Question

1. Consider the population model dP dt = 0.2P 1 − P 135 , where P(t) is the population at time t. (a) For what values of P is the population in equilibrium? (b) For what values of P is the population increasing? (c) What is the carrying capacity? (d) For which initial values of P does the population converge to the carrying capacity as t → [infinity]? 2. Consider the differential equation

Answer 1

Answer:

a

  $P = 0$  OR  $P = 135$

b

$P > 0$ and $P < 135$

OR

$P > 0$ and $P < 135$

c

Generally the carrying capacity is can be defined as the highest amount of population and environment can support for an unlimited duration or time period

d

$P = 67.5$

Step-by-step explanation:

From the question we are told that

The population model is $\frac{dP}{dt} = 0.2P(1 - \frac{P}{135} )$

Generally at equilibrium

$\frac{dP}{dt} = 0$

So

$0.2P = 0$

=> $P = 0$

Or

$(1 - \frac{P}{135} ) = 0$

=> $P = 135$

Thus at equilibrium P = 0 or P = 135

Generally when the population is increasing we have that

$\frac{dP}{dt} > 0$

So

$0.2P > 0$

=> $P > 0$

and

$(1 - \frac{P}{135} ) > 0$

$P < 135$

Now when the first value of P i.e $P< 0$ for $\frac{dP}{dt} > 0$

$P_2 > 135$

So when population increasing the values of P are

$P > 0$ and $P < 135$

OR

$P > 0$ and $P < 135$

So to obtain initial values of P where the population converge to the carrying capacity as $t \to [\infty]$

The rate equation can be represented as

$\frac{dP}{dt} = \frac{1}{5}P (1 - \frac{P}{135} )$

So we will differentiate the equation again we have that

$\frac{d^2 P}{dt^2} = \frac{(1 - \frac{P}{135} )}{5} - \frac{P}{675}$

Now as  $t \to [\infty]$

$\frac{d^2 P}{dt^2} \to 0$

So

   $\frac{(1 - \frac{P}{135} )}{5} - \frac{P}{675} = 0$      

=>    $\frac{(1 - \frac{P}{135} )}{5} = \frac{P}{675}$

=> $P = 67.5$