Respuesta :
Answer:
a) [tex] P(t=100) = 7 e^{0.03*100}=140.599 billion [/tex]
b) [tex] P(t=0) = 7 e^{0.03*0}=7 billion [/tex]
[tex] P(t=1) = 7 e^{0.03*1}=7.21 billion [/tex]
[tex] P(t=2) = 7 e^{0.03*2}=7.43 billion [/tex]
[tex] P(t=10) = 7 e^{0.03*10}=9.45 billion [/tex]
[tex] P(t=25) = 7 e^{0.03*25}=14.82 billion [/tex]
[tex] P(t=50) = 7 e^{0.03*50}=31.37 billion [/tex]
c) Figure attached
d) Figure attached
Explanation:
The proportional model on this case would be given by:
[tex] \frac{dP}{dt} = kP[/tex]
Where P is the population size, t the time on years and k a constant.
We can reorder this expression like this:
[tex] \frac{dP}{P} = k dt[/tex]
If we integrate both sides we got:
[tex] ln|P| = kt + C[/tex]
And using exponentials on both sides we got:
[tex] P(t) = e^{kt} e^C = P_o e^{kt}[/tex]
Where [tex]P_o=7 billion[/tex] represent the initial amount for the starting year t=0.
The rate on this case is given [tex] r =3\% = 0.03[/tex], so then our model would be given by:
[tex] P(t) = 7 e^{0.03t}[/tex]
Part a
For this case we just need to replace t=100 and we got:
[tex] P(t=100) = 7 e^{0.03*100}=140.599 billion [/tex]
Part b
For this case we have the following:
[tex] P(t=0) = 7 e^{0.03*0}=7 billion [/tex]
[tex] P(t=1) = 7 e^{0.03*1}=7.21 billion [/tex]
[tex] P(t=2) = 7 e^{0.03*2}=7.43 billion [/tex]
[tex] P(t=10) = 7 e^{0.03*10}=9.45 billion [/tex]
[tex] P(t=25) = 7 e^{0.03*25}=14.82 billion [/tex]
[tex] P(t=50) = 7 e^{0.03*50}=31.37 billion [/tex]
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.
To solve such problems we need to know about Exponential function.
Exponential Function
An exponential function is a function that shows the relationship in the form of [tex]y =a^x[/tex], where the independent variable x ranges over the entire real number line
The population of the world after 100 years is 140.6 billion.
Given to us
- The population of the world today = 7 billion,
- population grows at a constant rate = 3% per year,
- rate is almost certainly much faster than the future population growth rate.
Solution
The function for the population can be written as,
[tex]\dfrac{dP}{dt} = kP[/tex]
where,
P is the population,
k is the constant, and
t is the time in years.
rearranging the equation we get,
[tex]\dfrac{dP}{P} = kdt[/tex]
Integrating both sides,
[tex]\int\dfrac{dP}{P} = \int kdt\\\\ ln|P| = kt +C[/tex]
Taking antilog,
[tex]P(t) = e^{kt}+e^C[/tex]
[tex]P(t)=P_o\ e^{rt}[/tex]
where,
P(t) = Population of the world at time period t,
[tex]P_o[/tex] = Population of the world at the initial time,
t = time in years,
r = rate at which population grows.
Substituting all the values in the above function.
∵ [tex]P(t)=(7\ Billion)\ e^{0.03t}[/tex]
Thus, the human population can be calculated using the above exponential function at any point of time t.
Now, substitute the value of t in years to get the population at any point in time.
1.) t =0,
[tex]P(t)=(7\ Billion)\ e^{0.03t}\\\\ P(0)=(7\ Billion)\ e^{(0.03\times 0)}\\\\ P(0) = (7\ Billion)\ e^{(0)}\\\\ P(0) = (7\ Billion)\ \times 1\\\\ P(0) = (7\ Billion)[/tex]
Hence, the population of the world today is 7 billion.
2.) t =1,
[tex]P(t)=(7\ Billion)\ e^{0.03t}\\\\ P(1)=(7\ Billion)\ e^{(0.03\times 1)}\\\\ [/tex]
[tex]P(1)=7.2132\ billion[/tex]
Hence, the population of the world after 1 year is 7.2132 billion.
3.) t =2,
[tex]P(t)=(7\ Billion)\ e^{0.03t}\\\\ P(2)=(7\ Billion)\ e^{(0.03\times 2)}\\\\ [/tex]
[tex]P(2)=7.433\ billion[/tex]
Hence, the population of the world after 2 years is 7.433 billion.
4.) t =10,
[tex]P(t)=(7\ Billion)\ e^{0.03t}\\\\ P(10)=(7\ Billion)\ e^{(0.03\times 10)}\\\\ [/tex]
[tex]P(10)=9.45\ billion[/tex]
Hence, the population of the world after 10 years is 9.45 billion.
5.) t =25,
[tex]P(t)=(7\ Billion)\ e^{0.03t}\\\\ P(25)=(7\ Billion)\ e^{(0.03\times 25)}\\\\ [/tex]
[tex]P(25)=14.819\ billion[/tex]
Hence, the population of the world after 25 years is 14.819 billion.
6.) t =50,
[tex]P(t)=(7\ Billion)\ e^{0.03t}\\\\ P(50)=(7\ Billion)\ e^{(0.03\times 50)}\\\\ [/tex]
[tex]P(50)=31.372\ billion[/tex]
Hence, the population of the world after 50 years is 31.372 billion.
7.) t=100,
[tex]P(t)=(7\ Billion)\ e^{0.03t}\\\\ P(100)=(7\ Billion)\ e^{(0.03\times 100)}\\\\ [/tex]
[tex]P(100)=140.6\ billion[/tex]
Hence, the population of the world after 100 years is 140.6 billion.
Learn more about Exponential function:
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