22. Exponential Growth and Decay

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Transcript of 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 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