The logistic function is
y = a/(1+be^(-kx))
y(∞) = 300
==> a = 300
y(0) = 40
==> 300/(1+b) = 40
b = 15/2
The k just affects how slowly the population approaches the final value. See
http://www.wolframalpha.com/input/?i=plot+y%3D300%2F(1%2B7.5e%5E-x),+y%3D300%2F(1%2B7.5e%5E(-2x))+for+-5+%3C%3D+x+%3C%3D+5
I am having trouble figuring out where to start with this one.
Biologists stocked a lake with 40 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 300.
(a) Assume that the number of fish P(t) after t years grows according to the logistic equation
dP/dt = P(a − bP) = bP(a/b − P)
Here, according to Section 2.15 of the course notes, a/b is the carrying ca-pacity of the lake. Find the solution of this DE (with the given initial value at t = 0) as a separable equation. Your solution formula will contain b as an unknown parameter.
1 answer