I = ($1020)(0.08)(1/4) = $20.40

I = ($1040.40)(0.08)(1/4) = $20.81

I = ($1061.21)(0.08)(1/4) = $21.22

After 1 year, account contains : $1082.43

U(t)=T+(U-T)e^kt

Problem

**Chapter 5.6 - 5.7**

**Compound Interest and Growth and Decay**

Interest

Simple Interest Equation : I=Prt

I = Interest

r = rate

t = time

**Growth And**

**Decay**

Growth:A=Aoe^kt

(k>0)

Decay: A=Aoe^kt

(k<0)

Compound Interest

Cooling Problem

Example

10,000(0.5)(6)

per annum (year)

2 Types of Compound Interest

Compound Periodically

Compound Continuously

A=P(1+r/n)^nt

A=Pe^rt

Newton's Law of Cooling

A credit union pays interest of 8% per annum compounded quarterly on a certain savings plan. If $1000 is deposited in such a plan and the interest is left to accumulate, how much is in the account after 1 year?

A= Initial Value

K= Rate of growth/decay

T= Time

The temperature of any heated object will decrease exponentially over time towards the temperature of the environment.

k<0

An object is heated to 100 Celsius and is then allowed to cool in a room whose air temperature is 30 Celsius.

If the temperature of the object is 80 Celsius after 5 minutes, when will its temperature be 50 Celsius?

u(t)=T+(U-T)e^kt

u(t) = 30 + (100-30)e^kt = 30 + 70e^kt

u = 80 when t = 5

80 = 30 + 70e^k(5)

50 = 70e^5k

e^5k = 50/70

5k = ln(5/7)

k = (1/5)(ln)(5/7) = -0.0673

Estimating the Age of Ancient Tools

Traces of burned wood along with ancient stone tools in an archaeological dig in Chile were found to contain approximately 1.67% of the original amount of carbon 14.

If the half-life of Carbon-14 is 5600 years, approximately when was the tree cut and burned?

A(t) = A e^kt

o

A = Original amount of Carbon-14 present

A(5600) = 1/2(A)

o

o

1/2(A ) = A e^(k5600)

1/2 = e^(5600k)

5600k = ln(1/2)

k = (1/5600)ln(1/2) = -0.000124

0.0167A = A (e^-o.ooo124t)

0.0167 = e^-0.000124t

-0.000124t = ln(0.0167)

t = (1/-0.000124)ln(0.0167) = 33,000 years

$100 invested at 4% compounded

quarterly after a period of 2 years

P = present value of money

Radioactive Decay

t = years

r = per annum interest rate

The half-life of radium is 1690

years, if 10 grams is present

now, how much will be present

in 50 years?

A thermometer reads 60 F after 2 minutes, what will it read after 7 minutes?

Practice Problem

A = future value of money

n = the amount of times

compounded per year

o

o

o

o

A=P(1+r/n)^nt

P = present value of money

r = per annum interest rate

$100 invested at 4% compounded

quarterly after a period of 2 years

A=P(1+r/n)^nt

A=100(1+(0.04/4))^(4x2)

A=100(1+(0.04/4))^8

A=100(1+0.01)^8

A=100(1.01)^8

A=100(1.082857)

A=$108.29 after 2 years

A= Initial Value

K= Rate of growth/decay

T= Time

A(t)=Aoe^kt

0.5=e^k(1690)

ln0.5=1690k

(ln0.5)/1690=k

k=-0.00041014

A(t)=10e^(50)(-0.00041014)

A(t)=10e^-0.020507

A(t)=9.797

A = future value of money

n = the amount of times

compounded per year

A = future value of money

n = the amount of times

compounded per year

t = years

r = per annum interest rate

P = present value of money

A=P(1+r/n)^nt

A = future value of money

n = the amount of times

compounded per year

P = present value of money

t = years

r = per annum interest rate

Simple Interest Equation : I=Prt

I = Interest

r = rate

t = time