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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 interest-bearing 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 fixed-interest 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 half-life is the time required for half the material to decay.

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Half-life

0 012

kty y e

1

ln ln2

kte

ln1 ln 2 kt 0

ln 2 kt

ln 2tk

Half-life:

ln 2half-life

k

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Model: Carbon Dating

The radioactive elementC-14 (carbon-14) has ahalf-life of 5750 years.

The percentage of C-14present in the remains ofplants or animals can beused to determine theirage.

How old is a humanbone that has lost 25%of its C-14?

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In the carbon dating problem we are not given theinitial value, only that the half-life is 5750 years.

Exponential model has the form

Use half-life 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 half-life 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% C-14 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 real--world 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