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xe 2x dx Drill u x, dv e dx 2x 2x 2x e du dx, v 2 2x e e xe dx x dx 2 2 2x 2x xe e 2x xe dx C 2 4 2x Exponential Growth and Decay Lesson 6.4 Objectives • Students will be able to: – solve problems involving exponential growth and decay in a variety of applications. Definition: Separable Differential Equation A differential equation of the form dy f y g x dx is called separable. We separate the variable by writing it in the form dy g x dx. f y The solution is found by integrating each side with respect to its variable. Example Solve by Separation of Variables dy 1 2 Solve for y if when x = 1. 6 y x and y dx 25 dy 6 x dx 2 y 2 y dy 6 x dx y 2 dy 6 x dx 1 y 3x C 2 1 1 2 31 C 25 25 3 C 28 C 1 y 3x 28 2 Example Solve by Separation of Variables dy 1 2 Solve for y if when x = 1. 6 y x and y dx 25 1 2 3x 28 y 1 2 28 3 x y 1 y 2 28 3x 1 y 3x 28 2 Example Solve by Separation of Variables dy 2 Solve for y if xy and y 1 when x = 1. dx dy 2 x dx 2 y 2 2 y dy x dx y dy x dx 1 3 1 y x C 3 2 2 1 2 1 1 C 3 1 1 C 3 4 C 3 1 3 4 1 y x 3 3 1 Example Solve by Separation of Variables dy Solve for y if (xy ) 2 and y 1 when x = 1. dx 1 x 4 y 3 3 1 4 x y 3 3 y 3 4 x 3 y 1 3 x 4 3 3 The Law of Exponential Change • If y changes at a rate proportional to the amount present (that is, if dy/dt = kt), and if y = y0 (initial amount) when t = 0, then y = y0ekt The constant k is the growth constant if k > 0 or the decay constant if k< 0. Interest Formulas • Continuously Compounded Interest A(t) = A0ert A : initial amount t: time r: continuous interest rate • Compounded interest for n compounding periods A(t) = A0(1+r/n)nt Example Compounding Interest Continuously • Suppose you deposit $800 is an account that pays 6.3% annual interest. How much money will you have 8 years later if: – Compounded continuously? • A(t) = 800e.063(8) • $1324.26 – Compounded quarterly? • A(t) = 800(1+.063/4)4(8) • $1319.07 Half-Life • dy/dt: decay of a radioactive element over time. • dy/dt=-ky, k>0 • Half-life = ln 2/k, where k is a rate constant • Also, A(t) = A0(.5)t/h • t = time, h = half-life time period, A0 = original amount Example Half-Life • An isotope of neptunium (Np-240) has a half-life of 65 minutes. If the decay of Np-240 is modeled by the differential equation dy/dt = -ky, where t is measured in minutes, what is the decay constant k? • Half-life = ln 2/k • 65=ln2/k • 65k = ln2 • k=ln(2)/65=.01066 Example Choosing a Base • At the beginning of the summer, the population of a hive of a bald-faced hornets (which are actually wasps) is growing at a rate proportional to the population. From a population of 10 on May 1, the number of hornets grows to 50 in 30 days. If the growth continues to follow the same model, how many days after May 1 will the population reach 100? Example Solution 1 • Since dy/dt = -ky, the grown is exponential. • The population grows by a factor of 5 in 30 days: 10 X 5 = 50; therefore we model the growth in base 5 • y = 10(5)(t/30) • 100 = 10(5)t/30 • 10 = (5)t/30 Choosing a Base Solution 2 • Using the two points of (0, 10) and (30, 50), create an exponential equation to find b: • 50 = 10b30 • 5 = b30 • 1.055= b • 100=10(1.055)t • 10= (1.055)t Example Using Carbon-14 Dating Scientists who use carbon-14 dating use 5700 years for its half-life. Find the age of a sample in which 10% of the radioactive nuclei originally present have decayed. .90A0 A0 (.5) ln .90 ln(.5) ln .90 t ln .5 5700 t / 5700 .90 (.5) t / 5700 ln .90 ln(.5) ln .90 5700 t ln .5 t / 5700 t 5700 Using Newton’s Law of Cooling Example dT k (T Ts ) dt It’s solution, by the law of exponential T is temperature of the object at time t Ts is the surrounding temperature. change: (T Ts ) (T0 Ts )e kt where T0 is the temperature at t = 0. A hard-boiled egg at 980 C is put in a pan under running 180 C water to cool. After 5 minutes, the egg’s temperature is found to be 380 C. How much longer will it take to reach 200 C? 20 80e 5k (38 18) (98 18)e k 5 ln .25 5k ln e (20 18) (98 18)e .277t .25 e 5k ln .25 5k (1) 2 80e .277 t ln .025 13 .3 years t .277 ln .25 ln e 5k ln .25 .277 k 5 .025 e .277 t ln .025 ln e .277 t Homework • Page 357: 5, 8, 9, 10, 21, 23, 25, 31