Question 1027952: Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions the population growth follows a logistic growth model.
p(t)=d/1+ke^-ct
where c,d,and k are positive constants. For a certain fish population in a small pond d= 1200, k= 11, c = 0.2, and t is measured in years. The fish were introduced into the pond at time = 0.
a)
How many fish were originally put into the pond?
b)
Find the population of fish
after 10, 20, and 30 years.
c)
Evaluate P(t) for large values of t. What value does the population approach as
t→∞?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! it appears that the formula is:
p(t) = d/(1+ke^-ct)
that set of parentheses is very important because it changes the normal order of operations.
using the normal order of operations without parentheses, you would get:
p(t) = (d/1) + (ke^-ct)
that doesn't make a lot of sense because you would be getting no growth at all.
i'm assuming the formula is p(t) = d/(1+ke^-ct)
under that assumption, the population will grow to 1200.
when d = 1200 and k = 11 and c = .2, you get:
when t = 0, p(0) = 1200 / (1 + 11 * e^(-.2*0) which becomes p(0) = 1200 / (1 + 11 * e^0) which becomes p(0) = 1200 / (1 + 11) which becomes p(0) = 1200 / 12 which becomes p(0) = 100
the initial number of fish is therefore 100.
when t = 10, p(10) = 1200 / (1 + 11 * e^(-.2*10) which becomes p(10) = 1200 / (1 + 11 * e^-2) which becomes p(10) = 1200 / (1 + 1.488688116) which becomes p(10) = 1200 / (2.488688116) which becomes p(10) = 482 rounded to the nearest integer.
when t = 20, p(20) = 1200 / (1 + 11 * e^(-.2*20) which becomes p(20) = 1200 / (1 + 11 * e^-4) which becomes p(20) = 1200 / (1 + .2014720278) which becomes p(20) = 1200 / (1.2014720278) which becomes p(20) = 999 rounded to the nearest integer.
when t = 30, p(30) = 1200 / (1 + 11 * e^(-.2*30) which becomes p(30) = 1200 / (1 + 11 * e^(-6) which becomes p(20) = 1200 / (1 + .0024787522) which becomes p(30) = 1200 / 1.0024787522) which becomes p(30) = 1197 rounded to the nearest integer.
it is starting to become clear that, as t gets larger, e^(-.2*t) becomes smaller.
this leads to the conclusion as t approaches infinity, e^(-.2*t) will approach 0.
p(infinity) will therefore becomes 1200 / (1 + 0) which becomes 1200 / 1 which becomes 1200.
the number of fish in the pond will saturate at 1200, given that d = 1200 and k = 11 and c = .2 in the formula of p(t) = d / (1 + e^(-ct))
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