© 2010 pearson education inc.goldstein/schneider/lay/asmar, calculus and its applications, 12e–...

© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 1 of 47

Chapter 5

Applications of the Exponential and Natural

Logarithm Functions


Exponential Growth and Decay

Compound Interest

Applications of the Natural Logarithm Function to Economics

Further Exponential Models

Chapter Outline


§ 5.1

Exponential Growth and Decay


Exponential Growth

The Exponential Growth and Decay Model

Exponential Growth in Application

Exponential Decay

Exponential Decay in Application

Section Outline


Exponential Growth

Definition Example

Exponential Growth: A quantity, such that, at every instant the rate of increase of the quantity is proportional to the amount of the quantity at that instant

tetP 43


Exponential Growth & Decay Model



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

(World’s Population) The world’s population was 5.51 billion on January 1, 1993 and 5.88 billion on January 1, 1998. Assume that at any time the population grows at a rate proportional to the population at that time. In what year will the world’s population reach 7 billion?

Since the “oldest” information we’re given corresponds to 1993, that will serve as our initial time. Therefore the year 1993 will be the year t = 0 and the population at time t = 0 is 5.51 (measured in billions). Therefore, the year 1998 will be year t = 5 and the population at time t = 5 is 5.88 (measured in billions).Since the population grows at a rate proportional to the size of the population, we can use the exponential growth model P(t) = P0ekt to describe the population of the world.

Since P0 is the initial quantity, P0 = 5.51. Therefore, our formula becomes



Now we use the other given information (5.88 billion in 1998) to determine k.

This is our function so far.

CONTINUECONTINUEDD .51.5 ktetP

ktetP 51.5

When t = 5, the population is 5.88 billion people.

551.588.55 keP

Divide.507.1 ke

Rewrite in logarithmic form.k507.1ln

Solve for k.k013.0

Therefore, our formula to model this situation is .51.5 013.0 tetP

Now we can determine when the world’s population will be 7 billion.



CONTINUECONTINUEDD

This the derived function. tetP 013.051.5

Therefore, the world’s population will be 7 billion people about 18.36 years after our initial year, 1993. This would be the year 1993 + 18.36 = 2011.36. That is, around the year 2011. The graph is given below.

Replace P(t) with 7.te 013.051.57

Divide.te 013.027.1

Rewrite in logarithmic form.t013.027.1ln

Solve for t.t36.18



CONTINUECONTINUEDD

0

1

2

3

4

5

6

7

8

9

0 5 10 15 20 25 30

Time: t years after 1993

Wo

rld

Po

pu

lati

on

: P

(t)


Exponential Decay

Definition Example

Exponential Decay: A quantity, such that, at every instant the rate of decrease of the quantity is proportional to the amount of the quantity at that instant

tetP 43



EXAMPLEEXAMPLE

(Radioactive Decay) Radium-226 is used in cancer radiotherapy, as a neutron source for some research purposes, and as a constituent of luminescent paints. Let P(t) be the number of grams of radium-226 in a sample remaining after t years, and suppose that P(t) satisfies the differential equation

.120),(00043.0 PtPtP

(a) Find the formula for P(t).(b) What was the initial amount?

(c) What is the decay constant?

(d) Approximately how much of the radium will remain after 943 years?

(e) How fast is the sample disintegrating when just one gram remains? Use the differential equation.



(a) Since the function y = Cekt satisfies the differential equation y΄ = ky, the function P(t) = Cekt = Ce-0.00043t (where k = -0.00043).

Since for the function y = Cekt, C is always the initial quantity (at time t = 0), C = 12 (since P(0) = 12). Therefore, our function is

.12 00043.0 tetP

CONTINUECONTINUEDD(f) What is the weight of the sample when it is disintegrating at the rate of 0.004 grams per year?

(g) The radioactive material has a half-life of about 1612 years. How much will remain after 1612 years? 3224 years?

SOLUTIONSOLUTION



(c) Since our exponential decay function is , the decay constant, being the coefficient of t, is -0.00043.

(b) We were given P(0) = 12. Therefore the initial amount is 12 grams.

tetP 00043.012

CONTINUECONTINUEDD

(d) To determine approximately how much of the radium will remain after 943 years, we will evaluate the function at t = 943.

tetP 00043.012 This is the decay function.

94300043.012943 eP Evaluate the function at t = 943.

8943 P Simplify.

Therefore, after 943 years, there will be approximately 8 grams remaining.



CONTINUECONTINUEDD

This is the derivative function.

(e) To determine how fast the sample is disintegrating when just one gram remains, we must first recognize that this is a situation concerning the rate of change of a quantity, namely the rate at which the radium is disintegrating. This of course involves the derivative function. This function was given to us and is . Now we will determine the value of the derivative function at P(t) = 1 (when one gram remains).

)(00043.0 tPtP

)(00043.0 tPtP

Replace P(t) with 1. 00043.0100043.0 tP

So, when there is just one gram remaining, the radium is disintegrating at a rate of 0.00043 grams/year.

(f) To determine the weight of the sample when it is disintegrating at the rate of 0.004 grams per year, we must determine P(t) when P΄(t) = -0.004.



CONTINUECONTINUEDD

This is the derivative function. )(00043.0 tPtP

Replace P΄(t) with -0.004.)(00043.0004.0 tP

Solve for P(t).)(3.9 tP

So, the weight of the sample when it is disintegrating at the rate of 0.004 grams per year, is 9.3 grams.

(g) To determine how much of the radium will remain after 1612 years, that is one half-life, we will simply recognize that after one half-life, half of the original amount of radium will be disintegrated. That is, 12/2 = 6 grams will be disintegrated and therefore 6 grams will remain.

After 3224 years, two half-lives, half of what was remaining at the end of the first 1612 years (6 grams) will remain. That is, 6/2 = 3 grams. These results can be verified using the formula for P(t).


§ 5.2

Compound Interest


Compound Interest: Non-Continuous

Compound Interest: Continuous

Applications of Interest Compounded Continuously

Section Outline




mt

m

rPA

1

• P = principal amount invested

• m = the number of times per year interest is compounded

• r = the interest rate

• t = the number of years interest is being compounded

• A = the compound amount, the balance after t years


Compound Interest

Notice that as m increases, so does A. Therefore, the maximum amount of interest can be acquired when m is being compounded all the time - continuously.




rtPeA

• P = principal amount invested

•r = the interest rate

• t = the number of years interest is being compounded

• A = the compound amount, the balance after t years



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

(Continuous Compound) Ten thousand dollars is invested at 6.5% interest compounded continuously. When will the investment be worth $41,787?

We must first determine the formula for A(t). Since interest is being compounded continuously, the basic formula to be used is

.rtPeA

Since the interest rate is 6.5%, r = 0.065. Since ten thousand dollars is being invested, P = 10,000. And since the investment is to grow to become $41,787, A = 41,787. We will make the appropriate substitutions and then solve for t.

rtPeA This is the formula to use.

te 065.0000,10787,41 P = 10,000, r = 0.065, and A = 41,787.



Therefore, the $10,000 investment will grow to $41,787, via 6.5% interest compounded continuously, in 22 years.

te 065.01787.4 Divide by 10,000.

CONTINUECONTINUEDD

t065.01787.4ln Rewrite the equation in logarithmic form.

t22 Divide by 0.065 and solve for t.


Compound Interest: Present Value

Variables are defined the same as in Slide #21.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

(Investment Analysis) An investment earns 5.1% interest compounded continuously and is currently growing at the rate of $765 per year. What is the current value of the investment?

Since the problem involves a rate of change, we will use the formula for the derivative of That is, A΄ = rA. Since the investment is growing at a rate of $765 per year, A΄ = 765. Since the interest rate is 5.1%, r = 0.051.

.rtPeA

rAA This is the given function.

A051.0765 A΄ = 765 and r = 0.051.A000,15 Solve for A.

Therefore, the value of A for this situation is 15,000. We can now use this, and the present value formula, to determine P.



This is the present value formula.

Therefore, the current value is $14,254.18.

CONTINUECONTINUEDD

rtAeP

A = 15,000, r = 0.051 and t = 1 (since we were given the rate of growth per year).

1051.0000,15 eP

Simplify.18.254,14P


§ 5.3

Applications of the Natural Logarithm Function to Economics


Relative Rates of Change

Elasticity of Demand

Section Outline


Relative Rate of Change

Definition Example

Relative Rate of Change: The quantity on either side of the equation

is often called the relative rate of change of f (t) per unit change of t (a way of comparing rates of change for two different situations).

An example will be given immediately hereafter.

tf

tftf

dt

d ln



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

(Percentage Rate of Change) Suppose that the price of wheat per bushel at time t (in months) is approximated by

What is the percentage rate of change of f (t) at t = 0? t = 1? t = 2?

Since

.01.0001.04 tettf

,01.0001.0 tetf

we see that %.22.0

01.4

009.0

01.04

01.0001.0

0

0

f

f

%.65.0

0047.4

026.0

01.01001.04

01.0001.0

1

11

1

e

e

f

f



So at t = 0 months, the price of wheat per bushel contracts at a relative rate of 0.22% per month; 1 month later, the price of wheat per bushel is still contracting, but more so, at a relative rate of 0.65%. One month after that (t = 2), the price of wheat per bushel is contracting, but much less so, at a relative rate of 0.0087%.

%.0087.0

0034.4

00035.0

01.02001.04

01.0001.0

2

22

2

e

e

f

f

CONTINUECONTINUEDD



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

(Elasticity of Demand) A subway charges 65 cents per person and has 10,000 riders each day. The demand function for the subway is

(a) We must first determine E(p).

.902000 pq

(a) Is demand elastic or inelastic at p = 65?

(b) Should the price of a ride be raised or lowered in order to increase the amount of money taken in by the subway?

ppfq 902000 p

pfq

90

1000



Now we will determine for what value of p E(p) = 1.

p

p

p

pp

pf

pfppE

2180902000

90

1000

CONTINUECONTINUEDD

12180

p

p Set E(p) = 1.

pp 2180 Multiply by 180 – 2p.

1803 p Add 2p to both sides.

60p Divide both sides by 3.

So, p = 60 is the point at which E(p) changes from elastic to inelastic, or visa versa.



Through simple inspection, which we could have done in the first place, we can determine whether the value of the function E(p) is greater than 1 (elastic) or less than 1 (inelastic) at p = 65.

CONTINUECONTINUEDD

So, demand is elastic at p = 65.

13.1652180

6565

E

(b) Since demand is elastic when p = 65, this means that for revenue to increase, price should decrease.


§ 5.4

Further Exponential Models


Population Growth Equations

Exponential Models in Application

Section Outline


Population Growth Equations

ktePtf 1 PktBe

Py

1



EXAMPLEEXAMPLE

(Spread of News) A news item is spread by word of mouth to a potential audience of 10,000 people. After t days,

people will have heard the news. The graph of this function is found below.

te

tf4.0501

000,10

(a) Approximately how many people will have heard the news after 7 days?

(b) At approximately what rate will the news be spreading after 14 days?

(c) Approximately when will 7000 people have heard the news?(d) Approximately when will the news be spreading at the rate of 600 people per day?

(e) When will the news be spreading at the greatest rate?



(f) Use and to determine the differential

equation satisfied by f (t).

CONTINUECONTINUEDD

MktBe

My

1 yMkyy



CONTINUECONTINUEDD

SOLUTIONSOLUTION

(a) To determine approximately how many people will have heard the news after 7 days, we will evaluate f (7).

24750608.0501

000,10

501

000,10

501

000,107

8.274.0

ee

f

So, after 7 days, we would expect about 2475 people to have heard the news.

(b) To determine at approximately what rate the news will be spreading after 14 days, we will evaluate f ΄(14) (we use the derivative of the function since we seek a rate of change). We first use the quotient rule.

te

tf4.0501

000,10

This is the given function.



CONTINUECONTINUEDD

24.0

4.04.0

501

000,10504.005010t

tt

e

eetf

Use the quotient rule.

24.0

4.0

501

000,200t

t

e

etf

Simplify.

76.526404.1

57.739

501

000,20014 2144.0

144.0

e

ef Evaluate f ΄(14).

So, after 14 days, the news will be spreading at approximately 526.76 people per day.

(c) To determine approximately when 7000 people will have heard the news, we replace f (t) with 7000 and then solve for t.



CONTINUECONTINUEDD

tetf

4.0501

000,10


te 4.0501

000,107000

Replace f (t) with 7000.

000,105017000 4.0 te Multiply by the denominator.

000,10000,3507000 4.0 te Distribute.

000,3000,350 4.0 te Subtract.

0086.04.0 te Divide.

0086.0ln4.0 t Rewrite in logarithm form.

89.114.0

0086.0ln

t Divide.



CONTINUECONTINUEDDSo, 7000 people will have heard the news after approximately 11.89 days.(d) To determine approximately when the news will be spreading at the rate of 600 people per day, we need to replace f ΄(t) with 600. However, since this will be a long, messy process (a good algebraic exercise for you), we will just look at the given graph where f ΄(t) = 600.

As can be seen on the graph, the derivative has a value of 600 when t ≈ 6 or t ≈ 13.5.



CONTINUECONTINUEDD(e) The news will be spreading at the greatest rate when the rate (derivative) is greatest. That is,

at t = 10 (or when f ΄΄(t) = 0).



CONTINUECONTINUEDD(f) Since we know satisfies the differential equation

then we can rewrite f (t) in the form . Upon doing this, we will

have defined M, k, and B for our function. We can then use to create a differential equation satisfied by f (t).

MktBe

My

1 yMkyy

MktBe

My

1 yMkyy

te

tf4.0501

000,10


Mk 4.0

Since the number 10,000 is by itself in the numerator, it must be that M = 10,000. And since the number 50 is the only coefficient of e-0.4t, it must be that B = 50. So we must now rewrite -0.4 in the form –Mk to determine k. That is

Mk4.0

k000,104.0

.00004.0 k



CONTINUECONTINUEDDTherefore,

.501

000,10000,10

501

000,1000004.0000,1000004.0

4.04.0

tt ee

yyyMkyy

.501

000,10

501

000,1000004.0000,104.0 tt ee

tf

So, f (t) satisfies the differential equation

© 2010 pearson education inc.goldstein/schneider/lay/asmar, calculus and its applications, 12e–...

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