Q. 69 - Calculus

Question

 Suppose 100 rabbits are shipwrecked on a deserted island and their population P(t) after t years is determined by a logistic growth model, where the natural growth rate of the rabbits is k = 0.1 and the carrying capacity of the island is 1000 rabbits.

 (a) Set up a differential equation describing $\frac{dp}{dt}$ ,and solve it to get a formula for the population P(t) of rabbits on the island in t years.

 (b) Sketch a graph of the population P(t) of rabbits on the island over the next 100 years.

(c) It turns out that a population governed by a logistic model will be growing fastest when the population is equal to exactly half of the carrying capacity. In how many years will the population of rabbits be growing the fastest? 

Step-by-Step Solution

Answer

(a) $\mathrm{P}\left(\mathrm{t}\right)=\frac{1000}{1+9{\mathrm{e}}^{-0.1\mathrm{t}}}$

(b) Graph 

(c) The population of rabbit will be growing at fastest rate in about 21.97 years. 

Step 1. Given

100 rabbits are shipwrecked on a deserted island and their population P(t) after t years is determined by a logistic growth model, where the natural growth rate of the rabbits is k = 0.1 and the carrying capacity of the island is 1000 rabbits. 

Part(a) Step 2. Calculation

$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{Observe} \mathrm{that} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{scenario}, \mathrm{population} \mathrm{growth} \mathrm{is} \mathrm{bounded} \mathrm{by} \mathrm{the} \mathrm{carrying} \mathrm{capacity} \mathrm{L}. \\ \mathrm{So}, \mathrm{the} \mathrm{problem} \mathrm{is} \mathrm{represented} \mathrm{by} \mathrm{the} \mathrm{logistic} \mathrm{growth} \mathrm{model} \mathrm{of} \mathrm{the} \mathrm{form} \\ \frac{dP}{dt}=kP(1-\frac{P}{L}) \\ \mathrm{Use} \mathrm{the} \mathrm{given} \mathrm{data}, \mathrm{that} \mathrm{i}s, k=0.1,L=1000 \mathrm{in} \mathrm{the} \mathrm{above} \mathrm{equation} \mathrm{to} \mathrm{get} \mathrm{the} \mathrm{logistic} \mathrm{model} \mathrm{representing} \mathrm{the} \mathrm{growth} \mathrm{rate} \mathrm{as} \\ \frac{dP}{dt}=0.1 P(1-\frac{P}{10000}) \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{initial} \mathrm{population} \mathrm{of} \mathrm{the} \mathrm{rabbits} \mathrm{is} 100. \mathrm{This} \mathrm{means} \mathrm{that} P\left(0\right)=100. \mathrm{So}, \mathrm{the} \mathrm{model} \mathrm{representing} \mathrm{the} \mathrm{initial}-\mathrm{value} \mathrm{problem} \mathrm{is} \\ \frac{dP}{dt}=0.1 P(1-\frac{P}{10000}); P\left(0\right)=100...... \left(1\right) \\ \mathrm{Now}, \mathrm{proceed} \mathrm{to} \mathrm{solve} \mathrm{the} \mathrm{initial}-\mathrm{value} \mathrm{problem}. \\ \mathrm{Observe} \mathrm{that} \mathrm{the} \mathrm{differential} \mathrm{equation} \mathrm{does} \mathrm{not} \mathrm{involve} \mathrm{the} \mathrm{independent} \mathrm{variable} \mathrm{at} \mathrm{all}, \mathrm{so} \mathrm{solve} \mathrm{the} \mathrm{differential} \mathrm{equation} \mathrm{by} \mathrm{antidifferentiation} \mathrm{method}. \\ \frac{dP}{ P(1-\frac{P}{10000})}=0.1\int dt \\ \mathrm{The} \mathrm{integrand} \mathrm{on} \mathrm{the} \mathrm{left} \mathrm{side} \mathrm{needs} \mathrm{to} \mathrm{b} \mathrm{e} \mathrm{simplified} \mathrm{using} \mathrm{integration}(\mathrm{partial} \mathrm{fractions} ) \\ \mathrm{So}, \mathrm{first} \mathrm{resolve} \mathrm{the} \mathrm{fraction} \mathrm{into} \mathrm{partial} \mathrm{fraction} \mathrm{by} \mathrm{using} \mathrm{cover} \mathrm{up} \mathrm{rule} \\ \frac{1}{ P(1-\frac{P}{10000})}=\frac{A}{P}+\frac{B}{1-\frac{P}{10000}} \\ A=1 \\ B=\frac{1}{1000} \\ \frac{1}{ P(1-\frac{P}{10000})}=\frac{1}{P}+\frac{\frac{1}{1000}}{1-\frac{P}{10000}} \\ \mathrm{Hence}, \\ \int \frac{1}{P}dP+\frac{1}{1000}\int \frac{dP}{1-\frac{P}{10000}}=0.1\int dt \\ \mathrm{Simplifying} \mathrm{the} \mathrm{above} \mathrm{result} \mathrm{we} \mathrm{got} \\ \frac{1000\mathrm{P}}{1000-\mathrm{P}}={\mathrm{Ce}}^{0.1\mathrm{t}} \\ \frac{1000-\mathrm{P}}{1000\mathrm{P}}={\mathrm{Ae}}^{0.1\mathrm{t}} \mathrm{where} \mathrm{A}=\frac{1}{\mathrm{C}} \\ \frac{1000}{\mathrm{P}}=1+{\mathrm{Ae}}^{0.1\mathrm{t}} \\ \mathrm{now} \mathrm{take} \mathrm{P}=100,\mathrm{t}=0 \\ \mathrm{we} \mathrm{got} \mathrm{A}=9 \\ \mathrm{Hence} \mathrm{the} \mathrm{final} \mathrm{equation} \mathrm{is} \\ \mathrm{P}\left(\mathrm{t}\right)=\frac{1000}{1+9{\mathrm{e}}^{-0.1\mathrm{t}}} \end{gathered}$$

Part(b) Step 3. Graph

The graph of the population P(1) of rabbits on the island over the next 100 years, as

given below

Part(c) Step 4. Finding T

In this part it is required to find the time t in years when the population of rabbits will be growing at the fastest rate, given that the growth rate is fastest when the population has reached exactly half of the carrying capacity. So, take P(t) = 500 in equation (2) and solve for t

$$\begin{gathered} 500=\frac{1000}{1+9e^{-0.1t}} \\ 1+9e^{-0.1t}=2 \\ {e}^{0.1t}=9 \\ t=\frac{1}{0.1}\ln 9 \end{gathered}$$

Solve the above equation further to get t = 21.97 . Therefore, the population of rabbit will be growing at fastest rate in about 21.97 years.