Calc1231W7T1 - Population growth

Wednesday, 5 September 2018

12:14 PM

 

 

Untitled picture.png Machine generated alternative text:
In 1859 Thomas Austin released 24 rabbits on his property in Victoria, Australia. 
Let r(t) be the Australian rabbit population years t after the initial release of rabbits. Then suppose the population increases according to the differential equation: 
=0.1872 x r. 
Then 
With this initial condition, the solution to this differential equation is 
Hence, we can estimate (to the nearest rabbit): 
the rabbit population in 1860 to be 
the rabbit population in 1861 to be 
the rabbit population in 1950 to be 
29 
35 
600488369 

Untitled picture.png Machine generated alternative text:
After 1859, the Australian rabbit population spread quickly throughout the southern states. Queensland and Western Australia responded by constructing rabbit proof fences 
to protect their agricultural interests. The fences were not very effective. 
Suppose the Australian rabbit population is contained within a circle centred at Thomas Austin's house in Victoria with radius R(t) 
1859. Let's suppose also that the radius of this rabbit circle grows at a constant rate of 
— 40 kilometers per year. 
Initially O, so with this information we can solve the differential equation to find 
where t is the number of years after 
We can also express t as a function of R: 
R/40 
Using the last equation, the rabbit population 
• reaches Queensland, 1,240 kilometers away from Thomas Austin's house, after 
31 
• reaches Western Australia, 2,040 kilometers away from Thomas Austin's house, after 
O years, and 
O 
years. 

Untitled picture.png Machine generated alternative text:
After 1859, the Australian rabbit population spread quickly throughout the southern states. Queensland and Western Australia responded by constructing rabbit proof fences 
to protect their agricultural interests. The fences were not very effective. 
Suppose the Australian rabbit population is contained within a circle centred at Thomas Austin's house in Victoria with radius R(t) 
1859. Let's suppose also that the radius of this rabbit circle grows at a constant rate of 
— 40 kilometers per year. 
Initially O, so with this information we can solve the differential equation to find 
where t is the number of years after 
We can also express t as a function of R: 
R/40 
Using the last equation, the rabbit population 
• reaches Queensland, 1,240 kilometers away from Thomas Austin's house, after 
31 
• reaches Western Australia, 2,040 kilometers away from Thomas Austin's house, after 
O years, and 
O 
years. 

Untitled picture.png Machine generated alternative text:
The Australian rabbit population grew exponentially so long as resources were plentiful. But resources such as food and water place an environmental limit on the maximum 
population. As the population approaches this environmental limit, a different kind of growth is exhibited. This was first identified in 1845 by Pierre-Franqois Verhulst. 
Taking into consideration the environmental constraints, the rabbit population P (in millions now!!) is better described by the differential equation 
800 
This is a separable differential equation that can be solved by integrating 
800 
P(800 - P) 
The solution to this equation is given by 
800 
1+Ae—t/6 
where A is a constant determined by the initial conditions. If we let t be the number of years since 1950, then our initial condition becomes — 
— 600. From this we 
determine that
Untitled picture.png Machine generated alternative text:
The Australian rabbit population grew exponentially so long as resources were plentiful. But resources such as food and water place an environmental limit on the maximum 
population. As the population approaches this environmental limit, a different kind of growth is exhibited. This was first identified in 1845 by Pierre-Franqois Verhulst. 
Taking into consideration the environmental constraints, the rabbit population P (in millions now!!) is better described by the differential equation 
800 
This is a separable differential equation that can be solved by integrating 
800 
P(800 - P) 
The solution to this equation is given by 
800 
1+Ae—t/6 
where A is a constant determined by the initial conditions. If we let t be the number of years since 1950, then our initial condition becomes — 
— 600. From this we 
determine that 

Untitled picture.png 

Untitled picture.png Machine generated alternative text:
Furthermore if the environmental conditions were to remain unchanged, then (to the nearest million rabbits) the rabbit population in 1980 would be 
798 
million. 
Interestingly the rabbit population will approach the maximum sustainable population of 
lim P(t) = 
800 
00 
O million. 

Untitled picture.png Machine generated alternative text:
Consider the logistic differential equation 
This is the equation used to describe population growth in the presence of constraints first proposed in 1845 by Pierre-Franqois Verhulst, and it has remained the standard 
model for such population growth ever since. 
A key question faced by ecologists is: how to fit the logistic equation to population data? This essentially comes down to finding the best value for the constants K, r . 
Here is an example. Suppose an ecologist takes four measurements of an endangered penguin population P (in thousands) and the rate of change in population 
0.40 
0.50 
0.39 
is quadratic in P . So we instead consider 
We cannot immediately find a curve of best fit, as 
This quantity is linear in P . The data points are 
o 
0.25 
e 
0.13 
0.48 
e 
12 
The GeoGebra app below shows you the line of best fit (in black) to four data points (in red). The variable P is plotted on the horizontal axis, the variable 
P is Plotted 
on the vertical axis. 

Untitled picture.png Machine generated alternative text:
— = 0.47 — 0 1 p 
0.4) 
(2, 
(3, 
0.12) 
Move the points so that they match the data in the table above, then by reading off the equation of the line of best fit, deduce (to two decimal places) 
0.47 
4.7 

Untitled picture.png Machine generated alternative text:
— = 0.47 — 0 1 p 
0.4) 
(2, 
(3, 
0.12) 
Move the points so that they match the data in the table above, then by reading off the equation of the line of best fit, deduce (to two decimal places) 
0.47 
4.7 

Untitled picture.png Machine generated alternative text:
Let's use the logistic equation (which describes any population be it people, penguins or rabbits) to estimate the future human population of the world. Consider the following 
table of world population milestones, sourced from the World Population Clock: 
Year 
Population 
(Billions) 
1804 1927 1960 
2 
3 
0.0186 
1974 
4 
0.0194 
1987 
5 
0.0186 
1999 
6 
0.0129 
2012 
7 
0.0123 
In order to estimate the values of r and K in the expression 
we use the same technique from the previous question, and obtain 
From this we deduce that 
The solution to the logistic equation is then 
dt 
-r - r PæO.0259-O.0019P. 
0.0259 
O and 
13.63 
O (to two decimal places). 
where the constant A is determined by the initial conditions. If we let t be the number of years after 1999 then 
Hence, to two decimal places, 
6 
1.27 
From this we can predict that the world population in 2030 will be (to two decimal places) 
869 
O billion. 


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Untitled picture.png Machine generated alternative text:
Let's use the logistic equation (which describes any population be it people, penguins or rabbits) to estimate the future human population of the world. Consider the following 
table of world population milestones, sourced from the World Population Clock: 
Year 
Population 
(Billions) 
1804 1927 1960 
2 
3 
0.0186 
1974 
4 
0.0194 
1987 
5 
0.0186 
1999 
6 
0.0129 
2012 
7 
0.0123 
In order to estimate the values of r and K in the expression 
we use the same technique from the previous question, and obtain 
From this we deduce that 
The solution to the logistic equation is then 
dt 
-r - r PæO.0259-O.0019P. 
0.0259 
O and 
13.63 
O (to two decimal places). 
where the constant A is determined by the initial conditions. If we let t be the number of years after 1999 then 
Hence, to two decimal places, 
6 
1.27 
From this we can predict that the world population in 2030 will be (to two decimal places) 
869 
O billion. 

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