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6.2 Growth and Decay Law of Exponential Growth and Decay y Ce kt C = initial value k = constant of proportionality if k > 0, exponential growth occurs if k < 0, exponential decay occurs Solving a Differential Equation y' 2x Multiply both sides by y. y yy ' 2 x Integrate both sides with respect to x. yy 'dx y dy y 2 x dx dy = y’dx because dy/dx = y’ 2 x dx 2 x C 2 Multiply both sides by 2. 2 y 2x C 2 2 C = 2C Exponential Growth and Decay Model If y is a differential function of t such that y >0 and y’ = ky, for some constant k, then Y = Cekt C is the initial value of y, and k is the proportionality constant. Exponential growth occurs when k > 0, and exponential decay occurs when k < 0. Proof: y’ = ky y' k y y' 1 dy dt k dt dy k dt y y ln y kt C ye e kt C1 y' dt dy y ' dt y Ce kt So, all solutions of y’ = ky are in the form of y = Cekt Ex. The rate of change of y is proportional to y. When t = 0, y = 2. When t = 2, y = 4. What is the value of y when t = 3? Because y’ = ky we know that y = Cekt. We can find the values of C and k by applying the initial conditions. 2 = Ce0 4= C=2 k 2e2k 1 2 ln 2 .3466 y 2e When t = 3, the value of y is 2e0.3466(3) = 5.657 So, the model is 0.3466 t Newton’s Law of Cooling Let y represent the temperature (in oF) of an object in a room whose temperature is kept at a constant 60o. If the object cools from 100o to 90o in 10 minutes, how much longer will it take for its temperature to decrease to 80o? From Newton’s Law of Cooling, the rate of change in y is proportional to the difference between y and 60. y ' k y 60 dy dt k y 60 dy y 60 kdt Separate variables first. dy y 60 kdt ln y 60 kt C 1 ln y 60 kt C 1 y 60 e The model is: y 60 40 e kt C 1 y 60 Ce Take e to both sides. C e kt .02877 t C1 Using y = 100 when t = 0,solve for C. 100 60 Ce 40 = C k( 0) Since y = 90 when t = 10 90 60 40 e 30 40 e 60 C k (10 ) k (10 ) Finally, when y = 80, you obtain 80 60 40 e .02877 t t 24 .09 min So, it will require 14.09 more minutes k = -0.02877 for the object to cool to 80o. Ex. 2 Money is deposited in an account for which interest is compounded continuously. If the balance doubles in 6 years, what is the annual percentage rate? A = Pert 2P = Pert 2 = e6r ln 2 = 6r r . 1155 11 . 55 % Ex. 1 A sample contains 1 gram of radium. How much radium will remain after 1000 years? (Use a half-life of 1620 years.) 1 1e k 1 , 620 2 ln ln First we need to find k, the constant of proportionality. Take the ln of both sides. 1 2 1 2 ln e 1620 k k . 0004279 y 1e 1620 k . 0004279 (1000 ) y . 652 g Ex. Suppose that 10 grams of the plutonium isotope Pu-239 was released in the Chernobyl nuclear accident. How long will it take for the 10 grams to decay to 1 gram? Pu-239 has a half life of 24,100 years y = Cekt We know that C = 10 grams at time t = 0. First, find k. 5 = 10ek(24,100) 1 2 e 24 ,100 k k 1 24 ,100 ln 1 2 k .000028761 So, the model is To find the time it would take for y = 10e-.000028761t 10 grams to decay to 1 gram, solve for t in 1 = 10e-.000028761 t = 80,059 years