Calculus II Project # 2
Professor Janeba
May 4, 1997
[written by...]
[Student #1]
[Student #2]
[Student #3]
[Prof. Janeba says: Student's name are omitted for privacy reasons.]
The nation of Freedonia owns Big island, the land mass off the country's southern coast. As with many species, the native population of Farnsworth's Sparrows was almost hunted to extinction by Freedonians looking to make centerpieces. However, in the 1930's, a small population of Farnsworth's Sparrows was found to have survived on the mainland. Some of these birds were captured and transported to Big Island, so that the species could be re-established in its natural habitat. We have been commissioned to study the population trends of the Farnsworth's Sparrow, as reflected by the changing environment on Big Island. These changes include such situations as limited food supply and introduction of predators. As with many population models and situations where there is a rate of change, a differential equation is involved. The differential equation, a relationship between an unknown function and one or more of its derivatives, can be used to solve for quantities where one of the variables is not constant with relation to time. Thus we will be using differential equations to estimate and predict the population of Farnsworth's Sparrows on the island.
After their transport, due to a lack of natural predators and a moratorium on hunting, the population of Farnsworth's Sparrows grew at a rate jointly proportional to the number of birds on the island and the amount of food available to them. Big Island has food plants which the birds consume; the amount of those plants on the island is constant. However, many new sparrows cannot utilize that food because other sparrows have already taken it. If we let C equal the rate at which food becomes available in calories per year on the island (a constant rate) and P equal the number of sparrows on the island, the rate at which food becomes available is C - kP (where k is some unknown proportionality constant, representative of the amount of food in calories per year each sparrow eats). From this we know the rate of change of the sparrow population is proportional to the rate at which food becomes available (C - kP) and the population of sparrows, which is P. As a jointly proportional relationship, the rate of change of sparrow population is the above two terms multiplied together times a proportionality constant we called A. Thus we derive the equation
dP/dt = A(( C - kP) P).
The Audobon society has provided the following data: in 1935 the population of sparrows was estimated to be 1050, with the birthrate exceeding the death rate by 200 per year, while in 1939, the population was approximately 2100, with a growth rate of 300 birds per year. No more data was recorded until 1949, due to the war. We know from previous models and examples that an environment, such as Big Island, has a maximum population of organisms (in this case sparrows) it can support due to the limits on the resources available to them. In our case, the population of sparrows is dependent on the amount of food available. This maximum population is called the carrying capacity and is reached at one of the equilibrium positions on the graph, when dP/dt is zero. It is the number of sparrows that the population tends toward, given enough time, and is designated by L. Of course there is another equilibrium point, where the sparrow population is zero, but we are not interested in that and will ignore it. We began by solving for the constants in the previous equation by setting dP/dt equal to zero and combining some of the constants. First we divided through by P, resulting in:
(dP/dt)/P = A ( C - kP)
We know that (dP/dt)/P is the relative growth rate, and when the population growth rate equals zero, the relative growth rate is zero. If the growth rate is zero, the population will be at equilibrium, or L. Setting (dP/dt)/P equal to zero, we could combine some of the constants.
0 = A ( C - kP ).
0 = C - kP
Solving for P, we obtained P = C/k. Since we solved the equation as (dP/dt)/P equals zero, we know that C/k must be the carrying capacity, and therefore C/k = L and k = C/L. Substituting L back into our original equation we get:
dP/dt = A [ C - ((C/L) P) ] P.
Multiplying through, we get:
dP/dt = PAC - (P2AC)/ L.
Obtaining a common denominator L results in:
dP/dt = [LPAC - (P2AC)] / L.
Factoring out an APC gives us:
dP/dt = APC (L/L - P/L).
We combine the constants A and C to form W. This yields:
dP/dt = WP ( 1- P/L ).
Rearranging the equation to solve for W, we get:
W = (dP/dt)/ [P(1 - P/L)].
We know that W is a constant, so if we plug our two known data values in the equation, and set those equations equal to each other, we can solve for L. In one case, dP/dt = 200 and P = 1050, and in the other case dP/dt=330 and P=2100. This gives us:
200/ [1050 (1 - 1050/L)] = 330 / [ 2100 (1 - 2100/L)].
Solving for the carrying capacity, we find that L = 7050. This means that the Farnsworth's Sparrow population would stabilize at 7050 sparrows, given enough time. We can now solve for our other constant, W, by plugging in L along with one set of data values. Therefore,
W = 200/ [1050(1-1050/7050)].
So W equals 0.224. Plugging our constants back into the equation, we obtain a differential equation for the rate of change of the sparrow population: dP/dt = 0.224P (I - P/7050). We check our answer by plugging in the second set of sparrow population data and comparing our answer for dP/dt with the dP/dt given. This answer checks. We also check by graphing our differential equation and tracing to find the equilibrium point, which is 7050. The data shows an increase in the sparrow population that would be checked only by the availability of food resources. However, the Freedonian government sanctioned the development of Big Island, beginning in 1946. This development, in response to crowding on the mainland, was to have dire consequences for the sparrows. House cats, predators of sparrows, came to the island with the Freedonians. Within one year, the government banned cats from the island, but unfortunately some cats escaped the search. These now-wild cats compete for food, primarily Farnsworth's Sparrows. We let Q represent the population of feral cats, with dQ/dt being the rate of change of the cat population. Thus, the cat's growth rate is the sum of two terms: a negative number (which we denoted as B) times the cat population, Q, plus a second number which is proportional to catfood supply (or the number of sparrows available, P). We assign the constant of proportionality, R, to catfood supply, P. Thus, the equation can be stated as:
(dQ/dt) = -BQ + RP
Since the catfood supply, RP, is jointly proportional to the bird population and the cat population, we can represent the second term with:
(dQ/dt) = -BQ + RUPQ
Where U is another constant of proportionality for the bird population and the cat population. The term RUPQ can be replaced with ZPQ, where Z is the combination of the two constants of proportionality R and U. The first number, B, is negative because as the cat population increases, the growth rate decreases, and the second number is jointly proportional to the bird population and the cat population, which we denoted as ZPQ, because as the number of cats on the island grows, so does the number of lethal cat-bird encounters. Thus, our differential equation is:
dQ/dt = -BQ + ZPQ.
[Prof. Janeba says: The excerpt above is the first four pages of a student project report. The total project ran some 15 pages, including seven graphs. This project was much harder than introductory projects. I consider this writing to be excellent. I gave the project report an A overall.
There may be several typos here that are not the student's fault, caused when I scanned the report. The scanner often makes mistakes - I corrected many. If any remain, do not blame the student authors.
Note to non-Willamette folk who come across this web page: Freedonia, Farnsworth's Sparrow, and Big Island are all ficticious.]
Go back to the project assignment page
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