Logistic Growth

Question

The carrying capacity for deer in a particular small town is 2,200, and the rate of increase in their numbers is proportional to both the number, n, of deer and 2,200 - n. If there were 1,000 deer one month ago and 1,150 deer now, how many months will it take the deer to number 2,100?

Answer 1

The words describe the differential equation:

where "n" is the deer population,, "t" is time in months, "L" is the limit or "carrying capacity" and "k" is a constant pf proportionality. But since this problem is listed under precalculus and you have not as yet learned to solve this differential equation, I will assume they have given you the solution based upon the criteria noted. The solution is:

where "n" again is the deer population,, "t" is time in months, "L" is the limit or "carrying capacity" and this time there are two constants that need to be determined, both "b" and "k".

We have two data points given, (-1, 1000) and (0, 1150) so we can use those to find the missing constants:

Using (0, 1150):

1150 = 2200/(1 + be^-k(0))

1150 = 2200/(1 + b(1))

1150 = 2200/(1 + b)

1150(1 + b) = 2200

1150 + 1150b = 2200

1150b = 1050

b = 0.91304

So now the equation is:

n(t) = 2200/(1 + 0.91304e^-kt)

Using (-1, 1000):

1000 = 2200/(1 + 0.91304e^-k(-1))

1000 = 2200/(1 + 0.91304e^k)

1000(1 + 0.91304e^k) = 2200

1000 + 913.04e^k = 2200

913.04e^k = 1200

e^k = 1.3143

ln(e^k) = ln(1.3143)

k = 0.2733

So the final function is:

n(t) = 2200/(1 + 0.91304e^-0.2733t)

Now, solve for "t" when n = 2100

2100 = 2200/(1 + 0.91304e^-0.2733t)

2100(1 + 0.91304e^-0.2733t) = 2200

2100 + 1917.4e^-0.2733t = 2200

1917.4e^-0.2733t = 100

e^-0.2733t = 0.05215

ln(e^-0.2733t) = ln(0.05215)

-0.2733t = -2.9536

t = 10.8 months

Answer 2

I withdraw my answer because after some research I found:

the discrete time dynamic system logistical equation

y_n+1=ry_n((M-y_n)/M) must have equally spaced data

points spanning at least two generations, and M in the equation is equal to

the maximum population possible (not a stable maximum). r in the equation

is supposed to be the intrinsic growth rate for the population.

The data given indicates that William W.'s answer

of 10.8 months is likely the proper answer and method given problem data.

I disagree using the continuous growth method when

calculating the discrete logistical equation outcomes.

big reason is when the growth rate is higher than or equal to 3

in the logistic equation, the iterations become bifurcated

(population numbers oscillate between two values) and

greater than 3;.57 the numbers become chaotic.

The continuous method does not capture that, only

iterations do.

in this problem, month -1 is 1000 deer, month 0 is 1150 and carrying capacity

of 2200.

the logistic difference equation being y_n+1=ry_n((2200--y_n)/2200),,,r="intrinsic" rate of population growth.

The rate of growth calculates to 2.11. By month 3, the population

becomes stable at 1157 deer, and never reaches 2100 deer. I just plugged

the equation into my calculator and iterated.

let's say the growth rate was lower than the intrinsic the growth growth rate

(no predators, perfect breeding conditions) because of the starting

months. say the growth rate was 3.11 for deer.

mo.0 1150

mo.1 1707

mo.2 1896

mo.3 815,,,,bifurcation begins

say the growth rate was 2.95

mo.0 1150

mo.1 1619

mo.2 1261

mo.4 1587

mo.5 1305,,,"bifurcation" begins, should stabilize.

I think I the max population

that is stable is 2/3 of carrying capacity.

or 1467 deer in a logistic equation iteration.

If the population reached

2100/2200=.95 of the carrying capacity

unless the growth rate would need

to be 3.95 and the next value would

be 330 deer and population is going into chaotic

behavior.