If a population consists of 10,000 individuals at time t=0 years (P0), and the annual growth rate (excess of births over deaths) is 3% (GR), what will the population be after 1, 15 and 100 years (n)? Calculate the "doubling time" for this growth rate. Given this growth rate, how long would it take for this population of 10,0000 individuals to reach 1.92 million?
One equation that may be useful is:
Pt = Po * (1 + {GR/100})n
Apply the given formula,
P(t) = Po(1+r)n
where
r=0.03 is the annual growth rate in percent divided by a hundred.
n=number of years
Po=initial population=10000
For 1 year,
P(1) = 10000*(1.03)1
=10300
P(15)=10000*(1.03)15
= ______
P(100)=10000*(1.03)100
=_______
The doubling time is rougely 24 years using the rule of 72 (72 divided by the annual rate of interest).
The exact doubling time is log(2)/log(1.03)
=23.45 years.
Time required to grow from 10000 to 1920000 is
log(1920000/10000)/log(1.03)
=______ years
To calculate the population after a certain number of years, we can use the formula Pt = Po * (1 + (GR/100))^n, where Pt is the future population, Po is the initial population, GR is the growth rate, and n is the number of years.
Let's calculate the population after 1, 15, and 100 years using the given information:
1. After 1 year:
Pt = 10,000 * (1 + (3/100))^1
= 10,000 * (1 + 0.03)
= 10,000 * 1.03
= 10,300
So, after 1 year, the population will be 10,300.
2. After 15 years:
Pt = 10,000 * (1 + (3/100))^15
= 10,000 * (1 + 0.03)^15
= 10,000 * 1.488641
≈ 14,886
So, after 15 years, the population will be approximately 14,886.
3. After 100 years:
Pt = 10,000 * (1 + (3/100))^100
= 10,000 * (1 + 0.03)^100
= 10,000 * 2.704814
≈ 27,048
So, after 100 years, the population will be approximately 27,048.
Now, let's calculate the doubling time for this growth rate. The doubling time represents the number of years it takes for the population to double in size.
To find the doubling time, we can use the formula:
t_double = (log(2) / log(1 + (GR/100)))
Let's calculate the doubling time:
t_double = (log(2) / log(1 + (3/100)))
≈ (0.693147 / 0.02956)
≈ 23.451
So, the doubling time for this growth rate is approximately 23.451 years.
Finally, let's determine how long it would take for the population of 10,000 individuals to reach 1.92 million.
We need to find the number of years it takes for the population to grow from 10,000 to 1.92 million.
1.92 million can be expressed as 1,920,000.
We'll set up the equation and solve for n:
1,920,000 = 10,000 * (1 + (3/100))^n
Divide both sides of the equation by 10,000:
192 = (1.03)^n
Using logarithms, we can solve for n:
log(192) = n * log(1.03)
n = log(192) / log(1.03)
≈ 22.54
Therefore, it would take approximately 22.54 years for the population of 10,000 to reach 1.92 million.