Population growth problem

[nate9519]

1. The Problem

In the mythical country of Myland, the population recently reached 10 million. At that time, a popular news magazine predicted that with an average yearly growth rate of 2%, the population would be 20 million in 50 years. How does this value (18 million) compare to the value predicted by the model which assumes that the increase in population is proportional to the population at time t?

2. Homework Equations

dp/dt = kP
p(0) = 10,000,000
p(50) = 20,000,000

3. Attempt at solution

the 2% is throwing me off. I don't know where to put it. Could I use an interest formula and make the growth rate compound yearly at 2%?

 

[Quantum Defect]

1. The Problem

In the mythical country of Myland, the population recently reached 10 million. At that time, a popular news magazine predicted that with an average yearly growth rate of 2%, the population would be 20 million in 50 years. How does this value (18 million) compare to the value predicted by the model which assumes that the increase in population is proportional to the population at time t?

2. Homework Equations

dp/dt = kP
p(0) = 10,000,000
p(50) = 20,000,000

3. Attempt at solution

the 2% is throwing me off. I don't know where to put it. Could I use an interest formula and make the growth rate compound yearly at 2%?

Doubling time = ca. 70/% growth rate ==> Doubling time for 2% growth rate : 70/2 = 35 years -- a bit slower than 2% growth rate.

ln[P(50)/P(0)] = [k year^-1 * 50 years] ==> k = 0.014 yr^-1 ==> 1.4% per year

A very informative video of a U. Colorado physics professor talking about exponential growth:

 

[Mark44]

1. The Problem

In the mythical country of Myland, the population recently reached 10 million. At that time, a popular news magazine predicted that with an average yearly growth rate of 2%, the population would be 20 million in 50 years. How does this value (18 million) compare to the value predicted by the model which assumes that the increase in population is proportional to the population at time t?

Is there a typo here? "This value" should be 20 million, right? Otherwise, where did the 18 million figure come from?

 

[Chestermiller]

The 2% growth rate means that k = 0.02, if t is measured in years. So, after 50 years, what would the solution to your equation predict that the population would be? How does that compare with the 20 million?

Chet

 

[nate9519]

Is there a typo here? "This value" should be 20 million, right? Otherwise, where did the 18 million figure come from?

no that's suppose to be 18 million. I copied the problem down word for word

 

[Ray Vickson]

no that's suppose to be 18 million. I copied the problem down word for word

There is something seriously wrong with the problem statement. At a growth rate of 2% per year, the population in 50 years is nowhere near 18 million or 20 million.

The exact answer depends on how literally you interpret the phrase "2% per year". If this means exactly what the words say, then $k$ is not exactly $0.02$, although it is close; call this value $k_0$ (and I will let you figure out the value of $k_0$). On the other hand, if you (somewhat sloppily?) interpret this to mean that $k = k_1$ with $k_1 = 0.02$, then over 50 years the population values using $k=k_0$ and $k = k_1$ differ by a bit more than 1/4 million.

 

[Chestermiller]

There is something seriously wrong with the problem statement. At a growth rate of 2% per year, the population in 50 years is nowhere near 18 million or 20 million.

The exact answer depends on how literally you interpret the phrase "2% per year". If this means exactly what the words say, then $k$ is not exactly $0.02$, although it is close; call this value $k_0$ (and I will let you figure out the value of $k_0$). On the other hand, if you (somewhat sloppily?) interpret this to mean that $k = k_1$ with $k_1 = 0.02$, then over 50 years the population values using $k=k_0$ and $k = k_1$ differ by a bit more than 1/4 million.

Considering the first Relevant Equation, I think they meant for the students to take k = 0.02, but there certainly is ambiguity here. When the growth is described by the first equation, the term "continuous compounding" is typically used in the description.

Chet

 

[Ray Vickson]

Considering the first Relevant Equation, I think they meant for the students to take k = 0.02, but there certainly is ambiguity here. When the growth is described by the first equation, the term "continuous compounding" is typically used in the description.

Chet

Of course, and that is my point: using $k = 0.02$ (per year) yields an actual yearly growth is a bit more than 2%. More important, however, is the fact that using $k = 0.02$ produces a 50-year result much different from either 18 or 20 million.

 

[Chestermiller]

Of course, and that is my point: using $k = 0.02$ (per year) yields an actual yearly growth is a bit more than 2%. More important, however, is the fact that using $k = 0.02$ produces a 50-year result much different from either 18 or 20 million.

Ah, yes. If it were close to 0.012 rather than 0.02, that would give about 18 million in 50 years using continuous compounding. Typo?? As Quatum Defect indicated in post #2, to get 20 million, you would need 0.014.

Chet

 

[nate9519]

There is something seriously wrong with the problem statement. At a growth rate of 2% per year, the population in 50 years is nowhere near 18 million or 20 million.

The exact answer depends on how literally you interpret the phrase "2% per year". If this means exactly what the words say, then $k$ is not exactly $0.02$, although it is close; call this value $k_0$ (and I will let you figure out the value of $k_0$). On the other hand, if you (somewhat sloppily?) interpret this to mean that $k = k_1$ with $k_1 = 0.02$, then over 50 years the population values using $k=k_0$ and $k = k_1$ differ by a bit more than 1/4 million.

Considering the first Relevant Equation, I think they meant for the students to take k = 0.02, but there certainly is ambiguity here. When the growth is described by the first equation, the term "continuous compounding" is typically used in the description.

Chet

does the word average change the interpretation of this vague problem? could that somehow mean the yearly growth rates over the 50 year period were averaged and gave 2%?

 

[Chestermiller]

does the word average change the interpretation of this vague problem? could that somehow mean the yearly growth rates over the 50 year period were averaged and gave 2%?

Not in my opinion. The problem would still be the same with both interpretations.

Chet

 

[Quantum Defect]

does the word average change the interpretation of this vague problem? could that somehow mean the yearly growth rates over the 50 year period were averaged and gave 2%?

If the growth is exponential, you could use two interpretations of the 2% growth, which would change the growth constant (k), but would not change the underlying mathematics.

(1) 2% growth rate ==> k = 0.02 year^-1 : This is the interpretation that Chet used in post # 2. This is a reasonable interpretation.

(2) "2% growth rate per year" ==> ln (1.02/1.00) = k year^-1 ==> k = 0.0198 year^-1 (almost .02 year^-1) -- this is what Ray is getting at in posts #6 & #8 -- this will always give you 2% growth from year to year.

(3) actual growth rate ==> 18 million in 50 years? ==> k*50 years = ln (18/10) ==> k = 0.0118 year^-1
(4) actual growth rate ==> 20 million in 50 years? ==> k*50 years = ln(2) ==> k = 0.0139 year^-1

The difference in interpretation between (1) and (2) can yield noticeable differences in 50 years:
(1) pop(50 years) = 27.18 million; (2) pop(50 years) = 26.91 million

 

[Ray Vickson]

does the word average change the interpretation of this vague problem? could that somehow mean the yearly growth rates over the 50 year period were averaged and gave 2%?

There are two kinds of "growth rate":

(1) Percentage (geometric), which is what you have assumed, and which is what is normally intended when the term is used by "technical" people in such contexts as population growth, investment growth, etc.

(2) Absolute (arithmetic) growth, which is what is often meant in non-technical circles, such as by the writer of your newspaper account. In this view, the 2% per year means (0.02)(10 million) = 200,000 persons per year--year in and year out--for 50 years. That would yield an increase of 10 million, so would take the population up to 20 million over a 50-year period. I still can't see where the 18 million comes in.