Logistic Growth

Tags

Calculus

Cegep/2

Word count

242 words

Reading time

2 minutes

Growth rate of a function $y$ such that the derivative of $y$ is proportional to both $y$ and the difference between $y$ and the carrying capacity $M$

$$\begin{array}{rl}{y}^{\prime }& \propto y(M-y)\\ y^{\prime }& =ky(M-y)\\ y& =\frac{r}{M}y(M-y)\end{array}$$

Examples

A rumour is spreading in a population of 1700. Four people start the rumour and 3 days later 300 people know.

1. Find the function $P(t)$ that models the population after $t$ hours.
2. How many people will know after four days?

$$\begin{array}{rl}\frac{\mathrm{d}P}{\mathrm{d}t}& =\frac{r}{M}P(M-P)\\ \frac{\mathrm{d}P}{\mathrm{d}t}& =\frac{r}{1700}P(1700-P)\\ \int \frac{1700}{1700-P}\mathrm{d}P& =\int r\mathrm{d}t\\ \int \frac{1}{P}\mathrm{d}P+\int \frac{1}{1700-P}\mathrm{d}P& =rt+C\\ \ln |P|-\ln |1700-P|& =rt+C\\ \ln \left|\frac{P}{1700-P}\right|& =rt+C\\ \frac{P}{1700-P}>0⟹\frac{P}{1700-P}& =e^{rt+C}\\ \frac{P}{1700-P}& =\frac{1}{424}{e}^{rt}\\ P& =\frac{1700}{424}{e}^{rt}-\frac{P}{424}{e}^{rt}\\ P(1-\frac{1}{424}{e}^{rt})& =\frac{1700}{424}{e}^{rt}\\ P& =\frac{1700e^{rt}}{424+e^{rt}}\\ \\ 300& =\frac{1700e^{3r}}{424+e^{3r}}\\ 127200+300e^{3r}& =1700e^{3r}\\ r& \approx 1.5031\\ \therefore P(t)& =\frac{1700e^{1.5031t}}{424+e^{1.5031t}}\end{array}$$

$$P(4)=\frac{1700e^{1.5031\cdot 4}}{424+e^{1.5031\cdot 4}}\approx 834.135$$

Contributors

Ajitani Hifumi

Changelog

Last edited about 2 hours ago

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