6.5   Life and death Population    In 1965 the world population was three thousand million. Allowing for deaths, the growth rate is approximately 2% per year. What was the population in the year 2000? Solving this type of problem may help prevent over-population. The mathematics is simple, as it is known that the rate of change of size of a population is proportional to the population size. Thus if P = P(t) is the population size at time t, then      We claim that P0 is the population at time t = 0, and k is the growth rate. Equation P(t) = kP is a differential equation (d.e.).  Substituting in shows that the given P is a solution to the d.e.   To show the solution is unique, let Q be another solution.  Differentiating P/Q with respect to t gives the value 0, implying that P/Q = c, or P = cQ (c is a constant). Finally, we solve the original problem. Take k = 0.2, and let t = 0 in 1965. So when t = 35, P = 3 109 e(0.02)35      = 3 109 e0.7      = 3 109 2.01 = 6.03 109. So the world's population more than doubled by the year 2000. Population Half life Carbon Dating

www.vce.com  6.5   Life and death Half life   Strontium 90 (Sr 90) is a radioactive substance which occurs in the fall-out of the hydrogen bomb.  Sr 90 disintegrates at a rate proportional to its mass.  It has a half-life of 28 years: that is, after 28 years exactly half of the original amount remains. If we start with 10 g of Sr 90, how much will be present 8 years later? [Given: ln 2 0.69, e– 0.2 0.82.] Let A = A(t) be the amount of Sr 90 present at time t. Since A(t) t, where k is negative, as A is decreasing with time. Solving this equation as before, gives A = A0ekt. Now when t = 0, A = A0 = 10, and when t = 28, A = 5. Hence To solve the problem we set t = 8, giving                        A(8) = 10e– 0.2 8.2. Hence after 8 years there is still 8.2 g of Sr 90 present. This type of decrease is called exponential decay. Population Half life Carbon Dating

6.5   Life and death Carbon dating   Carbon 14 (C 14) is a rare radioactive form of carbon which has a half-life of about 5600 years. Growing plants take in C14 in the atmospheric proportion, but after the death of the plant, with no new C 14 coming in, the amount decreases due to radioactive decay. Assuming that the proportion of C 14 in the atmosphere has remained constant, we can date old objects made from plant material. The wood of an old coffin contains C14 in 0.88 of atmospheric proportion. How many years ago was the tree cut down to make the coffin? [Given: ln 2 0.69,  ln 0.88 0.12.] Take t = 0 when the tree was felled. Let A = A(t) be the amount of C14 present at time t. Arguing as before gives A = A0ekt.  Now when t = 5600, A = A0.  So,    0.5A0 = A0e5600k,  ln 0.5 = 5600k, Now we want A = 0.88A0.  So        0.88A0 = A0e– 0.00012t,  – 0.00012t = ln 0.88  –0.12, and so      t 1000 years Population Half life Carbon Dating