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Equation : I=P x r x t

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by

Jennifer Chan

on 12 December 2013

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Transcript of Equation : I=P x r x t

I = Prt = ($1000)(0.08)(1/4) = $20
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
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