Exponential Growth and Decay: Mathematical Model and Applications, Study notes of Law

Download Exponential Growth and Decay: Mathematical Model and Applications and more Study notes Law in PDF only on Docsity! Exponential Growth Many quantities grow or decay at a rate proportional to their size. I For example a colony of bacteria may double every hour. I If the size of the colony after t hours is given by y(t), then we can express this information in mathematical language in the form of an equation: dy/dt = 2y . A quantity y that grows or decays at a rate proportional to its size fits in an equation of the form dy dt = ky . I This is a special example of a differential equation because it gives a relationship between a function and one or more of its derivatives. I If k < 0, the above equation is called the law of natural decay and if k > 0, the equation is called the law of natural growth. I A solution to a differential equation is a function y which satisfies the equation. Annette Pilkington Exponential Growth Solutions to the Differential Equation dy(t) dt = ky(t) It is not difficult to see that y(t) = ekt is one solution to the differential equation dy(t) dt = ky(t). I as with antiderivatives, the above differential equation has many solutions. I In fact any function of the form y(t) = Cekt is a solution for any constant C . I We will prove later that every solution to the differential equation above has the form y(t) = Cekt . I Setting t = 0, we get The only solutions to the differential equation dy/dt = ky are the exponential functions y(t) = y(0)ekt Annette Pilkington Exponential Growth Population Growth: Example 1 Population GrowthExample The population of Mathland at the end the year 2000 was 500. The population increases (continuously or steadily) by approximately 10% per year. What is the function P(t), the size of the population after t years, using the exponential model above? I What differential equation does the function P(t) satisfy? dP(t) dt = kP(t) I What is the value of k? k = rate of growth = 0.1 I What is P(0)? P(0) = initial pop. size = 500 I Give a formula for P(t) P(t) = P(0)ekt = 500e0.1t I What will the population be at the end of the year 2050? At the end of the year 2050 we will have t = 50 and the population will be P(50) = 500e0.1(50) = 500e5 ≈ 74, 206 Annette Pilkington Exponential Growth Population Growth: Example 2 Example The population of Calculand was 700 in the year 2000 and was 3000 in the year 2010. Using the exponential model for population growth, find an estimate for the population of Calculand in 2015. I P(0) =? If we set t = 0 in the year 2000, therefore P(0) = 700 I P(10) =? When t = 10 the year is 2010, so P(10) = 3000. I Find the value of k. P(t) = P(0)ekt so P(10) = 700e10k . Therefore 3000 = 700e10k . To solve for k, we apply the logarithm to both sides to get ln(3000) = ln(700) + ln(e10k) = ln(700) + 10k. Therefore k = ln(3000)− ln(700) 10 = ln(30/7) 10 ≈ 0.147. I Find the formula for P(t) and use it to find P(15). We now have P(t) = P(0)ekt = 700e0.147t . To find the population in 2015, we find P(15) = 700e0.147×15 ≈ 6, 210. Annette Pilkington Exponential Growth Radioactive Decay Radioactive Decay Radioactive substances decay at a rate proportional to their mass. dm dt = km and m(t) = m0e kt , where m(t) denotes the mass of the substance at time t and m0 denotes the mass of the substance at time t = 0. The half-life of a radioactive substance is the time required for half of the quantity to decay. Carbon Dating the haf-life of Carbon-14 is approximately t1/2 = 5, 730 years (there is some variety in this depending on variables such as location). When a plant or animal dies, it stops taking in Carbon and the carbon it contains starts to decay. We can use this to figure out the age of artifacts by estimating the original mass of Carbon-14 in the object and the amount at present. We use the half-life to find the value of k above. I Example A bowl made of oak has about 40% of the carbon-14 that a similar quantity of living oak has today. Estimate the age of the bowl. Annette Pilkington Exponential Growth Interest Compounded Continuously Example If I invest $1000 for 5 years at a 4% interest rate with the interest compounded continuously, (a) how much will be in my account at the end of the 5 years? I We are given that A0 = 1000 and r = 0.04. I Because the interest is compounded continuously, we have A(t) = A0e 0.04t = 1000e0.04t I A(5) = 1000e0.04(5) = $1221.4. (b) How long before there is $2000 in the account? I We must solve for t in the equation 2000 = 1000e0.04t . I Dividing by 1000 and taking the natural logarithm of both sides, we get 2 = e0.04t → ln 2 = 0.04t → t = ln 2/0.04 ≈ 17.33yrs. Annette Pilkington Exponential Growth Interest compounded n times per year Sometimes interest is not compounded continuously. If I invest $A0 in an account with an interest rate of r × 100% per annum, the amount in the bank account after t years depends on the number of times the interest is compounded per year. In the chart below A0 = A(0) is the initial amount invested at time t = 0. A(t) is the amount in the account after t years. n = the number of times the interest is compounded per year. We Have A(t) = A0(1 + r n )nt Amt. after t years A(0) A(1) A(2) . . . A(t) n = 1 A0 A0(1 + r) A0(1 + r)2 . . . A0(1 + r)t n = 2 A0 A0(1 + r 2 )2 A0(1 + r 2 )4 . . . A0(1 + r 2 )2t n = 12 A0 A0(1 + r 12 )12 A0(1 + r 12 )24 . . . A0(1 + r 12 )12t . . . . . . . . . . . . n A0 A0(1 + r n )n A0(1 + r n )2n . . . A0(1 + r n )nt . . . . . . . . . . . . n → ∞ A0 lim n→∞ A0(1 + r n )n lim n→∞ A0(1 + r n )2n . . . lim n→∞ A0(1 + r n )nt (compounded continuously) = A0 = A0er = A0e2r . . . = A0ert Annette Pilkington Exponential Growth Examples Example If I borrow $50,000 at a 10% interest rate for 5 years with the interest compounded quarterly, how much will I owe after 5 years? I A(t) = A0(1 + r n )nt I A(t) = 50, 000(1 + .1 4 )4t I A(5) = 50, 000(1 + .1 4 )20 ≈ 81, 930.82 Annette Pilkington Exponential Growth