§7.6 Population growth and the logistic equation

Introduction

The earth's population

Solving the logistic differential equation

Summary

For $\frac{dP}{dt} = \frac{1}{2}P$, the slope field and typical solutions show that

Population data gives $P(-1) = 6.008$ and $P(1) = 6.159$ billion. The central difference estimate $P'(0)$ equals

Using $P(0) = 6.084$ and $P'(0) \approx 0.0755$ in the model $\frac{dP}{dt} = kP$ gives

With $k \approx 0.01241$ and $P(0) = 6.084$ billion, the natural growth model's solution is

For the logistic model $\frac{dP}{dt} = kP(N-P)$, the growth rate is greatest when

The model $\frac{dP}{dt} = P(0.025-0.002P)$ can be rewritten as $\frac{dP}{dt} = 0.002P(12.5-P)$. The population grows fastest when

With $p(0) = 0.1$ and $\frac{dp}{dt} = 0.2p(1-p)$, the solution is

Bacteria measured at $6000$ with per capita rate $3$ and at $9000$ with per capita rate $2$ obey a linear per capita law. The resulting differential equation is

Fish obeying $\frac{dP}{dt} = 0.1P(10-P)$ are harvested at $20\%$ per year, giving $\frac{dP}{dt} = 0.1P(10-P)-0.2P$. The new carrying capacity is

With $k \approx 0.01241$ and $P(0) = 6.084$ billion, the model $P = 6.084e^{0.01241t}$ predicts $P(10)$. Report the value, rounded to three decimals.

The exponential model $P = 6.084e^{0.012041t}$ reaches $12$ billion when $t = \frac{1}{0.012041}\ln\left(\frac{12}{6.084}\right)$. Report $t$, rounded to one decimal.

For the model $\frac{dP}{dt} = 0.002P(12.5-P)$, the population grows fastest at $P = \frac{12.5}{2}$. Report that value in billions.

The logistic model $P = \frac{12.5}{1.0546e^{-0.025t}+1}$ predicts the population in the year $2100$. Evaluate $P(100)$, rounded to three decimals.

The logistic model $P = \frac{12.5}{1.0546e^{-0.025t}+1}$ reaches $9$ billion when $t = \frac{1}{-0.025}\ln\left(\frac{12.5/9-1}{1.0546}\right)$. Report $t$, rounded to one decimal.

With $p(t) = \frac{1}{9e^{-0.2t}+1}$, the rumor spreads fastest when $p = 0.5$, which happens when $e^{-0.2t} = \frac{1}{9}$. Find that time in days, rounded to three decimals.

The rumor reaches $80\%$ of the group when $9e^{-0.2t} = \frac{1-0.8}{0.8} = \frac{1}{4}$, that is $e^{-0.2t} = \frac{0.25}{9}$. Find that time in days, rounded to two decimals.

With $b(t) = \frac{15000}{14e^{-5t}+1}$, the colony reaches $80\%$ of capacity when $e^{-5t} = \frac{1}{70}$. Find the time in days, rounded to three decimals.

Under harvesting, the fish population follows $P(t) = \frac{8}{-0.2e^{-0.8t}+1}$ thousand, starting from $P(0) = 10$. Compute $P(1)$, rounded to three decimals.

The pure exponential model $P = 6.084e^{0.01241t}$ predicts roughly $3012$ billion people in the year $2500$, exposing the need for a logistic correction.

For the logistic solution $P(t) = \frac{N}{\left(\frac{N-P_0}{P_0}\right)e^{-kNt}+1}$, direct substitution gives $P(0) = P_0$ and $\lim_{t\to\infty}P(t) = N$.

For the bacteria model $\frac{db}{dt} = \frac{1}{3000}b(15000-b)$, the colony grows fastest when $b = 7500$, halfway between the roots of the rate parabola.

In the harvested fish model, the population starts above the new capacity $8000$ and decreases toward it, staying above capacity at every finite time.