4.1.1.
Animal Population Dynamics
𝑁(𝑡+1)
Resources necessary for population growth are unlimited:
𝑁(𝑡) = = 𝑒
𝑟
𝑁(𝑡+1) and 𝑁(𝑡) = population after 𝑡 and 𝑡 + 1 years, respectively
where
𝑟 = the specific growth rate (net new organism per unit time)
Resources are limited:
𝑁(𝑡) = 𝐾𝑁0
𝑁
0 + (𝐾 − 𝑁 )𝑒
𝐾 is the carrying capacity, the numbers of individuals an area can support.
−𝑟𝑡
𝑁0 is thepopulation attime 𝑡 =
0years
Problem 2:
Assume that the population of the greater roadrunner in the Guadelope Desert
hectare at the beginning of 1999. If the carrying capacity, 𝐾, is 600 and 𝑟 = 0.
was 200 per
25/𝑦𝑒𝑎𝑟, what is the number of roadrunners one, five, and then ten years later?
What’ll happen when the number of roadrunner is equal to the carrying
capacity?
Solution:
Given: 𝑁0 = 200, 𝐾 = 600, and 𝑟 = 0.25/𝑦𝑟 Required: 𝑁1, 𝑁5,
and 𝑁10 Substitute all the known values,
𝑁𝑡 = (600)(200)
(200) + (600 − 200)𝑒−0.25𝑡
Evaluate the function at 𝑡 = 1, 𝑡 = 5, and 𝑡 = 10.
𝑁1 (200) + (600 − 200)𝑒−0.25(1) = 234.59 ~ 235
(600)(200)
= (600)(200)
(200) + (600 − 200)𝑒−0.25(5)
𝑁5
= 381.43 ~ 382
(600)(200)
(200) + (600 − 200)𝑒−0.25(1)
= = 515.39 ~ 516
𝑁10
=
The population will reach the carrying capacity is the resources are being
used faster than they are being replenished. If this occurs, the population
will then decrease in size because the resources are not enough to support
the life of the whole population.
4.1.2. Human Population Dynamics
Predicting the dynamics of human population is important to
environmental engineers because it is the basis for the determination
of design capacity for municipal and wastewater treatment systems
and for water reservoirs.
Population predictions are also important in the development of
resources and pollutant management plans.
Human population dynamics also depend on birth, death, gender ratio,
age structure, and dispersal.
In human populations, dispersal is referred to as an immigration and
emigration.
Assuming an exponential growth rate, the population can be predicted using
𝑃(𝑡) = 𝑃0𝑒𝑟𝑡
the equation:
𝑃(𝑡) = the population at time, 𝑡
where
𝑃𝑂 = population at time, 𝑡 = 0
𝑟
𝑡
= rate of growth
= time
The growth rate can be determined as a function of birth rate (𝑏), death rate
(𝑑),
immigration rate (𝑖), and emigration rate (𝑚):
𝑟=𝑏−𝑑+𝑖−𝑚
where the rates are all expressed as some value per unit time.
Problem 3:
A population of humanoids on the island of Huroth on the Planet Szacak has a
net birth rate (𝑏) of 1. 0 individuals / (individual—year) and a net death rate (𝑑)
of 0. 9 individuals / (individual— year). Assume that the net immigration rate is
equal to the net emigration rate. (a) How many years are required for the
population to double? (b) In year zero, the population on the island is 85, what is
the population 50 years later?
Solution:
Given 𝑏 = 1.0/𝑦𝑟, 𝑑 = 0.9/𝑦𝑟, 𝑖 = 𝑚
𝑟 = 𝑏 − 𝑑 + 𝑖 − 𝑚 = 1.0 − 0.9 + (0) = 0. 1/𝑦𝑟
Solve for the rate of growth
variable 𝑃0. The population doubled could be represented as 2𝑃0.
a.) If the initial population is not given, assume initial population as
Substitute these to the population growth equation
2𝑃 𝑂 = (
0.1
𝑃0 𝑒
)𝑡𝑦𝑟
Cancel out the 𝑃0 on both sides of the equation
2=𝑒
(0.1)𝑡
Take the natural logarithm of both sides of the equation to eliminate the
Euler’s number
Divide both sides by the coefficient of 𝑡.
ln 2 = 0. 1𝑡
𝑡= = 6. 93 𝑦𝑒𝑎𝑟𝑠
ln 2
0.1
b.) In this case, the initial population is 𝑃0 = 85.
It will take around 6. 93 years to double the population of the island.
𝑃(𝑡) = 85𝑒 ( 𝑦𝑒𝑎𝑟
0.1
)𝑡
Find the population after 𝑡 = 50 years.
𝑃(𝑡) = 85𝑒(0.1)(50) = 12,615.12 ~ 12, 616
After 50 years, the population on the island is 12, 616.
