Question

g Suppose that when a certain lake is stocked with fish, the birth and death rates ˇ and ı are both inversely propor- (a) Show that ????1 p????2 P.t/D 2ktC P0 ; where k is a constant. (b) If P0 D 100 and after 6 months there are 169 fish in the lake, how many will there be after 1 year?

Answer 1

k=1

P(12)=256

Explanation:

The model for population involving birth and death is given as:

`(dP)/(dt)=(b-d)P`

where b=birth rate and d=death rate.

If the birth and death rate are inversely proportional to
`√(P)`

`b=(A)/(√(P) ) \\d=(B)/(√(P) )` where A and B are constants of variation.

Substituting b and d into our model

`(dP)/(dt)=((A)/(√(P) )-(B)/(√(P) ))P\\(dP)/(dt)=(k)/(√(P) )P\\` where A-B=k, another constant

Simplifying using indices

`(dP)/(dt)={k}P^{1-(1)/(2) }\\(dP)/(dt)={k}P^(1)/(2) \\`

Next, we Separate Variables and Integrate both sides

`(dP)/(√(P) )={k}dt\`

`\int(dP)/(√(P) )=\int{k}dt\`

`2P^(1/2) =kt+C` where C is the constant of integration

`P(t) =((kt)/(2) +C)^2`

When t=0, P(t)=
`P_0`, C=
`√(P_0)`

`P(t) =((kt)/(2) +√(P_0))^2` as required.

(b)If
`P_0`=100, t=6 months, P(t)=169

`P(t) =((kt)/(2) +√(P_0))^2`

`169 =((6k)/(2) +√(100))^2\\√(169) =3k+100\\13=3k+10\\3k=13-10=3\\k=1`

Since we have found the constant k, we can then calculate the population after 1 year. Note that we use 12 months since we used month earlier to get k.

`P(t) =((kt)/(2) +√(P_0))^2`

`P(12) =((1X12)/(2) +√(100))^2\\=(6+10)^2=256`

Therefore the population after a year is 256.