Exponential Growth and Decay - Lecture Slides | MAT 021B, Study notes of Calculus

Download Exponential Growth and Decay - Lecture Slides | MAT 021B and more Study notes Calculus in PDF only on Docsity! •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit 7.2 Exponential Growth and Decay •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit The Law of Exponential Change In modeling many real-world situations, a quantity y increases or decreases at a rate proportional to its size at a given time. Examples of such quantities: • the size of a population; • funds earning interest in a bank account • the amount of a decaying radioactive material •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Mathematically, we need to solve dy dt = ky, y(0) = y0 where k is the rate constant. Now, let us solve this initial value problem. 1 y dy dt = k ⇒ ∫ y′(t) y dt = ∫ kdt ⇒ ln |y| = kt + C ⇒ |y| = ekt+C = eCekt ⇒ y = ±eCekt, C is an arbitrary constant ⇒ y = Aekt, A is an arbitrary constant Substitute y(0) = y0, we have A = y0. Therefore, y = y0e kt •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Remark: If the derivative is proportional to itself, then it must be y0ekt. This model have MANY APPLICATIONS in real-world! •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Unlimited Population Growth Under ideal conditions (unlimited space, adequate food supply, immunity to disease, etc), the rate of increase of a population P at time t is proportional to the size of the population at time t. That is, P ′(t) = kP (t) for some constant k > 0, called the growth constant. Thus, P (t) = P (0)ekt It grows exponentially fast. •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit 9 = 4.5e0.0164t ⇒ e0.0164t = 2 ⇒ t = ln 2 0.0164 ≈ 42.27 i.e. the year 1980 + 42 = 2022. • In the year 1998, the world population is estimated to be 5.9 billion. What does the model tells us? •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit 9 = 4.5e0.0164t ⇒ e0.0164t = 2 ⇒ t = ln 2 0.0164 ≈ 42.27 i.e. the year 1980 + 42 = 2022. • In the year 1998, the world population is estimated to be 5.9 billion. What does the model tells us? P (18) = 4.5e0.0164·18 = 4.5e0.2952 ≈ 6.045 Difference is 6.045− 5.9 = 0.145. •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Radioactive decay Radioactive substances decay at a rate proportional to the amount of the sub- stance present. Let A(t) be the amount of the radioactive substance at time t. Then, A′(t) = −kA(t) for some decay constant k > 0. Thus, A(t) = A(0)e−kt •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Example: A year ago, there were 5 grams of a radioactive substance. Now, there are 4 grams. How much will remain 3 years from now? What is its half-life? R(t) = R(0)e−kt Given: R(−1) = 5, R(0) = 4 Want: R(3). 5 = 4ek k = ln 5 4 . R(3) = 4e−3 ln 5 4 = 4 · (5 4 )−3 = 44 53 = 256 125 = 2.048 The half-life is ln 2 ln 54 = 3.1063 year •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Continuously Compounded Interest If you invest an amount A0 of money at a fixed annual interest rate r and if interest is added to your account k times a year, the formula for the amount of money you have at the end of t years is At = A0(1 + r k )kt. The interest might be added (“compounded”, bankers say) •monthly (k = 12), •weekly (k = 52), • daily (k = 365), or even more frequently, say by the hour or by the minute. By taking the limit as interest is compounded more and more often, we arrive at the following formula for the amount after t years, •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit lim k→∞ At = lim k→∞ A0(1 + r k )kt = lim x→0 A0(1 + x) rt x by setting x = r k = A0[ lim x→0 (1 + x) 1 x]rt = A0e rt •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Example: Suppose that $1000 is deposited in Wells Fargo that pay 5% com- pounded continuously. How much will be in the account after 3 years, and what is the interest earned during this period? What happens after 30 years? Here, A(0) = 1000, r = 0.05. So, A(t) = 1000e0.05t After 3 years, A(3) = 1000e0.15 ≈ 1161.83 Interest earned is $161.83. After 30 years, A(30) = 1000e1.5 = 4481.69 •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Heat Transfer: Newton’s Law of cooling When an object (e.g. a cup of hot soup) is hotter than the surrounding medium, it will cool to the temperature of the surrounding medium. Observation (Newton’s Law of Cooling): The rate at which an object’s temperature is changing at any given time is roughly proportional to the difference between its temperature and the temperature of the surrounding medium. This observation is described by the differential equation dH dt = −k(H −HS) where •H is the temperature of the object at time t •HS is the constant surrounding temperature. •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Let y = H −HS, then dy dt = dH dt = −k(H −HS) = −ky The solution is: y(t) = y0e −kt Substitute H −HS for y, we have H −HS = (H0 −HS)e−kt This is the equation for Newton’s Law of Cooling. Example: A hard-boiled egg at 98◦C is put in a sink of 18◦C water. After 5 minutes, the egg’s temperature is 38◦C. Assuming that the water has not warmed appreciably, how much long will it take the egg to reach 20◦C?