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

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

[tex] P = 67.5 [/tex]

Step-by-step explanation:

From the question we are told that

The population model is [tex]\frac{dP}{dt} = 0.2P(1 - \frac{P}{135} )[/tex]

Generally at equilibrium

[tex]\frac{dP}{dt} = 0 [/tex]

So

[tex]0.2P = 0[/tex]

=> [tex] P = 0[/tex]

Or

[tex](1 - \frac{P}{135} ) = 0 [/tex]

=> [tex] P = 135[/tex]

Thus at equilibrium P = 0 or P = 135

Generally when the population is increasing we have that

[tex]\frac{dP}{dt} > 0 [/tex]

So

[tex]0.2P > 0[/tex]

=> [tex]P > 0[/tex]

and

[tex](1 - \frac{P}{135} ) > 0 [/tex]

[tex] P < 135 [/tex]

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

[tex] P_2 > 135 [/tex]

So when population increasing the values of P are

[tex]P > 0[/tex] and [tex] P < 135 [/tex]

OR

[tex]P > 0[/tex] and [tex] P < 135 [/tex]

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

The rate equation can be represented as

[tex]\frac{dP}{dt} = \frac{1}{5}P (1 - \frac{P}{135} )[/tex]

So we will differentiate the equation again we have that

[tex]\frac{d^2 P}{dt^2} = \frac{(1 - \frac{P}{135} )}{5} - \frac{P}{675}[/tex]

Now as [tex]t \to [\infty][/tex]

[tex]\frac{d^2 P}{dt^2} \to 0 [/tex]

So

[tex] \frac{(1 - \frac{P}{135} )}{5} - \frac{P}{675} = 0[/tex]

=> [tex] \frac{(1 - \frac{P}{135} )}{5} = \frac{P}{675}[/tex]

=> [tex] P = 67.5 [/tex]