In Tuesday May 17 s class we considered the stochastic popul

In Tuesday May 17 s class we considered the stochastic population model where p_n(t) is the probability that the population has size n at time t. This is an infinite system of differential equations, one for every value of n. Hence, it\'s a pretty difficult system of equations to solve! We instead used this equation to derive a differential equation for the mean (or \"expected\") value of n. We did so by multiplying Equation by n and Slimming over all values of n. By changing indices in the sums involving p_n, i (t) and p_n i(t) to make all sums involve pn{t), we found that several terms cancelled out, giving

Solution

6 a.      2e^3t + 1 = 102 or 2e^3t = 12-1 = 101 or, e^3t = 101/2 = 50.5. Then on taking natural logarithms of both sides, we have 3t ln e = ln 50.5 or 3t = 3.921973336 so that t = 3.921973336/3 = 1.307324445 or 1.31 ( on rounding off to 2 decimal places) ( as log a^b = b log a and ln e = 1)

b.        50 e^-0.1x = 12.52 or, e^-0.1x   = 12.52/50 = 0.2504. Then on taking natural logarithms ofboth sides, we have -0.1x ln e = ln 0.2504 or, -0.1x = - 1.38469564 so that x = - 1.38469564/-0.1 = 13.8469564 or, 13.85 ( on rounding off to 2 decimal places).

c. 1000 = 3200/(1 +40e^-0.4t) or, 1 +40e^-0.4t = 3200/1000 = 3.2 or, 40e^-0.4t = 3.2 – 1 = 2.2 or, e^-0.4t = 2.2/40 =

0.055. Then on taking natural logarithms ofboth sides, we have, -0.4tln e = ln 0.055 or, -0.4t = -2.900422094 so that t = -2.900422094/-0.4 = 7.251055234 or, 7.25 (on rounding off to 2 decimal places).