Given that the population of the earth was 6.0 billion people in 1999 (t = 0) and 6.9 billion people in 2009 (t = 10), find: a) an exponential growth function for the world’s population that fits that two data points. b) Find the doubling time for the world population using the model in part a). c) Find the instantaneous growth rate of the population in 2014 (t = 15).
Since we're interested in doubling, let's start with
p(t) = 6.0*2^(kt)
Now we know that p(10) = 6.9, so
6.0*2^(10k) = 6.9
2^(10k) = (6.9/6.0) = 1.15
10k = log1.15/log2
k = 0.02
So, p(t) = 6.0*2^(t/50)
The doubling time is 50 years
If you insist on using base e, then that would be 6.0e^(ln2/50 t)
dp/dt = 6.0 * ln2 * 1/50 * 2^(t/50) = 0.083*2^(t/50)
at t=15, that is 0.102
a) To find an exponential growth function for the world's population, we can use the formula:
P(t) = P0 * e^(kt)
Where:
P(t) is the population at time t
P0 is the initial population at time t=0
e is the base of natural logarithms (approximately 2.71828)
k is the constant growth rate
t is the time
We can use the given data points to form two equations:
P(0) = 6.0 billion
P(10) = 6.9 billion
Using the first data point (t = 0):
6 = P0 * e^(0 * k)
6 = P0
So we have P0 = 6 billion.
Using the second data point (t = 10):
6.9 = 6 * e^(10k)
e^(10k) = 6.9/6
e^(10k) ≈ 1.15
10k ≈ ln(1.15)
k ≈ ln(1.15) / 10
Therefore, the exponential growth function for the world's population is:
P(t) = 6 * e^(ln(1.15)/10 * t)
b) To find the doubling time for the world population using the model in part a), we need to solve the equation P(t) = 2P0, where P0 is the initial population.
2P0 = 6 * e^(ln(1.15)/10 * t)
e^(ln(1.15)/10 * t) = 2/6
ln(1.15)/10 * t ≈ ln(2/6)
t ≈ ln(2/6) * 10 / ln(1.15)
Therefore, the doubling time for the world population using the given model is approximately ln(2/6) * 10 / ln(1.15).
c) To find the instantaneous growth rate of the population in 2014 (t = 15), we can differentiate the exponential growth function:
P'(t) = 6 * ln(1.15)/10 * e^(ln(1.15)/10 * t)
Substituting t = 15:
P'(15) = 6 * ln(1.15)/10 * e^(ln(1.15)/10 * 15)
Evaluate the expression to find the instantaneous growth rate of the population in 2014.