Question

n many population growth problems, there is an upper limit beyond which the population cannot grow. Many scientists agree that the earth will not support a population of more than 16 billion. There were 2 billion people on earth at the start of 1925 and 4 billion at the beginning of 1975. If yy is the population, measured in billions, tt years after 1925, an appropriate model is the differential equation

Answer 1

Answer:

$\frac{dP}{dt} = rP(1 - \frac{P}{K}) = 0.017P(1 - \frac{P}{16})$

Step-by-step explanation:

The logistic function of population growth, that is, the solution of the differential equation is as follows:

$P(t) = \frac{KP_{0}e^{rt}}{K + P_{0}(e^{rt} - 1)}$

We use this equation to find the value of r.

In this problem, we have that:

$K = 16, P_{0} = 2, P(50) = 4$

So we find the value of r.

$P(t) = \frac{KP_{0}e^{rt}}{K + P_{0}(e^{rt} - 1)}$

$4 = \frac{16*2e^{50r}}{16 + 2*(e^{50r} - 1)}$

$4 = \frac{32e^{50r}}{14 + 2e^{50r}}$

$56 + 8e^{50r} = 32e^{50r}}$

$24e^{50r} = 56$

$e^{50r} = 2.33$

Applying ln to both sides of the equality

$50r = 0.8459$

$r = 0.017$

So

The differential equation is

$\frac{dP}{dt} = rP(1 - \frac{P}{K}) = 0.017P(1 - \frac{P}{16})$