# help please (logistic equation)

#### jellybean

##### New member
biologists stocked a lake with 400 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake? to be 10,000. the number of fish tripled in the first year.

(a)assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after t years.

(b)how long will it take for the population to increase to 5000?

#### skeeter

##### Elite Member
dP/dt = kP(M - P)

let P be in units of thousands of fish ...

dP/dt = kP(10 - P)

separate variables ...

dP/[P(10 - P)] = kdt

use partial fractions ...

(1/10)[1/P + 1/(10 - P)]dP = kdt

[1/P + 1/(10 - P)]dP = 10kdt

integrate ...

ln(P) - ln(10 - P) = 10kt + C

ln[P/(10 - P)] = 10kt + C

change to an exponential function ...

P/(10 - P) = Ce<sup>kt</sup>

when t = 0, P = 0.4 ...

0.4/(10 - 0.4) = C = 1/24

P/(10 - P) = (1/24)e<sup>kt</sup>

P = (10 - P)(1/24)e<sup>kt</sup>

24P = (10 - P)e<sup>kt</sup>

24P = 10e<sup>kt</sup> - Pe<sup>kt</sup>

24P + Pe<sup>kt</sup> = 10e<sup>kt</sup>

P(24 + e<sup>kt</sup>) = 10e<sup>kt</sup>

P = 10e<sup>kt</sup>/(24 + e<sup>kt</sup>)

P = 10/(1 + 24e<sup>-kt</sup>)

number of fish tripled in the first year, so P = 1.2 when t = 1

substitute those values in for P and t and solve for k ... you're done.