topic 2- exponential models

8/4/2019 Topic 2- Exponential Models

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LSP 120: Quantitative Reasoning and

Technological Literacy

Topic 2: Exponential Models

Prepared by Ozlem Elgun 1


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Exponential Change

Like linear relationships, exponential relationships describe a

specific pattern of change between a dependent and independentvariable

Recognizing and modeling this pattern allows us to makepredictions about future values based on an exponential rate ofchange.

While in linear relationships, y changes by a fixed absolute amountfor each change in x (such as adding 3 each time),

In exponential relationships, y changes by a fixed relative amountfor each change in x (such as adding 3% each time).

An exponential relationship is one in which for a fixed change in x,

there is a fixed percent change in y.Prepared by Ozlem Elgun 2


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HOW TO CALCULATE PERCENT CHANGE

We can use Excel to determine if a relationship is exponential

by filling the neighboring column with the percent change

from one Y to the next. As with linear, do not put this formulain the cell next to the first pair of numbers but in the cell next

to the second.

oldoldnew

originaldifferencegeercentChan

P

x y Percent change

0 1921 96 =(B3-B2)/B2

2 48

3 24



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How to Format Cells in Excel

Right click on the cell you want to format

Click on Format Cells

Select the Number Tab Choose the appropriate format you want

i.e. If the cell contains a number, select Number

Display at least two decimal points



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You can display percent change in decimal or percentformat. For descriptive or visual representations you wantto use the percent format. For calculations you want to

use the decimal format.

Above, I explain how to change the format of your cells.An easy way to format your cells to display percentages isto click on the % icon on Excel.

If the column is constant, then the relationship is

exponential. So this function is exponential.

x y Percent change

0 192

1 96 -50%

2 48 -50%

3 24 -50%



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Which of the following are

exponential?

x y

0 0

1 1

2 4

3 9

x y

0 3

1 5

2 9

3 13

x y

0 .5

5 1.5

10 4.5

15 13.5



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Exponential function equation As with linear, there is a general equation for exponential functions. The

equation for an exponential relationship is:

y = P*(1+r)x P = initial value (value ofywhenx= 0),

ris the percent change (written as a decimal),

xis the input variable (usually units of time),

Y is the output variable (i.e. population)

The equation for the above example would be

y = 192 * (1- 0.5)x

Ory = 192 * 0.5x.

We can use this equation to find values for y if given an x value.

Write the exponential equation for other examples.



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Why are exponential relationships important?

Where do we encounter them?

Populations tend to growth exponentially not linearly

When an object cools (e.g., a pot of soup on the dinner table), the temperaturedecreases exponentially toward the ambient temperature (the surroundingtemperature)

Radioactive substances decay exponentially

Bacteria populations grow exponentially

Money in a savings account with at a fixed rate of interest increases exponentially

Viruses and even rumors tend to spread exponentially through a population (atfirst)

Anything that doubles, triples, halves over a certain amount of time

Anything that increases or decreases by a percent



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If a quantity changes by a fixedpercentage,

it grows or decaysexponentially.



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How to increase/decrease a number

by a percent There are 2 ways to do this:

N = P + P * r

or

N= P * (1 + r)

According to the distributive property, these two formulas are thesame. For the work we will be doing later in the quarter, thesecond version is preferred.

Similarly to the formula above, N is the ending value (new value), Pis the starting value (initial, reference, old value) and r is the

percent (written in decimal form). To write a percent in decimalform, move the decimal 2 places to the left. Remember that ifthere is a percent decrease, you will be subtracting instead ofadding.



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Example1

Increase 50 by 10%

N= 50 + 50 * .1

= 50 + 5

= 55

OR

N = 50 * (1+.10)

= 50 * 1.10

= 55

Reminder

formulas:



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Example 2

Sales tax is 9.75%. You buy an item for $37.00.What is the final price of the article?

N = 37 + 37 * .0975 = 40.6075or

N = 37 * (1+.0975) = 40.6075

Since the answer is in dollars, roundappropriately to 2 decimal places.

The final price of the article is $40.61

Reminder

formulas:



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Example 3

In 1999, the number of crimes in Chicago was 231,265.Between 1999 and 2000 the number of crimes decreased5%. How many crimes were committed in 2000?

N = 231,265 - 231,265 * .05 = 219,701.75or

N = 231,265 * (1-.05) = 219,701.75

Since the answer is number of crimes, round appropriatelyto the nearest whole number. The number of crimes in2000 was 219,701.

Reminder

formulas:


E ti l th d i i i


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Exponential growth or decay is increasing or

decreasing by same percent over and over

If a quantity P is growing by r% each year, after one year there will be

P*(1 + r)

So, P has been multiplied by the quantity 1 + r.

If P*(1 + r) is in turn increased by rpercent, it will be multiplied by (1 + r) again.

So after two years, P has become:

P*(1+r)*(1+r) = P*(1 + r)2

So after 3 years, you have P*(1+r)3, and so on. Each year, the exponent increasesby one since you are multiplying what you had previously by another (1+r).

If a quantity P is decreasing by r%, then by the same logic, the formula is P*(1 r).



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Example: A bacteria population is at 100 and is growing by 5% per minute. 2 questions:

1. How many bacteria cells are present after one hour?

2. How many minutes will it take for there to be 1000 cells.

If the population is growing at 5% a minutes, this means it is being multiplied by 1.05 each

minute. Using Excel, we can set up a table to calculate the population at each minute. Notethat the time column begins with 0.

The formula in the second cell of the column is the cell above it multiplied by 1 + the percent

written as a decimal. (Note that you do not enter the exponent in the formula) Filling the

column will give us the population at each minute.

X

minutes

Y

population

0 100

1 =B2*(1+.05)2

3

4

5

Reminder formula for calculating the

new population after each period:

y=P*(1 + r)

where

P= starting (reference, old) value

r= rate (percent change in decimals)

y= the new value



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A th l C t A h d l ti f 125 illi i 1995 It l ti


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Another example: Country A had population of 125 million in 1995. Its population was

growing 2.1% a year. Country B had a population of 200 million in 1995. Its population

was decreasing 1.2% a year. What are the populations of the countries this year? Has

the population of country A surpassed the population of B? If not, in what year will the

population of country A surpassed the population of B?

Create your own Excel table. The formula in cell B3 is =B2*(1+0.021) and in C3 is

=B2*(1-.012). Then each column is filled. Since the As population was not larger than

Bs in 2006, the columns needed to be extended down farther.

year A B1995 125 200

1996 127.625 197.6

1997 130.3051 195.2288

1998 133.0415 192.8861

1999 135.8354 190.5714

2000 138.6879 188.2846

2001 141.6004 186.0251

2002 144.574 183.79282003 147.6101 181.5873

2004 150.7099 179.4083

2005 153.8748 177.2554

2006 157.1061 175.1283

2007 160.4054 173.0268

2008 163.7739 170.9505

2009 167.2131 168.8991

2010 170.7246 166.8723Prepared by Ozlem Elgun 18

S l i E i l E i


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Solving Exponential Equations

Remember that exponential equations are in the

form:

y = P(1+r)x

P is the initial (reference, old) value r is the rate, a.k.a. percent change (and it can be

either positive or negative)

x is time (years, minutes, hours, seconds decades

etc)

Y is the new value



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Solving for rate(percent change) and the initial value

Solving for the rate(percent change):

We already know that we solve for percent change

between two values that are one time period

apart:

oldoldnew

originaldifferencegeercentChan

P

year population (in millions) percent change (in decimal format) percent change (%)

1995 1251996 127.625 (127.625-125)/125= .021 2.1%



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Solving for the rate(percent change)What if we had to solve for the rate (percent change) but the new value is not one time period but

several time periods ahead? How do we solve for the rate then?

Example: The population of country A was 125 million in 1995. In the year 2010 its population was170.7246 million people. If we assume that the percent change (rate of population growth) wasconstant, at what rate did the population grew?

reminderour formula is y = P*(1+r)x

In this case:

y= 170.7246 million

P= 125 millionx= number of time periods from the initial value to the new value=2010-1995=15 years

Solve for r!

170.7246 = 125 *(1+r)15

170.7246/ 125 = (1+r)15

1.3657968 = (1+r)15

1.36579681/15

= (1+r)1.020999994 =1+r

1.020999994 -1 = r

0.020999994 = r

0.021 = r 2.1%

The population grew 2.1 percent every year in country A from 1995 to 2010.Prepared by Ozlem Elgun 21


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Solving for the initial value (a.k.a. old value, reference value)

What if we had to solve for the initial value?

Example: The population of a country was 125 million in 1995. If we knew that the percent change (rate of

population growth) was constant, and grew at a rate of 2.1 percent every year, what was its population

in year 1990?

reminderour formula is y = P*(1+r)x

In this case:

y= 125 millionx= number of time periods from the initial value to the new value=1995-1990=5 years

r=2.1% (0.021 in decimals)

Solve for P!

125 = P*(1+0.021)5

125 = P*(1.021)5

125= P*1.109503586

125/1.109503586= P

112.6629977=P

In year 1990, there were 112.66 million people in country A.Prepared by Ozlem Elgun 22


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Important reminder

You use y = P*(1+r)x

formula ifr is increasingyearly/monthly etc. and you are asked to calculate avalue over a number of time periods.

Example: the population is growing at 0.01% yearly. Thepopulation is 1 million now; what will it be in 10 years?

You use y = P*(1+r) formula (without x) if you are givenan r value that does not grow annually/monthly etc.but represents overall percent change across time.

Example: The population is was 1 million in 2000. It grew15% in 10 years. What was the population in 2010?


S l i f ti ( ) i E l


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Solving for time (x) using ExcelExample: The population of Bengal tigers was 4700 in 2010. If we assume that the rate of

population growth will remain constant, and will be at a rate of 2.3 percent every year, how

many years have to pass for Bengal tiger population to reach 10,000?

year

bengal tiger

population

percent

change

2010 4700

0.023



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Solving for time (using logarithms)

To solve for time, you can get an approximation by

using Excel. To solve an exponential equationalgebraically for time, you must use logarithms.

There are many properties associated withlogarithms. We will focus on the following property:

logax

= x * loga for a>0

This property is used to solve for the variable x

(usually time), where x is the exponent.Prepared by Ozlem Elgun 26

Sol ing time ( ) ith logarithms


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Solving time (x) with logarithms:

Example: The population of Bengal tigers was 4700 in 2010. If we

assume that the rate of population growth will remain constant, and

will be at a rate of 2.3 percent every year, how many years have to

pass for Bengal tiger population to reach 10,000?

Start with : Y= P * (1 + r)X.

Fill the variables that you know. To use logarithms, x (time) must be

your unknown quantity.

y= 10000

P= 4700

R=0.023

Solve for x!

The equation for this situation is: 10000 = 4700 * (1+0.023)X


Solving time (x) with logarithms: Continued.


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g ( ) g

We need to solve for x:

Step 1: divide both sides by 4700

10000/4700 = (1+0.023)X

2.12766 = (1+0.023)X

USE THE LOG property you learned earlier logax = x * loga for a>0

Step 2: take the log of both sides

log(2.12766) = log (1+0.023)X

Step 3: bring the x down in front

log(2.12766) = x * log (1+0.023)

Step 4: divide both sides by log (1+.023)

log(2.12766) /log(1+.023) = x to get 33.20316

Step 5: Write out your answer in words:

If we assume that the rate of population growth will remain constant, and will be at a rateof 2.3 percent every year 33.2 years have to pass for Bengal tiger population to reach10,000.



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Application of Exponential Models:

Carbon Dating

A radioisotope is an atom with an unstable nucleus, which is a nucleus

characterized by excess energy which is available to be imparted either to a

newly-created radiation particle within the nucleus, or else to an atomic

electron. The radioisotope, in this process, undergoes radioactive decay,

and emits a gamma ray(s) and/or subatomic particles. These particles

constitute ionizing radiation. Radioisotopes may occur naturally, but canalso be artificially produced.

Radiocarbon dating, or carbon dating, is a radiometric dating method that

uses the naturally occurring radioisotope carbon-14 (14C) to determine the

age of carbonaceous materials up to about 58,000 to 62,000 years

One of the most frequent uses of radiocarbon dating is to estimate the age

of organic remains from archaeological sites.



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The Dead Sea Scrolls are a collection of 972 documents, including texts from

the Hebrew Bible, discovered between 1946 and 1956 in eleven caves in and

around the ruins of the ancient settlement of Khirbet Qumran on the

northwest shore of the Dead Sea in the West Bank.

We date the Dead Sea Scrolls which have about 78% of the normally

occurring amount of Carbon 14 in them. Carbon 14 decays at a rate of about

1.202% per 100 years. I make a table of the form.

Using excel and extending the table we find that the Dead Sea Scrolls would

date from between 2100 to 2000 years ago.

Years (after death) % Carbon remaining percent change percent change (%)0 100

100 98.798 -0.01202 -1.202%

200

.

.


Years (after death) % Carbon remaining percent change percent change (%)


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( ) g p g p g ( )

0 100

100 98.798 -0.01202 -1.202%

200 97.61044804 -0.01202 -1.202%

300 96.43717045 -0.01202 -1.202%

400 95.27799567 -0.01202 -1.202%500 94.13275416 -0.01202 -1.202%

600 93.00127845 -0.01202 -1.202%

700 91.88340309 -0.01202 -1.202%

800 90.77896458 -0.01202 -1.202%

900 89.68780143 -0.01202 -1.202%

1000 88.60975405 -0.01202 -1.202%1100 87.54466481 -0.01202 -1.202%

1200 86.49237794 -0.01202 -1.202%

1300 85.45273956 -0.01202 -1.202%

1400 84.42559763 -0.01202 -1.202%

1500 83.41080194 -0.01202 -1.202%

1600 82.4082041 -0.01202 -1.202%

1700 81.41765749 -0.01202 -1.202%

1800 80.43901725 -0.01202 -1.202%

1900 79.47214026 -0.01202 -1.202%

2000 78.51688513 -0.01202 -1.202%

2100 77.57311217 -0.01202 -1.202%2200 76.64068337 -0.01202 -1.202%



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Use logarithms to solve for time

When were the dead sea scrolls created?78 = 100*(1-0.01202)X

.78=(0.98798) X

Log (.78)= x*log (0.98798)

-0.107905397= x*(-0.005251847)

-0.107905397/-0.005251847=x

20.546=x

Since x is in units of 100 years

Dead Sea Scrolls date back 2054.6 years

(Current estimates are that a 95% confidence interval fortheir date is 150 BC to 5 BC)


Activity 3, Question 1:


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y , QBeryllium-11 is a radioactive isotope of the alkaline metal Beryllium. Beryllium-11

decays at a rate of 4.9% every second.

a) Assuming you started with 100%, what percent of the beryllium-11 would be remaining after10 seconds? Either copy and paste the table or show the equation used to answer the question.

y = 100*(1-0.049)10

=100*(0.951)10

Y=60.51

60.61 % Beryllium-11 remains after 10 seconds.

seconds % Beryllium 11 remainig percent change percent change (%)

0 100

1 95.1 -0.049 -4.9%

2 90.4401 -0.049 -4.9%

3 86.0085351 -0.049 -4.9%

4 81.79411688 -0.049 -4.9%

5 77.78620515 -0.049 -4.9%

6 73.9746811 -0.049 -4.9%

7 70.34992173 -0.049 -4.9%

8 66.90277556 -0.049 -4.9%

9 63.62453956 -0.049 -4.9%

10 60.50693712 -0.049 -4.9%


Activity 3 Question 1


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Activity 3, Question 1Beryllium-11 is a radioactive isotope of the alkaline metal Beryllium. Beryllium-11

decays at a rate of 4.9% every second.

b) How long would it take for half of the beryllium-11 to decay? This time

is called the half life. (Use the "solve using logs" process to answer the

question) Show your work.

50 =100*(1-0.049) X

.50 =(0.951) X

Log (.50) = x*log (0. 951)

-0.301029996= x*(-0.021819483)-0.301029996 /-0.021819483=x

13.796=x

It would take 13.796 seconds for Beryllium-11 to reach its half life.


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