Ppt Applicaton of EXF

8/10/2019 Ppt Applicaton of EXF

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The best thing about exponentialfunctions is that they are so

useful in real world situations

Applications of Exponential Functions


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Exponential functions

are used to :

1) model populations2) carbon date artifacts3) help coroners determine time of death

4) compute investments5) as well as many other applications


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1. Population

Many times scientists will start with a certain

number of bacteria or animals and watchhow the population grows. For example, ifthe population doubles every 5 days, this canbe represented as an exponential function.

Most population models involve using thenumber e. To learn more about e, click here(link to exp-log-e and ln.doc)


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Population

Modelscan occur

two ways :

1st way: if we are givenan exponential function

2ndway: involves coming up with

an exponential equationbased on information given


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(1) The population of a city is P = 250,342e0.012twhere t =

0 represents the population in the year 2000.a. Find the population of the city in the year 2010.b. Find the population of the city in the year 2015.c. Find when the population will be 320,000.

Let's Practice:


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c. We know the population in the year 2015 is almost 300,000 from ourwork 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 tosolve is 320,000 = 250,342e 0.012(t)

So it will takebetween 20 and

21 years for thepopulation toreach 320,000.This meansbetween theyears 2020 and

2021 thepopulation will be320,000.


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Before we do the next example, lets look at ageneral form for population models. Most of thetime, we start with an equation that looks like:

P = Poekt

P represents the population after a certain amountof time

Porepresents the initial population or the populationat 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 tous in the problem.

Summary:


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(1) A scientist starts with 100 bacteria in an experiment.

After 5 days, she discovers that the population hasgrown to 350.

a. Determine an equation for this bacteria population.

b. Use the equation to find out the population after 15

days.c. Use the equation to find out when the population is

1000.

Let's Practice this once more:


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(1) a. To find the equation, we need to know values for Poand k.Remember the equation is in the form P = Poekt. where P, e, and t are allparts of the equation we will come up with. We only need values for Poand k. Pois given by the amount the scientist starts with which is 100.Finding k requires a little more work.We know that Pois 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 backto our general form of andreplace Po and k. So ourequation isP = 100e 0.25055t

Answers :


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b. 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.

c. We will set our equation equal to 1000 to get 1000 = 100e 0.25055tandsolve.

So between 9 and 10days, the bacteriapopulation will be1000


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2. Exponential Decay

Solving an exponential decay problem is very

similar to working with population growth. Infact, certain populations may decreaseinstead of increase and we could still use thegeneral formula we used for growth. But in

the case of decrease or decay, the value of kwill be negative.


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Let's Practice:

(1)The number of milligrams of a drug in a personssystem after t hours is given by the function D =20e-0.4t.

a.Find the amount of the drug after 2 hours.

b. Find the amount of the drug after 5 hours.

c. When will the amount of the drug be 0.1

milligram (or almost completely gone from thesystem)?


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(1) a. 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.

b. 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.

c. We need to let D = 0.1 and solve the equation 0.1 = 20e-0.4t

Answer :

After approximately 13 hours and 15minutes, the amount of the drug will

be almost gone with only 0.1milligrams remaining in the body.


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3. Compound Interest

The formula for interest that is compounded is

A represents the amount of money after a certain amount of timeP represents the principle or the amount of money you start with

r represents the interest rate and is always represented as a decimaln is the number of times interest is compounded in one year

if interest is compounded annually then n = 1if interest is compounded quarterly then n = 4if interest is compounded monthly then n = 12

t represents the amount of time in years


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(1) Suppose your parents invest $1000 in a savingsaccount for college at the time you are born. The

average interest rate is 4% and is compoundedquarterly. How much money will be in the collegeaccount when you are 18 years old?

(2) Suppose your parents had invested that same $1000 ina money market account that averages 8% interest

compounded monthly. How much would you have forcollege after 18 years?

Let's Practice:


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(1)We will use our formula and letP = 1000, r = 0.04, n = 4 and t = 18.

(2)P = 1000, r = 0.08, n = 12 and t = 18

Answer :

http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_ExponentsApps.xml

Ppt Applicaton of EXF

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