128

Exponential Decay x f (x)  ab , a  0 , 0  b  1. Exponential Decay x f (x)  ab When a > 0 and 0 < b < 1, the exponential function is often referre...
Author: Donald Riley
6 downloads 0 Views 4MB Size
Exponential Decay x f (x)  ab , a  0 , 0  b  1. Exponential Decay x f (x)  ab When a > 0 and 0 < b < 1, the exponential function is often referred to as an exponential “decay” function. What this means, graphically, is that as we look at the function from left to right, the function(the y-value that is) is decreasing(or decaying) quickly at an exponential rate(the y’s are multiplied by the same number that is a fraction). An example of an exponential decay x  1 f (x)     2  , shown below is

If we continued the table, we see that the y-value is going down very fast. x 4 5 6 7 y

1/16

1/32

1/64

1/128

How does this compare to the decay of other functions we are familiar with? Look at the y-values from left to right in this table. x

1

2

3

4

5

6

7

8

9

10

11

1 y = 2x

1/2

1/4

1/6

1/8

1/10

1/12

1/14

1/16

1/18

1/20

1/22

1/81

 1   y =  x

2

1

1/4

1/9

1/16

1/25

1/36

1/49

1/64

1/100

1/121

 1   2

x

1/2

1/4

1/8

1/16

1/32

1/64

1/128

1/256 1/512 1/1024

1/2048

y=

We can see this graphically as well if we look at the graph of these 3 functions below,

What is the domain and range for each of these functions? Domain: All real numbers Range: y > 0

Are there any asymptotes on the graph? Yes, y = 0 ( x - axis ). So, from the table and graph, we see that an exponential function decays much faster than other functions we have seen. If we look at an x-y table that demonstrates exponential decay, the y-value is always multiplied by the same fractional number when the x-value increases by the same amount. For example, the table to the right is an x 0 2 4 6 exponential function with y 6 3/2 3/8 3/32 b = 1/2 and a = 6. A decreasing exponential function gets this special “decay” name because it is used to model many real-life situations. Some of the many examples of situations that we can model using an exponential decay function are: 1.) Molecular decay of a radioactive substance 2.) Bacteria population decay 3.) Depreciation of the value of cars, office equipment, etc. 4.) Level of a drug in a person’s bloodstream as time passes 5.) Drop in population of a city due to migration patterns x f (x)  ab Note(Important): Notice in the function rule , if we let x = 0, then f(x) = a, which represents the initial amount of a substance. If you are given no initial amount of a substance, then just pick an easy value to start with like 10.

In many exponential functions( or models ) we will investigate, the x will often be changed to “t” which represents the time elapsed since you started something such as measuring the time since you started owning a

car or the time since you initially exposed a radioactive substance to the atmosphere. We should be able to look at the rule for an exponential function and determine whether it is a decay example and then graph it using an x-y table. *Don’t forget to have negative values for x in your table. If the variable is time( t ), then you probably will not need negative values. Now, you try to do this with the following functions: x x  1  1 f (x)  12   f (x)  4    3  and  2 . 1.) Fill out an x-y table going from -3 to 3. *You can use a calculator to fill out your x-y tables. Hint: To get from one yvalue to the next one on the table, just multiply by the b in f (x)  ab x . 2.) Use your x-y table to get a graph of the function.

Solutions:  1 f (x)  12    3

x

 1 f (x)  4    2

x

1.) x

-3

-2

-1

0

1

y

32 4

10 8

36

12

4

2

3

4/3 4/9

x

-3

-2

-1

0

1

2

3

y

32

16

8

4

2

1

1/2

2.)

Looking at the graphs above, what is the domain and range? Intercepts? Asymptotes?

Here is an example of exponential decay: With the advent of cd’s, the sale of vinyl record albums each year declined at an exponential rate. This decline from 1982 to 1993 can be t modeled using the equation N(t)  265(0.4) where N is the number of albums sold in millions each year since 1982. 1.) What is the initial amount, decay factor, and annual percent of decrease? The initial amount sold would be a = 265 million, the same number we get if we let t = 0 in our model. The decay factor is 40%. The annual percent of decrease is therefore 60%. 2.) Graph the model,

t

0

1

2

3

4

5

6

N

265

106

42.4

17

6.8

2.7

1.1

*Calculator Note: When graphing exponential functions, make sure to adjust your window settings( x-min/x-max/y-min/y-max ) in such a way so that you can see the graph. Sometimes the graphs of these functions will not fit the usual scaling. A common way to use exponential decay functions is to speak of how many percent a quantity decreases in a given time interval such as a second, minute, month, year, etc. When we use exponential decay in this way, we write the function rule in a slightly different way: Exponential Decay Function(when decay is a percent): f (t)  a(1 r)t where a is the initial amount of some quantity, 1 − r is the decay factor(or decay rate), and t is the time since you started observing the data.

Note: We usually let the beginning time be t = 0, which is when the amount of the quantity will be equal to “a”. Let’s take a look at some real-world applications problems that use exponential decay as a percent.

Example A: You buy a new car for $24,000. The value of the car decreases by 16% each year. 1.) Write an exponential decay function which models this situation. Let V = value of the car( in dollars ). Let t = the time since you first purchased the car ( in years ). V(t)  24,000(1 0.16)t    w(t)  24,000(0.84)t .

2.) Use the graph of this function to predict when the car will have a value of $12,000. What scaling should we use for this graph? The scaling looks like it should be x-min = 0, x-max = 9, x-scl = 1; y-min = 0, y-max = 26,000, y-scl = 2,000. Let’s do this graph on the graphing calculator. From the graph, we estimate that the car will have a value of $12,000 after about 4 years since it was purchased.

Example B : In a particular city, there are 40,000 homes. Each year, 10% of the homes are expected to get rid of their individual septic systems and instead use the city provided sewer system. If all homes in the city initially had individual septic systems, estimate the number of homes that will still have individual septic systems after 5 years. 1.) Write an exponential decay function which models this situation. Let N = number of individual septic systems in this city( in thousands ). Let t = the time since the residents of the city first started converting to the city provided sewer system ( in years ). N(t)  40,000(1 0.1)t    w(t)  40,000(0.9)t .

2.) Use this model to predict the number of homes that will still have individual septic systems after 5 years. Plugging 5 in for t, we estimate that the number of homes that will still have individual septic systems after 5 years will be approximately 23,620 homes. 3.) Use the graph of this function to predict when the number of homes that will still have individual septic systems will fall to about 17,200. What scaling should we use for this graph? The scaling looks like it should be x-min = 0, x-max = 20, x-scl = 2; y-min = 0, y-max = 45,000, y-scl = 5,000. Let’s do this graph on the graphing calculator.

From the graph, we estimate that the number of homes that will still have individual septic systems will fall to about 17,200 in approximately 8 years.

Now, you try to do the following problem: A new 3-D printer costs $23,000. The value of the printer will decrease by 15% each year. t V(t)  a(1 r) 1.) Using , write an exponential function which models this situation. Let V = the value of the printer t years after you have purchased it.

2.) Use your model to estimate how much the printer will be worth after 3 years. 3.) Use the graph of this function to predict when the value of the printer will be half of its original value. t t V(t)  23,000(1 .15)    c(t)  23,000(.85) Answers: 1.)

2.) Approximately $14,125. 3.) Using scaling of x-min = 0, x-max = 10, x-scl = 1 and y-min = 0, y-max = 25,000, y-scl = 5,000. Using the trace function, we find that when y = 11,518, x ≈ 4.25 which means that it will take approximately 4.3 years until the value of the printer will be worth half of its original value.

Another common example of exponential decay is calculating Amortization and/or Depreciation of loans. Amortization usually refers to spreading an intangible asset’s cost over that asset's useful life. For example, a doctor’s office may purchase a piece of medical equipment that usually has a life of 17 years. The cost involved with operating the medical equipment is spread out over that 17 years, and each year the doctor’s office will “write off” that amount on their taxes. Depreciation on the other hand, refers to prorating a tangible asset’s cost over that asset's life. For example, an office building can be purchased and each year, its value will depreciate. The amount of property taxes owed on a building will depend on the current value of the building. Therefore, the owner of the building will need to calculate the actual value of the building each year so that he/she will pay the proper amount of taxes on the property. Let’s take a look at some other “fun” formulas used in loans and/or investments at the following website: http://oakroadsystems.com/math/loan.htm#Formulas Let’s use one of these formulas( using different, more descriptive variables ) in a problem. A new automobile worth $18,354 has a very high loan interest rate of 17% per year. The owner makes monthly payments on the loan of $400. What would the payoff amount be 24 months after owning the car? To solve this problem, we use the following formula:

m  m n  A(n)   A0  1 r /12      r /12 r /12 Loan Payoff Formula: where A = the original amount of the loan, m = the monthly payment, r = the annual interest rate usually expressed as a percent, n = the number of monthly payments made so far 0

Plugging in the numbers: 400  400 24  A(24)   18, 354  1 0.17 /12     A(24)  $14, 385     0.17 /12  0.17 /12

So, after making 2 years worth of payments ( 2 x 12 x $400 ) or $9600, the customer has only paid off ( $18,354 − $14,385 ) or $3,969 of the car loan. *This is why it is always a good idea to pay off any loan as quickly as possible to avoid the effects of interest on the loan. Aside: No, you will not need to use the Loan Payoff formula above in any homework problem you do. *Hmwk: From the problems below, do 1-24 all, 35-41 odd, 43-46 all, 50-55 all. *For problems 4-9 and 35-41 odd, make sure you have the correct scaling on the y-axis. *For problems 43-45, use the DESCRIPTIVE variables given in the problem when you are writing your function/model. *For problems 50-55 you may use a graphing calculator; you do not need to actually show the graph in your homework. *You may use a graphing calculator to check any problem that involves graphing.