Drill

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 is called separable. We separate the variable by writing it in the form The solution is found by integrating each side with respect to its variable.

Example Solve by Separation of Variables Solve for y if and when x = 1.

Example Solve by Separation of Variables Solve for y if and when x = 1.

Example Solve by Separation of Variables Solve for y if and when x = 1.

Example Solve by Separation of Variables Solve for y if and when x = 1.

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 Choosing a Base Solution 1 Solution 2 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 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 42.92 days

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. 866.42 years

Example Using Newton’s Law of Cooling It’s solution, by the law of exponential change: where T0 is the temperature at t = 0. T is temperature of the object at time t Ts is the surrounding temperature. 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?

Homework Page 357: 5, 8, 9, 10, 21, 23, 25, 31