modeling our world

Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 9, Unit C, Slide 1

Modeling Our World9


Unit 9C

Exponential Modeling


Exponential Functions

An exponential function grows (or decays) by the same relative amount per unit time. For any quantity Q growing exponentially with a fractional growth rate r,

Q = Q0 (1+r)t

whereQ = value of the exponentially growing quantity at time tQ0 = initial value of the quantity (at t = 0)

r = fractional growth rate for the quantityt = time

Negative values of r correspond to exponential decay.Note that the units of time used for t and r must be the same.While an exponential growing quantity has a constant relative growth rate, its absolute growth rate increases.


ExampleThe 2010 census found a U.S. population of about 309 million, with an estimated growth rate of 0.9% per year. Write an equation for the U.S. population that assumes exponential growth at this rate. Use the equation to predict the U.S. population in 2100.

Solution

Initial value is the 2010 population, Q0 = 309 million. Growth rate is P% = 0.9% per year

Fractional growth rate is r = P/100 = 0.009 per year

Q = Q0 × (1 + r)t = 309 million × (1 + 0.009)t

= 309 million × (1.009)t


Example (cont)

Note that, because the units of r are per year, t must be measured in years. The year 2100 is t = 90 years after 2010.

Our exponential function therefore predicts a 2100 population of

Q = 309 million × (1.009)90 ≈ 692 million


To graph an exponential function, use points corresponding to several doubling times (or half-lives, in the case of decay). Start at the point (0,Q0), the initial value at t = 0.

For an exponentially growing quantity, the value of Q is 2Q0 (double the initial value) after one doubling time (Tdouble), 4Q0 after two doubling times (2Tdouble), 8Q0 after three doubling times (3Tdouble), and so on.

For an exponentially decaying quantity, the value of Q falls to Q0/2 (half the initial value) after one half-life (Thalf), Q0/4 after two half-lives (2Thalf), Q0/8 after three half-lives (3Thalf), and so on.

Graphing Exponential Functions


Exponential Growth

To graph exponential growth, first plot the points (0,Q0), (Tdouble,2Q0), (2Tdouble,4Q0), (3Tdouble,8Q0), and so on. Then fit a curve between these points, as shown tothe right.


Exponential Decay

To graph exponential decay, first plot the points (0,Q0), (Thalf,Q0/2), (2Thalf,Q0/4), (3Thalf,Q0/8), and so on. Then fit a curve between these points, as shown to the right.


If given the growth or decay rate r, use the form

If given the doubling time Tdouble, use the form

If given the half-life Thalf, use the form

Forms of the Exponential Function


Example: China’s Coal Consumption

China’s rapid economic development has lead to an exponentially growing demand for energy, and China generates more than two-thirds of its energy by burning coal. During the period 2000 to 2012, China’s coal consumption increased at an average rate of 8% per year, and the 2012 consumption was about 3.8 billion tons of coal.

a. Use these data to predict China’s coal consumption in 2020.

If t = 0 represents 2012 Q0 = 3.8, r = 0.08, and t = 8 years.

Q = Q0 (1+r)t = 3.8 (1 + 0.08)8 = 3.8 (1.08)8 ≈ 7.0

China’s predicted coal consumption is about 7 billion tons.


Example (cont)

b. Make a graph projecting China’s coal consumption through 2050. Discuss the validity of the model.

There are several ways to make the graph, but let’s do it by finding the doubling time. Using the exact doubling time formula (see box, p. 486), we find

10 102

10 10

log 2 log 29.0

log (1 ) log (1.08)

T

r


Example (cont)

Given China’s serious problems with pollution from coal burning and concerns about the impact of coal burning on global warming, it seems unlikely that such an enormous increase in coal consumption will really occur.


Example

Consider an antibiotic that has a half-life in the bloodstream of 12 hours. A 10-milligram injection of the antibiotic is given at 1:00 p.m. How much antibiotic remains in the blood at 9:00 p.m.? Draw a graph that shows the amount of antibiotic remaining as the drug is eliminated by the body.

Solution

Q0 = 10 milligrams is the initial dose; t = 0 Q is the amount of antibiotic in the blood t hours later. Half-life is 12 hours.

/121

102

t

Q


Example (cont)

At 9:00 p.m., which is t = 8 hours after the injection, the amount of antibiotic remaining is

8/12

2/3

110

2

110

2

6.3 mg

Q


Example (cont)

Eight hours after the injection, 6.3 milligrams of the antibiotic remain in the bloodstream. Graphing this exponential decay function up to t = 100 hours, we see that the amount of antibioticdecreases steadily toward zero.


ExampleThe famous Allende meteorite lit up the skies of Mexico as it fell to Earth on February 8, 1969. Laboratory studies have shown that potassium-40 decays into argon-40 with a half-life of about 1.25 billion (1.25 × 109) years and that all the argon-40 in the meteorite must be a result of such decay. By comparing the amounts of the two substances in the meteorite samples, scientists determined that only 8.5% of the potassium-40 originally present in the rock remains today (the rest has decayed into argon-40). How old is the rock that makes up the Allende meteorite?


Example (cont)Our goal is to find t, which is the age of the rock. We are given that the half-life of potassium-40 is Thalf = 1.25 × 109 years and that 8.5% of the original potassium-40 remains, which means Q/Q0 = 0.085.

half/

0

1

2

t TQ

Q

half/

10 10 0log 1/ 2 log /t T

Q Q

10 10 0half

log (1/ 2) log /t

Q QT

10 0

half10

log /

log (1/ 2)

Q Qt T

9 10

10

log 0.0851.25 10 yr

log (1/ 2)t

94.45 10 yrt

The Allende meteorite is about 4.45 billion years old


Exponential growth functions have rates of change that increase.

Exponential decay functions have rates of change that decrease.

Linear functions have straight line graphs and constant rates of change. Exponential functions have graphs that rise or fall steeply and have variable rates of change.

Changing Rates of Change

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