Logistic Growth Models

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What is the logistic model of population growth?

Logistic Population Model

The logistic population $P (t)$ $P(t)$ can be expressed by

$P (t) = \frac{K P_{0}}{P_{0} + (K - P_{0}) e^{- r t}}$ $P(t)={KP_0}/{P_0+(K-P_0)e^{-rt}}$ ,

where $P_{0}$ $P_0$ is an initial population, $K$ $K$ is a carrying capacity, and $r$ $r$ is a growth constant.

Wataru · · Sep 28 2014
What is carrying capacity ?

The carrying capacity of a species is the maximum population of that species that the environment can sustain indefinitely, given available resources. It acts as an upper limit on population growth functions.

On a graph, assuming that the population growth function is depicted with the independent variable (usually $t$ $t$ in cases of population growth) on the horizontal axis, and the dependent variable (the population, in this case $f (x)$ $f(x)$ ) on the vertical axis, the carrying capacity will be a horizontal asymptote.

In the normal course of events, barring extreme circumstances, the population will not surpass the carrying capacity. However, some extreme circumstances (such as the sudden influx of more members of the population from external areas, along with certain natural cyclic variations) can cause the population to temporarily exceed the carrying capacity. This leads to a sharp decrease in the population (a "population crash") as resources become more scarce, leading to starvation and dehydration, as well as deaths caused by fighting over the now-scarce resources.

Psykolord1989 . · 1 · Sep 20 2014
How does logistic growth occur?

Logistic growth of population occurs when the rate of its growth is proportional to the product of the population and the difference between the population and its carrying capacity $M$ $M$ , i.e.,

$\frac{d P}{d t} = k P (M - P)$ ${dP}/{dt}=kP(M-P)$ , where $k$ $k$ is a constant,

with initial population $P (0) = P_{0}$ $P(0)=P_0$ .

As you can see above, the population grows faster as the population gets larger; however, as the population gets closer to its carrying capacity $M$ $M$ , the growth slows down.

by separating variables and integrating,

$\Rightarrow \int \frac{1}{P (M - P)} d P = \int k d t$ $Rightarrow int 1/{P(M-P)}dP=int kdt$

by Partial Fraction Decomposition,

$\Rightarrow \frac{1}{M} \int (\frac{1}{P} + \frac{1}{M - P}) d P = \int k d t$ $Rightarrow 1/M int(1/P+1/{M-P})dP=int kdt$

by multiplying by $M$ $M$ ,

$\Rightarrow \int (\frac{1}{P} + \frac{1}{M - P}) d P = \int k M d t$ $Rightarrow int(1/P+1/{M-P})dP=int kMdt$

$\Rightarrow ln | P | - ln | M - P | = k M t + C_{1}$ $Rightarrow ln|P|-ln|M-P|=kMt+C_1$

$\Rightarrow ln ∣ ∣ ∣ \frac{P}{M - P} ∣ ∣ ∣ = k M t + C_{1}$ $Rightarrow ln|P/{M-P}|=kMt+C_1$

$\Rightarrow ∣ ∣ ∣ \frac{P}{M - P} ∣ ∣ ∣ = e_{1}^{k M t + C_{1}} = e^{k M t} \cdot e_{1}^{C_{1}}$ $Rightarrow |P/{M-P}|=e^{kMt+C_1}=e^{kMt}cdot e^{C_1}$

$\Rightarrow \frac{P}{M - P} = \pm e_{1}^{C_{1}} e^{k M t} = C e^{k M t}$ $Rightarrow P/{M-P}=pm e^{C_1}e^{kMt}=Ce^{kMt}$

Since $P (0) = P_{0}$ $P(0)=P_0$ ,

$\frac{P_{0}}{M - P_{0}} = C e^{k M (0)} = C$ $P_0/{M-P_0}=Ce^{kM(0)}=C$

So, the equation becomes

$\frac{P}{M - P} = \frac{P_{0}}{M - P_{0}} e^{k M t}$ $P/{M-P}=P_0/{M-P_0}e^{kMt}$

by solving for $P$ $P$ , we have the logistic equation

$\Rightarrow P (t) = \frac{M}{1 + (\frac{M}{P_{0}} - 1) e^{- k M t}}$ $Rightarrow P(t)=M/{1+(M/P_0-1)e^{-kMt}}$ .

I hope that this was helpful.

Wataru · 1 · Oct 16 2014

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Applications of Definite Integrals

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Logistic Growth Models

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Applications of Definite Integrals