Question

Suppose the world population today is 7 billion, and suppose this population grows at a constant rate of 3% per year from now on. (This rate is almost certainly much faster than the future population growth rate)a. What would the population equal 100 years from now?b. Compute the level of the population for t =0, t =1, t =2, t =10, t =25, and t =50.Today is t = 0.c. Make a graph of population versus time (on a standard scale).d. Now make the same graph on a log scale

Answer 1

Answer:

a) $P(t=100) = 7 e^{0.03*100}=140.599 billion$

b) $P(t=0) = 7 e^{0.03*0}=7 billion$

$P(t=1) = 7 e^{0.03*1}=7.21 billion$

$P(t=2) = 7 e^{0.03*2}=7.43 billion$

$P(t=10) = 7 e^{0.03*10}=9.45 billion$

$P(t=25) = 7 e^{0.03*25}=14.82 billion$

$P(t=50) = 7 e^{0.03*50}=31.37 billion$

c) Figure attached

d) Figure attached

Explanation:

The proportional model on this case would be given by:

$\frac{dP}{dt} = kP$

Where P is the population size, t the time on years and k a constant.

We can reorder this expression like this:

$\frac{dP}{P} = k dt$

If we integrate both sides we got:

$ln|P| = kt + C$

And using exponentials on both sides we got:

$P(t) = e^{kt} e^C = P_o e^{kt}$

Where $P_o=7 billion$  represent the initial amount for the starting year t=0.

The rate on this case is given $r =3\% = 0.03$, so then our model would be given by:

$P(t) = 7 e^{0.03t}$

Part a

For this case we just need to replace t=100 and we got:

$P(t=100) = 7 e^{0.03*100}=140.599 billion$

Part b

For this case we have the following:

$P(t=0) = 7 e^{0.03*0}=7 billion$

$P(t=1) = 7 e^{0.03*1}=7.21 billion$

$P(t=2) = 7 e^{0.03*2}=7.43 billion$

$P(t=10) = 7 e^{0.03*10}=9.45 billion$

$P(t=25) = 7 e^{0.03*25}=14.82 billion$

$P(t=50) = 7 e^{0.03*50}=31.37 billion$

Part c

The graph is on the first figure attached.

Part d

If we take a log-log scale we have the following values

We need to exclude the point t=0 since the natural log for 0 is not defined.

ln 1 =0 , ln 2= 0.693, ln 10=2.30, ln 25 =3.22, ln 50= 3.91

The result would be the figure 2 attached. And we see a better result for the graph.