# Exponential Growth and Decay Exponential Growth and Decay Chapter 7.4 Separable Differential Equations DEFINITION A differential equation of the form is called separable. We separate the variables by writing it in the form The solution is found by antidifferentiating each side with respect to its isolated

variable. 2 Example 1: Solving by Separation of Variables Solve for y if and when . 3

Example 1: Solving by Separation of Variables Now, find the indefinite integral for each side of the equation. 4

Example 1: Solving by Separation of Variables 5 Example 1: Solving by Separation of Variables

Apply the initial condition that when to find . The solution is on 6 Law of Exponential Change In exponential growth, the amount present at any given time (whether it be bacteria, people, or compounded interest on money) is

proportional to the rate of growth Likewise, in exponential decay, the amount present at any given time (say, radioactive material) is proportional to the rate of decay Both statements are captured by the differential equation The constant k is the growth constant (positive) or the decay constant (negative) 7 Law of Exponential Change

Before continuing, think about what this equation says What is ?

This is the rate of change of some amount of substance with respect to time What is y on the right side of the equation? y is the amount of substance What does the solution to this differential equation represent? The solution is a function showing the amount of stuff at any time t 8 Law of Exponential Change

The differential equation is a separable equation, so we can solve it as shown in Example 1 9 Law of Exponential Change 10

Law of Exponential Change Note that is just a constant so we can assign and our equation becomes This equation shows that the only growth function that results in a growth rate proportional to the amount present is exponential The amount is the amount present at

11 Law of Exponential Change LAW OF EXPONENTIAL CHANGE If y changes at a rate proportional to the amount present (that is, if ), and if when , then The constant k is the growth constant if or the decay constant if 12

Continuously Compounded Interest Suppose that is an amount invested at a fixed annual interest rate of , If interest is added to the account k times per year, the amount of money present after t years is The value of k can be 12 (monthly), 52 (weekly) 365 (daily) or any period of time at all 13

Continuously Compounded Interest Suppose that is an amount invested at a fixed annual interest rate of , If interest is added to the account k times per year, the amount of money present after t years is The value of k can be 12 (monthly), 52 (weekly) 365 (daily) or any period of time at all We say that the interest is compounded discretely

14 Continuously Compounded Interest Suppose that, rather than compounding in discrete amounts represented by k, compounding takes place continuously We can represent this situation as We can also investigate this by assuming that the interest is added continuously at a rate proportional to the amount in the account That is, if and

15 Continuously Compounded Interest Solving this separable differential equation gives 16 Continuously Compounded Interest

Solving this separable differential equation gives Applying the initial value gives us So our formula for continuously compounded interest is 17 Example 2: Compounding Interest Continuously

Suppose you deposit \$800 in an account that pays 6.3% annual interest. How much will you have 8 years later if the interest is (a) compounded continuously (b) compounded quarterly ()? 18 Example 2: Compounding Interest Continuously a) Using the formula for continuous compounding

b) With quarterly compounding 19 Radioactivity When an atom in a radioactive element decays, it emits part of its structure (for example, an electron) as radiation and the atom becomes a new element

Experiments show that this decay occurs at a rate that is proportional to the amount of radioactive nuclei present That is, if y is the amount of radioactive nuclei, then Therefore, as before 20 Radioactivity Therefore, as before

Note that we use the convention that in order to use the negative sign in the exponent, indicating that the amount present is the result of decay rather than growth The half-life of a radioactive element is the time required for half of the radioactive nuclei present in a sample to decay The next example shows that the half-life depends only on the substance, not on the amount 21

Example 3: Finding Half-Life Find the half-life of a radioactive substance with decay equation and show that the half-life depends only on k. 22 Example 3: Finding Half-Life You are asked to find the time t after which half of the original sample has decayed. That is

Solve for t 23 Example 3: Finding Half-Life This shows that the time t after which half of some original amount of sample remains is dependent only on the constant k. In turn, k depends on the particular

element (or rather, the particular isotope of the element) and is determined experimentally. The half-life does not depend on the initial amount of substance. 24 Modeling Growth with Other Bases Solving the differential equation results in an exponential function in e However, it is always possible to change from base e to any other

positive base If , then That is, where 25 Modeling Growth with Other Bases In changing from base to some other base we must consider: The factor n (i.e., a multiple) by which an initial amount increases (or

decreases) over a period of time from to How to represent the new constant h in 26 Modeling Growth with Other Bases Begin with a system of growth or decay Suppose that from time to time , growth (or decay) occurs in some system that has an initial value of

During this interval of time, the value of A increases (or decreases) by a factor of n That is, We will choose our base b equal to the factor n 27 Modeling Growth with Other Bases We have Take the natural logarithm of each side to remove e

Therefore But recall that So we get 28 Example 4: Choosing a Base At the beginning of the summer, the population of a hive of bald-faced hornets 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 thirty days. If the growth continues to follow the same model, how many days after May 1 will the population reach 100? 29 Example 4: Choosing a Base

By what factor does the population increase over the given period of time? It grows from 10 to 50, a factor of 5; let

What is the period of time over which this occurs? Thirty days, so What is the value of our constant h? We have What equation will we use to answer the question? The equation is 30

Example 4: Choosing a Base The equation is After how many days t does the population reach 100? 31 Example 5: 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. 32 Example 5: Using Carbon-14 Dating Since we are dealing with half-life, it is convenient to use as b, and since half-of a sample of carbon-14 decays in 5700 years, then . Our equation is . Since 10% of has decayed and the formula determines how much is left, then

Therefore 33 Example 5: Using Carbon-14 Dating 34

Newtons Law of Cooling Suppose that a heated object is left to cool in some surrounding medium (like the air) that is large enough for the temperature to be considered constant Experimental evidence suggests that the rate at which the objects temperature is changing at any given time is roughly proportional to the difference between its temperature and the temperature of the surrounding medium That is, , where is the objects temperature at time t and is the

temperature of the surrounding medium 35 Newtons Law of Cooling Although this looks different from the differential equations you have seen, the difference is very small Suppose we let with initial conditions Then becomes

36 Example 6: Using Newtons Law of Cooling A hard-boiled egg at C is put in a pan under running C water to cool. After 5 minutes the eggs temperature is found to be C. How much longer will it take the egg to reach C? 37

Example 6: Using Newtons Law of Cooling We must first determine the value of the constant k. Since we know that the eggs temperature after 5 minutes is C and that its initial temperature was C, then Solve for k in 38 Example 6: Using Newtons Law of Cooling Now, given that , find the time t that it takes the egg to reach a temperature of C.

It takes additional minutes to reach C. 39