Exponential Growth and Decay (6

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Name ____________________
AP Calculus AB
Exponential Growth and Decay (6.4)
Law of Exponential Change: Suppose we are told that a quantity, y, increases at a rate
proportional to the amount present. If we know the amount present at t = 0 (say the
amount is yo) we can find y as a function of t by solving the following initial value
problem:
dy
 ky
dt
Initial condition: y = yo when t = 0
Law of exponential change: If y changes at a rate proportional to the amount present (
dy
 ky ) and y = yo when t =0, then y  y o e kt where k>0 represents growth and k<0
dx
represents decay. K is called the rate constant of the equation.
Name ____________________
AP Calculus AB
Three examples of exponential change:
Compound Interest
Suppose you invest $10,000 in a bank account that pays 5% in interest a year.
The interest can be given to you once at the end of the year, or at intervals throughout the
year. When given at intervals throughout the year it is called compound interest. The
advantage of compound interest is that you earn interest on the interest.
Example: $10,000 at 5%.
1) Not compounded: After one year you’ll have $10,500
2) Compounded every three months (quarterly)
.05
* 10,000 )
1st Quarter - $10,125
(
4
2nd Quarter - $10,125 + $126.56 = $10,251.56
3rd Quarter – $10,251.56 + $128.14 = $10,379.70
4th Quarter - $10,379.70 + $129.75 = $10,509.45
You get an extra $9.45 over not compounding
Formula for compound interest is:
r
A(t )  Ao (1  ) kt
k
Where:
Ao  original amount invested
A(t )  Amount at any time t
r  annual interest rate
k  # of times compounded per year
What amount will we have after 1 year on our $10,000 (5% rate) if we compound:
.05 12
)  $10,511.52
12
.05 52
Every week : A(1)  10,000(1 
)  $10,512.46
52
:
.05 365
Every day : A(1)  10,000(1 
)  $10,512.67
365
.05 8760
Every hour : A(1)  10,000(1 
)
 $10,512.70
8760
Every month : A(1)  10,000(1 
Name ____________________
AP Calculus AB
Now let’s say that instead of being added at discreet intervals let’s say the interest is
being compounded continuously, i.e. k   . We have:
r kt
A(t )  Ao lim(1  )
k 
k
r k t
A(t )  Ao lim((1  ) )
k 
k
k
Let m 
r
1 mr t
A(t )  Ao lim[(1  ) ]
m 
m
1 m rt
A(t )  Ao [ lim (1  ) ]
m 
m
1 m
Recall that lim (1  )  e
m 
m
A(t )  Ao e rt
A(1)  10,000e.05  $10,512.71
Name ____________________
AP Calculus AB
Another way to think about continuous compound interest is with the following
differential equation:
dA
 rA Where the rate of change in the amount is proportional to the amount
dt
and where r = interest rate, and A is the amount currently in the bank. Also, the initial
condition is that at t=0 A=Ao.
In this case: A(t )  Ao e rt
Example: Suppose you have $1,000,000 deposited in an account that pays 10% interest
annually. How much money will you have in ten years if:
a) The bank compounds monthly
b) The bank compounds continuously
Name ____________________
AP Calculus AB
Radioactive Decay – Occurs when atoms shed some of their mass and
transform into a new element.
The decay of a radioactive element is roughly proportional to the amount present.
It can be modeled by:
dy
 ky  y  y o e  kt where y is the mass at any time t, yo is the initial amount of the
dt
element, and k is a constant.
Half-life: The time required for half of a radioactive element to decay.
Example: Find the half-life of a radioactive substance with decay equation: y  y o e  kt and
show that the half-life depends only on k.
y  y o e  kt
1
y o  y o e  kt
2
Name ____________________
AP Calculus AB
Newton’s Law of Cooling: The rate at which an object’s
temperature is changing is proportional to the difference between its temperature and the
temperature of the surrounding medium.
dT
 k (T  TS )
dt
T  TS  (T0  TS )e  kt
Where TO  Initial temperatu re
TS  Surroundin g temperatu re
T  Temperatur e of substance at any given time t.
Name ____________________
AP Calculus AB
Example: A hardboiled egg at 98oC is put in a pan under running 18oC water to cool. After 5
minutes, the egg’s temperature is found to be 38oC. How much longer will it take the egg to
reach 20oC?
Name ____________________
AP Calculus AB
Resistance proportional to velocity: The resistance
(air) an object experiences is proportional to its velocity. The slower an object moves the less
its forward progress is resisted by the air through which it passes.
F  kv
V  VO e
k
t
m
Where V = Velocity at any given time t
Vo = Initial velocity
m = Mass of object
k = Constant
Name ____________________
AP Calculus AB
Example: For a 50kg ice skater the k is about 2.5 kg/s
a) How long will it take for the skater to coast from 7 m/s to 1 m/s?
b) How far will the skater coast until s/he comes to a complete stop?
Problem Set:
Day 1: p338 1-10,
Day 2: p338 11, 13, 15, 19, 23 + handout
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