Q. 17 - Calculus

Question

Suppose a population P(t) of animals on a small island grows according to a logistic model of the form $\frac{dP}{dt}=kP(1-\frac{P}{100})$for some constant $K$.

 

(a) What is the carrying capacity of the island under this model? 

(b) Given that the population $P\left(t\right)$ is growing and that $0<P\left(t\right)<500$, is the constant k positive or negative, and why? 

(c) Explain why $\frac{dP}{dt}\approx kP$for small values of $P$.

 (d) Explain why $\frac{dP}{dt}\approx 0$for values of $P$ that are close to the carrying capacity

Step-by-Step Solution

Answer

Ans:  

(a)   The carrying capacity of the island is $500$.

(b)   Constant $k$ is positive.

(c)    Approximated by,    $\begin{array}{r}\frac{dP}{dt}\approx kP(1-0)\\ \approx kP\end{array}$

(d)    

 $\begin{array}{r} \frac{dP}{dt}\approx kP\cdot 0\\ \approx 0\end{array}$ 

The growth rate is zero when the population approaches carrying capacity.

Step 1. Given information.

given,

        $\frac{dP}{dt}=kP\left(1-\frac{P}{100}\right)$

Step 2. (a) Remember that the standard equation of the population growth model with carrying capacity is given by

 $\frac{dP}{dt}=kP\left(1-\frac{P}{L}\right)$

find that $L=500$. Therefore, the carrying capacity of the island is $500$.

Step 3. (b) About constant:

It is given that the population $P\left(t\right)$ is growing. This means that the rate of growth of the population is positive. Mathematically, it implies that $\frac{dP}{dt}$ is positive or $\frac{dP}{dt}>0$. So, the left-hand side of equation (1) is positive. Now, on the right-hand side $P$ is positive, and P is less than $500$, the factor $1-\frac{P}{500}$ is positive and less than $1$ . That is $0<1-\frac{P}{500}<1$ Hence, for the right-hand side to be positive the constant $k$ must be positive. Therefore, in the growing population model (1), the constant $k$ is positive.

Step 4. (c) When P is small,

The quality $\frac{P}{500}$ will be negatively small. This implies the differential equation can be approximated by

    $\begin{array}{r}\frac{dP}{dt}\approx kP(1-0)\\ \approx kP\end{array}$

Therefore, for a small value of $P, \begin{array}{r}\frac{dP}{dt}\approx kP\\ \end{array}$

Step 5. (d) For value of P close to the carrying capacity P &#8776; 500 . &#160;So,&#160; P 500 &#8776; 1 and then 1 - P 500 &#8776; 0 .

In such case

    $\begin{array}{r}\frac{dP}{dt}\approx kP\cdot 0\\ \approx 0\end{array}$

That is the growth rate is zero when the population approaches carrying capacity.