copyright © 2015, 2008, 2011 pearson education, inc. section 4.5, slide 1 chapter 4 exponential...

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.5, Slide 1

Chapter 4Exponential Functions


4.5 Using Exponential Functions to Model Data


Exponential model, exponentially related, approximately exponentially related

Definition

An exponential model is an exponential function, or its graph, that describes the relationship between two quantities for an authentic situation. If all of the data points for a situation lie on an exponential curve, then we say the independent and dependent variables are exponentially related.


Exponential model, exponentially related, approximately exponentially related

Definition

If no exponential curve contains all of the data points, but an exponential curve comes close to all of the data points (and perhaps contains some of them), then we say the variables are approximately exponentially related.


Example: Modeling with an Exponential Function

Suppose that a peach has 3 million bacteria on it at noon on Monday and that one bacterium divides into two bacteria every hour, on average.



Let B = f(t) be the number of bacteria (in millions) on the peach at t hours after noon on Monday.

1. Find an equation of f.2. Predict the number of bacteria on the peach at noon on Tuesday.


Solution

1. We complete a table of value of f based on the assumption that one bacterium divides into two bacteria every hour.


Solution

1. As the value of t increases by 1, the value of B changes by greater and greater amounts, so it would not be appropriate to model the data by using a linear function. Note, though, that as the value of t increases by 1, the value of B is multiplied by 2, so we can model the situation by using an exponential model of the form f(t) = a(2)t. The B-intercept is (0, 3), so f(t) = 3(2)t.


Solution

1. Use a graphing calculator table and graph to verify our work.


Solution

2. Use t = 24 to represent noon on Tuesday. Substitute 24 for t in our equation of f:

f(24) = 3(2)24 = 50,331,648

According to the model, there would be 50,331,648 million bacteria. To omit writing “million,” we must add six zeroes to 50,331,648 – that is 50,331,648,000,000. There would be able 50 trillion bacteria at noon on Tuesday.


Exponential Model y = abt

If y = abt is an exponential model where y is a quantity at time t, then the coefficient a is the value of that quantity present at time t = 0.

For example, the bacteria model f(t) = 3(2)t has coefficient 3, which represents the 3 million bacteria that were present at time t = 0.



A person invests $5000 in an account that earns 6% interest compounded annually.

1. Let V = f(t) be the value (in dollars) of the account at t years after the money is invested. Find an equation of f.2. What will be the value after 10 years?


Solution

1. Each year, the investment value is equal to the previous year’s value (100% of it) plus 6% of the previous year’s value. So, the value is equal to 106% of the previous year’s value.


Solution

1. As the value of t increases by 1, the value of V is multiplied by 1.06. So, f is the exponential function f(t) = a(1.06)t.

Since the value of the account at the start is $5000, we have a = 5000. So, f(t) = 5000(1.06)t.


Solution

2. To find the value in 10 years, substitute 10 for t:

f(10) = 5000(1.06)10 ≈ 8954.24

The value will be $8954.24 in 10 years.


Half-life

Definition

If a quantity decays exponentially, the half-life is the amount of time it takes for that quantity to be reduced to half.



The world’s worse nuclear accident occurred in Chernobyl, Ukraine, on April 26, 1986. Immediately afterward, 28 people died from acute radiation sickness. So far, about 25,000 people have died from the accident, mostly due to the release of the radioactive element cesium-137 (Source: Medicine Worldwide)



Cesium-137 has a half-life of 30 years. Let P = f(t) be the percent of the cesium-137 that remains at t years since 1986.

1. Find an equation of f.2. Describe the meaning of the base of f.3. What percent of the cesium-137 will remain in 2014?


Solution

1. We discuss two methods of finding an equation of f.

Method 1 The table shows the values of P at various years t. The data can be modeled well with an exponential function. So,

301

( ) 1002

t

f t


Solution

1. We can use a graphing calculator table and graph to verify our equation. Write this equation in the form f(t) = abt:

Since we can write

P = f(t) = 100(0.977)t

30301

1 1( ) 100 100

2 2

tt

f t 1 30

10.977,

2


Solution

1. Method 2 Instead of recognizing a pattern from the table, we can find an equation of f by using the first two points on the table, (0,100) and (30, 50). Since the P-intercept is (0, 100), we have

P = f(t) = 100bt


Solution1. To find b, we substitute the coordinates of (30, 50)

into the equation: 30105 00 b30100 50b 30 50 1

100 2b

1 3012

b

0.977b

So, an equation of f is f(t) = 100(0.977)t, the same equation we found using Method 1.


Solution

2. The base of f is 0.977.Each year, 97.7% of the previous year’s cesium-137 is present. In other words, the cesium-137 decays by 2.3% each year.

3. Since 2014 – 1986 = 28, substitute 28 for t:

f(28) = 100(0.977)28 ≈ 52.13

In 2014, about 52.1% of the cesium-137 will remain.


Meaning of the Base of an Exponential Model

If f(t) = abt, where a > 0, models a quantity at time t, then the percent rate of change is constant. In particular,

• If b > 1, then the quantity grows exponentially at a rate of b – 1 percent (in decimal form) per unit of time.• If 0 < b < 1, then the quantity decays exponentially at a rate of 1 – b percent (in decimal form) per unit of time.



The number of severe near collisions on airplane runways has decayed approximately exponentially from 67 in 2000 to 16 in 2010 (Source: Federal Aviation Administration). Predict the number of severe near collisions in 2018.


Solution

Let n be the number of severe near collisions on airplane runways in the year that is t years since 2000. Known values of t and n are shown in the table below.


Solution

Because the variables t and n are approximately exponentially related, we want an equation in the form n = abt. The n-intercept is (0, 67). So, the equation is of the form

n = 67bt


SolutionTo find b, substitute (10, 16) into the equation and solve for b:

1016 67b1067 16b

10 1667

b

1 101667

b 0.867b


Solution

Substitute 0.867 for b in the equation n = 67bt:

n = 67(0.867)t

Finally, to predict the number of severe near collisions in 2018, substitute 2018 – 2000 = 18 for t in the equation and solve for n:

n = 67(0.867)18 ≈ 5.13


Solution

The model predicts that there will be about 5 severe near collisions in 2018. We use a graphing calculator table to check that each of the three ordered pairs (0, 67), (10, 16), and (18, 5.13) approximately satisfies the equation n = 67(0.867)t.


Four-Step Modeling ProcessTo find a model and then make estimates and predictions,

1. Create a scattergram of the data. Decide whether a line or an exponential curve comes close to the points.2. Find the equation of your function.3. Verify your equation by checking that the graph comes close to all of the data points.4. Use your equation of the model to draw conclusions, make estimates, and/or make predictions.

copyright © 2015, 2008, 2011 pearson education, inc. section 4.5, slide 1 chapter 4 exponential...

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