The best thing about exponential functions is that they are so useful in real world situations. Exponential functions are used to model populations, carbon date artifacts, help coroners determine time of death, compute investments, as well as many other applications.
We will discuss in this lesson three of the most common applications: population growth
, exponential decay
, and compound interest
. For other applications, consult your textbook or ask your teacher for additional examples. 1. Population
Many times scientists will start with a certain number of bacteria or animals and watch how the population grows. For example, if the population doubles every 5 days, this can be represented as an exponential function. Most population models involve using the number e. To learn more about e, click here (link to exp-log-e and ln.doc) 2. Exponential Decay
Population models can occur two ways. One way is if we are given an exponential function. The second way involves coming up with an exponential equation based on information given. Let’s look at each of these separately.
- The population of a city is P = 250,342e0.012t where t = 0 represents the population in the year 2000.
Summary: Before we do the next example, let’s look at a general form for population models. Most of the time, we start with an equation that looks like
- Find the population of the city in the year 2010.
To find the population in the year 2010, we need to let t = 10 in our given equation.
P = 250,342e0.012(10) = 250,342e0.12 = 282,259.82 Since we are dealing with the population of a city, we normally round to a whole number, in this case 282,260 people.
- Find the population of the city in the year 2015.
To find the population in the year 2015, we need to let t = 15.
P = 250,342e0.012(15) = 250,342e0.18 = 299,713.8 We’ll round this answer to 299,714 people.
- Find when the population will be 320,000.
We know the population in the year 2015 is almost 300,000 from our work in part (b). So it makes sense that the answer has to be higher than 2015. Remember that P in the equation represents the population value, which we are given to be 320,000. Only now we do not know the time value t. The equation we need to solve is
320,000 = 250,342e0.012(t) To review solving exponential equation, click here.
So it will take between 20 and 21 years for the population to reach 320,000. This means between the years 2020 and 2021 the population will be 320,000.
P = Poekt
Let's Practice this once more:
- P represents the population after a certain amount of time
- Po represents the initial population or the population at the beginning
- k represents the growth (or decay) rate
- t represents the amount of time
- Remember that e is not a variable, it has a numeric value. We do not replace it with information given to us in the problem.
- A scientist starts with 100 bacteria in an experiment. After 5 days, she discovers that the population has grown to 350.
- Determine an equation for this bacteria population.
To find the equation, we need to know values for Po and k. Remember the equation is in the form P = Poekt. where P, e, and t are all parts of the equation we will come up with. We only need values for Po and k. Po is given by the amount the scientist starts with which is 100. Finding k requires a little more work.
We know that Po is 100 and after t = 5 days the population P is 350. We can use this information to find k.
Now that we know k, we go back to our general form of and replace Po and k. So our equation is
P = 100e0.25055t
- Use the equation to find out the population after 15 days.
We will substitute the value of 15 for t in P = 100e0.25055t.
P = 100e0.25055(15) = 100e3.75825 = 4287.33 or approximately 4287 bacteria after 15 days.
- Use the equation to find out when the population is 1000.
We will set our equation equal to 1000 to get 1000 = 100e0.25055t and solve.
So between 9 and 10 days, the bacteria population will be 1000.
Solving an exponential decay problem is very similar to working with population growth. In fact, certain populations may decrease instead of increase and we could still use the general formula we used for growth. But in the case of decrease or decay, the value of k will be negative. 3. Compound Interest
- The number of milligrams of a drug in a persons system after t hours is given by the function D = 20e-0.4t.
- Find the amount of the drug after 2 hours.
To solve the problem we let t = 2 in the original equation.
D = 20e-0.4(2) = 20e-0.8 = 8.987 After 2 hours, 8.987 milligrams of the drug are left in the system.
- Find the amount of the drug after 5 hours.
Replace t with 5 in the equation to get
D = 20e-0.4(5) = 20e-2.0 = 2.707 After 5 hours, 2.707 milligrams remain in the body.
- When will the amount of the drug be 0.1 milligram (or almost completely gone from the system)?
We need to let D = 0.1 and solve the equation 0.1 = 20e-0.4t
After approximately 13 hours and 15 minutes, the amount of the drug will be almost gone with only 0.1 milligrams remaining in the body.
The formula for interest that is compounded is
- A represents the amount of money after a certain amount of time
- P represents the principle or the amount of money you start with
- r represents the interest rate and is always represented as a decimal
- n is the number of times interest is compounded in one year
if interest is compounded annually then n = 1
if interest is compounded quarterly then n = 4
if interest is compounded monthly then n = 12
- t represents the amount of time in years
- Suppose your parents invest $1000 in a savings account for college at the time you are born. The average interest rate is 4% and is compounded quarterly. How much money will be in the college account when you are 18 years old?
We will use our formula and let P = 1000, r = 0.04, n = 4 and t = 18.
- Suppose your parents had invested that same $1000 in a money market account that averages 8% interest compounded monthly. How much would you have for college after 18 years?
P = 1000, r = 0.08, n = 12 and t = 18