bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§9.1b§9.1bThe Base The Base ee



Review §Review §

Any QUESTIONS About• §9.1 → Exponential Functions, base a

Any QUESTIONS About HomeWork• §9.1 → HW-42

9.1 MTH 55



Compound Interest Compound Interest Terms Terms

INTEREST ≡ A fee charged for borrowing a lender’s money is called the interest, denoted by I

PRINCIPAL ≡ The original amount of money borrowed is called the principal, or initial amount, denoted by P• Then Total AMOUNT, A, that accululates in

an interest bearing account if the sum of the Interest & Principal → A = P + I




TIME: Suppose P dollars is borrowed. The borrower agrees to pay back the initial P dollars, plus the interest amount, within a specified period. This period is called the time (or time-period) of the loan and is denoted by t.

SIMPLE INTEREST ≡ The amount of interest computed only on the principal is called simple interest.




INTEREST RATE: The interest rate is the percent charged for the use of the principal for the given period. The interest rate is expressed as a decimal and denoted by r.

Unless stated otherwise, it is assumed the time-base for the rate is one year; that is, r is thus an annual interest rate.



Simple Interest FormulaSimple Interest Formula

The simple interest amount, I, on a principal P at a rate r (expressed as a decimal) per year for t years is



Example Example Calc Simple Interest Calc Simple Interest

Rosarita deposited $8000 in a bank for 5 years at a simple interest rate of 6%

a) How much interest will she receive?

b) How much money will she receive at the end of five years?

SOLUTION a) Use the simple interest formula with:

P = 8000, r = 0.06, and t = 5



Example Example Calc Simple Interest Calc Simple Interest

SOLUTION a) Use Formula

I Prt

I $8000 0.06 5 I $2400

SOLUTION b) The total amount, A, due her in five years is the sum of the original principal and the interest earned

AP IA$8000 $2400

A$10, 400



Compound Interest FormulaCompound Interest Formula

AP 1 rn

nt

A = $-Amount after t years P = Principal (original $-amount) r = annual interest rate (expressed as a

decimal) n = number of times interest is compounded

each year t = number of years



Compare Compounding PeriodsCompare Compounding Periods

One hundred dollars is deposited in a bank that pays 5% annual interest. Find the future-value amount, A, after one year if the interest is compounded:

a) Annually.

b) SemiAnnually.

c) Quarterly.

d) Monthly.

e) Daily.




SOLUTIONIn each of the computations that follow, P = 100 and r = 0.05 and t = 1. Only n, the number of times interest is compounded each year, is changing. Since t = 1, nt = n∙1 = n.

a) AnnualAmount: AP 1

r

n

n

A100 1 0.05 $105.00




b) Semi Annual Amount:

AP 1r

n

n

A100 10.05

2

2

$105.06

AP 1r

4

4

A100 10.05

4

4

$105.09

c) Quarterly Amount:




d) Monthly Amount: AP 1

r

12

12

A100 10.05

12

12

$105.12

AP 1r

365

365

A100 10.05

365

365

$105.13

e) Daily Amount:



The Value of the Natural Base The Value of the Natural Base ee The number e, an irrational number, is

sometimes called the Euler constant. Mathematically speaking, e is the fixed

number that the expression 1

1

h

h

approaches e as h gets larger & larger

The value of e to 15 places:

e = 2.718281828459045



Continuous Compound InterestContinuous Compound Interest

The formula for Interest Compounded Continuously; e.g., a trillion times a sec.

APert A = $-Amount after t years P = Principal (original $-amount) r = annual interest rate (expressed as a

decimal) t = number of years



Example Example Continuous Interest Continuous Interest

Find the amount when a principal of $8300 is invested at a 7.5% annual rate of interest compounded continuously for eight years and three months.

SOLUTION: Convert 8-yrs & 3-months to 8.25 years. P = $8300 and r = 0.075 thenuse Formula

APert

A$8300e 0.075 8.25

A$15, 409.83



Compare Continuous CompoundingCompare Continuous Compounding

Italy's Banca Monte dei Paschi di Siena (MPS), the world's oldest bank founded in 1472 and is today one the top five banks in Italy

If in 1797 Thomas Jefferson Placed a Deposit of $450k the MPS bank at an interest rate of 6%, then find the value $-Amount for the this Account Today, 213 years Later



Compare Continuous CompoundingCompare Continuous Compounding

SIMPLE Interest

AP Prt P 1 rt A$450,000 1 0.06 213 A$6.201 million.

YEARLY Compounding

AP 1 r t $450,000 1 0.06 213

A$1.105 1011

A$110.5 million.



Account $Value for $450k invested at 6% Interest for 213 Years

159.80

145.30

0.11

0.01

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170

Co

nti

nu

ou

sQ

ua

rte

rly

Ye

arl

yS

imp

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Inte

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Co

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ou

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ing

Account Value ($B)M55_Sec9_1_Compare_Compounding_0810.xls



The NATURAL Exponential FcnThe NATURAL Exponential Fcn

The exponential function

with base e is so prevalent in the sciences that it is often referred to as THE exponential function or the NATURAL exponential function.

f x ex



Compare 2Compare 2xx, , eexx, 3, 3xx

SeveralExponentialsGraphically



Example Example Xlate Xlate eexx, Graphs, Graphs Use translation

to sketch the graph of

SOLUTION Move ex graph:• 1 Unit RIGHT

• 2 Units UP

21 xexg



Example Example Graph Exponential Graph Exponential

Graph f(x) = 2 − e−3x

SOLUTIONMake T-Table,Connect-Dots

22

1.951

1

−18.09

−401.43

y = f(x)

0

−1

−2

x



Exponential Growth or DecayExponential Growth or Decay

Math Model for “Natural” Growth/Decay:

A t A0ekt

A(t) = amount at time t

A0 = A(0), the initial amount

k = relative rate of • Growth (k > 0); i.e., k is POSITIVE

• Decay (k < 0); i.e., k is NEGATIVE

t = time



Exponential Exponential GrowthGrowth

An exponential GROWTH model is a function of the form

00 keAtA kt

where A0 is the population at time 0, A(t) is the population at time t, and k is the exponential growth rate • The doubling time is the amount of time

needed for the population to double in size

A0

A(t)

t

2A0

Doubling time

kteAA 0



Exponential Exponential DecayDecay

An exponential DECAY model is a function of the form

00 keAtA kt

where A0 is the population at time 0, A(t) is the population at time t, and k is the exponential decay rate • The half-life is the amount of time needed

for half of the quantity to decay

A0

A(t)

t½A0

Half-life

kteAA 0



Example Example Exponential Growth Exponential Growth

In the year 2000, the human population of the world was approximately 6 billion and the annual rate of growth was about 2.1 percent.

Using the model on the previous slide, estimate the population of the world in the years

a) 2030

b) 1990




SOLUTION a) Use year 2000 as t = 0 Thusfor 2030 t = 30

A0 6

k 0.021

t 30

A t 6e 0.021 30

A t 11.265663

The model predicts there will be 11.26 billion people in the world in the year 2030




SOLUTION b) Use year 2000 as t = 0 Thusfor 1990 t = −10

The model postdicted that the world had 4.86 billion people in 1990 (actual was 5.28 billion).

A0 6

k 0.021

t 10

A t 6e 0.021 10

A t 4.8635055



WhiteBoard WorkWhiteBoard Work

Problems From §9.1 Exercise Set• 40, 58, 63

Calculating

e



All Done for TodayAll Done for Today

WorldPopulation



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22

bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

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