predicting current growth rate w/ a logistical model

[BigBuck01]

Had a question in class today that I could use a little help with! Where do I start??? here's the question!

- Use the 1960's annual growth rate of 4.2% and population of 4 million to predict the current growth w/ a logistical model. Current country population is 8 million. Carrying capacity is 20 million.

 

[Deleted member 4993]

Had a question in class today that I could use a little help with! Where do I start??? here's the question!

- Use the 1960's annual growth rate of 4.2% and population of 4 million to predict the current growth w/ a logistical model. Current country population is 8 million. Carrying capacity is 20 million.

Start with definitions - How is logistic model defined?

Then define your unknowns - dependant and independant variables.

Then tell us excatly where you are stuck ...

 

[soroban]

Hello, BigBuck01!

Please check the wording of the problem.
As stated, it doesn't make sense.

Use the 1960's annual growth rate of 4.2% and population of 4 million
. . to predict the current growth with a logistical model.
Current country population is 8 million. .??
Carrying capacity is 20 million. . What does this mean?

$\displaystyle \text{The growth formula is: }\;P \;=\;P_o(1+r)^n$

. . $\displaystyle \text{where: }\;\begin{Bmatrix}P &=& \text{present population} \\ P_o &=& \text{initial population} \\ r &=& \text{annual growth rate} \\ n &=& \text{number of years} \end{Bmatrix}$


$\displaystyle \text{We are given: }\:\begin{array}{ccc}P_o &=& 4,\!000,\!000 \\ r &=& 0.042 \\ \end{array}$

$\displaystyle \text{So we have: }\;P \;=\;4,\!000,\!000(1.042)^n$


$\displaystyle \text{We are told that the current population (now) is 8 million.}$
. . $\displaystyle \text{When is "now"?}$

$\displaystyle \text{We have: }\:8,\!000,\!000 \:=\:4,\!000,\!000(1.042)^n \quad\Rightarrow\quad 1.042^n \:=\:2$

$\displaystyle \text{Take logs: }\;\ln(1.042^n) \:=\:\ln 2 \quad\Rightarrow\quad n\ln1.042 \:=\:\ln 2$

. . $\displaystyle n \;=\;\frac{\ln2}{\ln1.042} \;=\;16.84677015 \;\approx\;17\text{ years}$


$\displaystyle \text{Evidently "now" is somewhere in the late 1970's or early 1980's.}$

$\displaystyle \text{So what is the question?}$

 

[BigGlenntheHeavy]

Current country population is 8 million.

What is current? 2010, 2000, 1990, etc.

The population in 1960 was 4 million, in what year had the population grown to 8 million?

If you got this problem out of a book, what is it latest copywrite? That would be current.

 

[BigGlenntheHeavy]

$\displaystyle \frac{dP}{dt} \ = \ kP\bigg(1-\frac{P}{K}\bigg), \ common \ generic \ logistic \ model \ with \ k \ = \ annual \ growth \ rate$

$\displaystyle and \ K \ equals \ carrying \ capacity.$

$\displaystyle Hence, \ \int\frac{dP}{P\bigg(1-\frac{P}{K}\bigg)} \ = \ \int kdt$

$\displaystyle -ln\bigg|\frac{P-K}{P}\bigg| \ = \ kt+C$

$\displaystyle ln\bigg|\frac{P-K}{P}\bigg| \ = \ -kt-C$

$\displaystyle \bigg|\frac{P-K}{P}\bigg| \ = \ (e^{-kt})(e^{-C})$

$\displaystyle \frac{P-K}{P} \ = \ Ae^{-kt}, \ A \ = \ \pm \ e^{-C}$

$\displaystyle 1-\frac{K}{P} \ = \ Ae^{-kt}$

$\displaystyle \frac{K}{P} \ = \ 1-Ae^{-kt}$

$\displaystyle \frac{P}{K} \ = \ \frac{1}{1-Ae^{-kt}}$

$\displaystyle Hence, \ P(t) \ = \ \frac{K}{1-Ae^{-kt}}$

$\displaystyle Now \ K \ = \ 20 \ and \ k \ = \ .042, \ ergo \ P(t) \ = \ \frac{20}{1-Ae^{-.042t}}$

$\displaystyle Let \ 1960 \ = \ 0, \ then \ P(0) \ = \ 4 \ = \ \frac{20}{1-A}, \ \implies \ A \ = \ -4$

$\displaystyle Hence, \ P(t) \ = \ \frac{20}{1+4e^{-.042t}}, \ is \ this \ what \ you \ wanted?$

$\displaystyle Note: \ It \ will \ take \ about \ 23 \ years \ (1983) \ to \ reach \ a \ population \ of \ 8 \ million$

$\displaystyle and \ the \ \lim_{t\to\infty} P(t) \ = \ 20 \ million, \ the \ carrying \ capacity.$