Biologists stocked a lake with 2200(half of the carrying capacity)? It will take

Question

Biologists stocked a lake with 2200(half of the carrying capacity)? It will take years. P(t) =. (b) How long will it take for the population to increase to k = t, years. k, and then solve the equation to find an expression for the size of the population after dp/dt = (kP(1-p/k), determine the constant 4400. The number of fish doubled in the first year. (a) Assuming that the size of the fish population satisfies the logistic equation 400 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be

Explanation / Answer

The formula (logistic equation) you have to use is as follows:

P(t) = Po*K/(Po + (K - Po)e^-kt)
Note: Notice the capitol K and the small k (they represent different things)

K = max carrying capacity
Po = initial population

Given:
Po = 300
P1 = 300 * 3 = 900.....since the population is doubled
K = 6000

Thus, plug into the equation:
so you get,
P(t) = (300*6000)/(300+5700*e^-kt)

Now we know tht P1 is 900 (after the population is tripled)
thus, we can plug in P(t) = 900 and t=1 (t=1, because the population triples after 1 year)

so, we get,

900 = (300*6000)/(300(1+19*e^-k(1)))
thus, 300 cancels out

we are left with:
900 = 6000/(1+19*e^-k)

now solve for k (small k):
1+19*e^-k = 6000/900
19*e^-k = (60/9)-1
e^-k = ((60/9)-1)/19

now take ln of both sides, thus,

-k = ln(0.298)
-k = -1.210
k = 1.210

now, plug this k value into the equation, and you have your answer!

so, the final answer is:
P(t) = 6000/(1+19*e^-1.210t)

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