22. Exponential Growth and Decay

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Farallon Islands , California

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The number of great white sharks in a population
increases at a rate that is proportional to the number ofsharks present (at least for awhile.)
So does any population of living creatures. Other thingsthat increase or decrease at a rate proportional to theamount present include radioactive material and money in
an interestbearing account.
If the rate of change is proportional to the amount present,
the change can be modeled by:
dyky
dt

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dy kydt
1 dy k dt y
1dy k dt
y
ln y kt C
Rate of change is proportionalto the amount present.
Divide both sides byy.
Integrate both sides.

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1dy k dt
y
ln y kt C
Integrate both sides.
Exponentiate both sides.
When multiplying like bases, sumexponents. Summed exponentscan be written as multiplication.
ln y kt Ce e
C kty e e

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ln y kt Ce e
C kty e e
Exponentiate both sides.
When multiplying like bases, sumexponents. Summed exponentscan be written as multiplication.
C kty e e
kty Ae Since is a constant, let .Ce Ce A

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C kty e e
kty Ae Since is a constant, let .Ce Ce A
At , .0t 0y y0
0
ky Ae
0y A
1
0
kty y e This is the solution to our original initial
value problem.

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0
kt
y y eExponential Change:
If the constant kis positive, then the equationrepresents growth.
Ifkis negativenegative, then the equation represents decaydecay.
Note: This lecture will talk about exponential change
formulas and where they come from. The problems inthis section of the book mostly involve using thoseformulas. There are good examples in the book, which I
will not repeat here.

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Continuously Compounded Interest
If money is invested in a fixedinterest account where theinterest is added to the account ktimes per year, at an
annual interest rate r, the amount present after tyears is:
0 1kt
rA t A
k
If the money is added back more frequently, you will makea little more money.
Adding interest continuouslygives the optimal result.

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Find the amount that results from the investment:
$50 invested at 6% compounded monthly after aperiod of 3 years.
Example
)212.060
= 59.83

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Investing $1,000 at a rate of 10% compounded annually,Investing $1,000 at a rate of 10% compounded annually,quarterly, monthly, and daily will yield the followingquarterly, monthly, and daily will yield the followingamounts after 1 year:amounts after 1 year:
A = P(1 + r) = 1,000(1 + .1) = $1100.00A = P(1 + r) = 1,000(1 + .1) = $1100.00
Comparing Compounding Periods
1103.814.1000 4
1104.7112.1000 12
1105.16365.1
000365

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Of course, the bank does not employ some clerk tocontinuously calculate your interest with an adding machine.
We could calculate: 0lim 1
kt
k
rA
k
but we wont learn how to find this limit for a bit.
Since the interest is proportional to the amount present,the equation becomes:
Continuously CompoundedInterest:
0
rtA A e
You may also use:
rtA Pe
which is the same thing.

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Investing $1,000 at a rate of 10%, compoundeddailydaily yields :
1105.16365.1000365
Investing $1,000 at a rate of 10%, compoundedcontinuouslycontinuously yields :
A = 1000 eA = 1000 e.1.1 = $1105.17= $1105.17
Comparing Compounding Periods

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What amount will result from investing $100What amount will result from investing $100at 12%, compounded continuously, after aat 12%, compounded continuously, after a
period ofperiod of years.years.43
A = PeA = Pertrt
A = 100 eA = 100 e.12(3.75).12(3.75)
A = $156.83A = $156.83
Example

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Radioactive Decay
The equation for the amount ofa radioactive element left after
time tis:
0
kty y e
This allows the decay constant, k,to be positive.
The halflife is the time required for half the material to decay.

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Halflife
0 012
kty y e
1
ln ln2
kte
ln1 ln 2 kt 0
ln 2 kt
ln 2tk
Halflife:
ln 2halflife
k

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Model: Carbon Dating
The radioactive elementC14 (carbon14) has ahalflife of 5750 years.
The percentage of C14present in the remains ofplants or animals can beused to determine theirage.
How old is a humanbone that has lost 25%of its C14?

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In the carbon dating problem we are not given theinitial value, only that the halflife is 5750 years.
Exponential model has the form
Use halflife to find k.
Model: Carbon Dating
tkeytf
0)(
5750
)0()0(2
1 keff
5750
2
1
k
e

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Model: Carbon Dating
Use halflife to find k :5750
2
1 ke
0
57502ln1ln k
57502ln k
k 0.000125750
2ln
5750ln21ln
ke

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Exponential model: 25% C14 lost; 75% remains.
Thus
t
eytf
00012.0
0)(
teff 00012.0)0()0(75.0
Model: Carbon Dating
5.239712
28770
00012.0
0.2877
00012.0
75.0ln
t
t
e
00012.0
75.0
The bone is roughly
2,397 years old.

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Solve the related equation graphically
te 00012.075.0
75.000012.0 tey
Model: Carbon Dating

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Newtons Law of Cooling
Espresso left in a cup will cool to the temperature of thesurrounding air. The rate of cooling is proportional to thedifference in temperature between the liquid and the air.
(It is assumed that the air temperature is constant.)
If we solve the differential equation: sdT
k T Tdt
we get:Newtons Law of Cooling
0kt
s sT T T T e
where is the temperatureof the surrounding medium,which is a constant.
sT

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AlgorAlgorMortisMortis:
The cooling rate of the body after death. Body cools byBody cools by
RadiationRadiation
(the higher the body temperature, the more heat lost)(the higher the body temperature, the more heat lost)
Conduction depends on surface contactConduction depends on surface contact
faster if in water because enhanced contactfaster if in water because enhanced contact ConvectionConvection
Wind cools fasterWind cools faster
Rate of cooling of body after deathRate of cooling of body after death 1.51.5 F per hour underF per hour under normal conditionsnormal conditions
No realNo realworld conditions areworld conditions are normalnormal

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AlgorAlgorMortisMortis
Ambient temperature
Newtons Law of Cooling
TTis body temperature,tt is time
The bigger the temperature difference, the fasterthe cooling rate
Outdoors, temperature varies a lotmust
correct formula by varying TTambientambient
tkeTTTT ambientambient
0

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If the surrounding / ambient temperature is
constant, Newtons Law of Cooling is easy to
solve
Measure temperature at two different times
without moving the body to find k
0
( ) ( ) kt
ambient ambient
T t T T T e
AlgorAlgorMortisMortis

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ExampleExample
Suppose the temperatureof a homicide victim is
85 when it is discovered
and the ambient
temperature is 68.
If, after two hours, the
corpses temperature is
74, determine the time ofdeath. (Assume constant
ambient temperature.)

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ExampleExample
tkeTTTtT ambientambient
0
;68;74;85;6.98 210
ambientTTTT;6.30686.980 ambientTT
kte6.306885
26.306874 tke 26.306 tke
tke 6.3017
E l

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ExampleExample
tktke
2
6
17
ke 26
17
0.52076
17ln
2
1
6
17ln2
kk
tk
e
6.3017 2
6.306
tk
e

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ExampleExample
0.52076
17ln
2
1
6
17ln2
kk
Recall:
tk
e6.3017 t5207.06.30
17
ln
hrst 13.1
prior to discovery