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Drupady [299]
3 years ago
5

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 th

e 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
Mathematics
1 answer:
Kryger [21]3 years ago
5 0

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

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