The survival of ancient manuscripts can be modeled by a logistic equation. The n

Question

The survival of ancient manuscripts can be modeled by a logistic equation. The number of copies of a particular manuscript was found to approach a limiting value over the five centuries after its publication. Let g(t) represent the proportion of manuscripts known to exist after t centuries out of the limiting value, so that m = 1, For this manuscript, it was that k = 3.5 and G_0 = 0.00332. Complete a through e. a. Find the growth function G(t) for the proportion of copies of the manuscript found. G(t) = __ b. Find the proportion of manuscripts and rate of growth after 1 century. The proportion of manuscripts after 1 century is ___. (Type an integer or decimal rounded to four decimal places as needed.) c. Find the proportion of manuscripts and rate of growth after 2 centuries. The proportion of manuscript after 2 centuries is __. (Type an integer or decimal rounded to four decimal places as needed.) d. Find the proportion of manuscripts and rate of growth after 3 centuries. The proportion of manuscripts after 3 centuries ___. (Type an integer or decimal rounded to four decimal places as needed.) The rate of growth after 3 centuries is __ per century. (Type an integer or decimal rounded to four decimal places as needed.) e. what happens to the rate of growth over time? At first, the rate of growth However, over time, the rate of growth appears to be because the values of are as t

Explanation / Answer

a) Logistic equation is given by :

dG/dt = kG(1-G/K)

Here , k is a constant, G is proportion of manuscript known to exist after t centuries out of the limiting value

K is carrying capacity

Analytic solution to this equation is given by :

G(t) = K/(1+Ae-kt); A = (K-G0)/G0

We are given in problem, m = carrying capacity =K=1

G0 = 0.00332

k = 3.5

Hence G(t) = 1/(1+(1-0.00332)/0.00332))e-3.5t

=> G(t) = 0.00332/(0.00332+0.99668e-3.5t) or 1/1+300.2e-3.5t

(b) G(1) = 1/1+300.2e-3.5 = 0.099

dG/dt = 3.5(0.099)(1-0.099) = 0.312

(c) G(2) = 1/1+300.2e-7 = 0.785

Rate of growth = dG/dt = kG(1-G/K)

We have G = 0.785, K =1; k = 3.5

=> dG/dt =3.5(0.785)(1-0.785) = 0.591

(d) At t=3, G(3) = 1/1+300.2e-10.5 = 0.992

dG/dt = 3.5(0.992)(1-0.992) =0.0278

(e) At first the rate of growth increases and then decreases after 2 centuries becauise value when t=2 is greater than the value at t=3

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