Using t=(1/k) ln (N/No), determine when at the growth rate of 1.24% per year, the worlds population of 6.16 billion in 2001 would grow to occupy the earth at a density of 1 person per square meter (1 person/m^2) of dry land. The earth has 1.31 x 10^14 m^2 of dry land. Thus this number also represents the population size that would fill the earth if only one person would occupy the area of one meter.
To determine when the world's population would reach a density of 1 person per square meter of dry land, we can use the formula for exponential growth:
N = No * e^(kt)
Where:
N = final population size (unknown)
No = initial population size (6.16 billion)
k = growth rate per year (1.24% or 0.0124)
t = time in years (unknown)
First, let's divide the density of 1 person per square meter by the total dry land area to find the population size that would fill the entire earth with 1 person per square meter:
Population density = 1 person/m^2
Dry land area = 1.31 x 10^14 m^2
Population size = Population density * Dry land area
Population size = 1 * (1.31 x 10^14)
Population size = 1.31 x 10^14
Now we can set up an equation using the exponential growth formula:
1.31 x 10^14 = 6.16 billion * e^(0.0124t)
To solve for t, we need to rearrange the equation:
e^(0.0124t) = (1.31 x 10^14) / (6.16 billion)
Now take the natural logarithm (ln) of both sides to isolate t:
ln(e^(0.0124t)) = ln((1.31 x 10^14) / (6.16 billion))
Using the properties of logarithms, ln(e^(0.0124t)) simplifies to 0.0124t:
0.0124t = ln((1.31 x 10^14) / (6.16 billion))
Now divide both sides by 0.0124 to solve for t:
t = ln((1.31 x 10^14) / (6.16 billion)) / 0.0124
Using a scientific calculator or programming language with a natural logarithm (ln) function, you can calculate the value of t to determine when the world's population would reach the desired density.