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

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`

Explanation:

From the question we are told that

The population model is
`(dP)/(dt) = 0.2P(1 - (P)/(135) )`

Generally at equilibrium

`(dP)/(dt) = 0`

So

`0.2P = 0`

=>
`P = 0`

Or

`(1 - (P)/(135) ) = 0`

=>
`P = 135`

Thus at equilibrium P = 0 or P = 135

Generally when the population is increasing we have that

`(dP)/(dt) > 0`

So

`0.2P > 0`

=>
`P > 0`

and

`(1 - (P)/(135) ) > 0`

`P < 135`

Now when the first value of P i.e
`P< 0` for
`(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

`(dP)/(dt) = (1)/(5)P (1 - (P)/(135) )`

So we will differentiate the equation again we have that

`(d^2 P)/(dt^2) = ((1 - (P)/(135) ))/(5) - (P)/(675)`

Now as
`t \to [\infty]`

`(d^2 P)/(dt^2) \to 0`

So

`((1 - (P)/(135) ))/(5)  - (P)/(675) = 0`

=>
`((1 - (P)/(135) ))/(5)   =  (P)/(675)`

=>
`P = 67.5`