life of Plutonium-239 is 24,000 years?
9. Argon-37 is produced from the decay of Calcium-40 whose half-life is 35 days. How long before 1/5 of the original amount remains?
10. As aortic valves close, aortic pressure at timet,P(t), is modeled byP(t) =95e−0.491t. What is the initial aortic pressure? The pressure att=0.1 sec? When will the aortic pressure reach 80 mm Hg?
11. In Carbon-14 decay what remains of 2.5 g after 10,000 years?
12. From 1967 to 1980 the California gray whale population,N(t), was described by the model N(t) =N0e0.025t. In fact, their growth to an estimated 21,000 whales in this period and into the early 1990s was such that in 1993 gray whales were removed from an endangered species list. Assuming the given model applies, estimate the number of California gray whales for 1967(t=0).
APPLICATIONS TO FINANCE 129