Lesson 16: Exponential Growth and Decay

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Section3.4ExponentialGrowthandDecay

V63.0121.027, CalculusI

October27, 2009

Announcements

I Quiz3thisweekinrecitation

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Outline

Recall

Theequation y′ = ky

Modelingsimplepopulationgrowth

ModelingradioactivedecayCarbon-14Dating

Newton’sLawofCooling

ContinuouslyCompoundedInterest

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Derivativesofexponentialandlogarithmicfunctions

y y′

ax (ln a)ax

ln x1x

loga x1ln a

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Outline

Recall

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DefinitionA differentialequation isanequationforanunknownfunctionwhichincludesthefunctionanditsderivatives.

Example

I Newton’sSecondLaw F = ma isadifferentialequation,where a(t) = x′′(t).

I Inaspring, F(x) = −kx, where x isdisplacementfromequilibriumand k isaconstant. So

−kx = mx′′ =⇒ x′′ +km

I Themostgeneralsolutionis x(t) = A sinωt+ B cosωt, whereω =

√k/m.

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Example

−kx = mx′′ =⇒ x′′ +km

√k/m.

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Example

−kx = mx′′ =⇒ x′′ +km

√k/m.

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Example

−kx = mx′′ =⇒ x′′ +km

√k/m.

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Example

I Find a solutionto y′(t) = y(t).I Findthe mostgeneral solutionto y′(t) = y(t).

Solution

I A solutionis y(t) = et.I Thegeneralsolutionis y = Cet, not y = et + C.

(checkthis)

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Example

Solution

I A solutionis y(t) = et.

I Thegeneralsolutionis y = Cet, not y = et + C.

(checkthis)

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Example

Solution

I A solutionis y(t) = et.I Thegeneralsolutionis y = Cet, not y = et + C.

(checkthis)

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Ingeneral

Example

I Findasolutionto y′ = ky.I Findthegeneralsolutionto y′ = ky.

Solution

I y = ekt

I y = Cekt

RemarkWhatis C? Plugin t = 0:

y(0) = Cek·0 = C · 1 = C,

so y(0) = y0, the initialvalue of y.

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Ingeneral

Example

Solution

I y = ekt

I y = Cekt

y(0) = Cek·0 = C · 1 = C,

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Ingeneral

Example

Solution

I y = ekt

I y = Cekt

y(0) = Cek·0 = C · 1 = C,

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ExponentialGrowth

I Itmeanstherateofchange(derivative)isproportionaltothecurrentvalue

I Examples: Naturalpopulationgrowth, compoundedinterest,socialnetworks

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Outline

Recall

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Bacteria

I Sinceyouneedbacteriatomakebacteria, theamountofnewbacteriaatanymomentisproportionaltothetotalamountofbacteria.

I Thismeansbacteriapopulationsgrowexponentially.

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BacteriaExample

ExampleA colonyofbacteriaisgrownunderidealconditionsinalaboratory. Attheendof3hoursthereare10,000bacteria. Attheendof5hoursthereare40,000. Howmanybacteriawerepresentinitially?

SolutionSince y′ = ky forbacteria, wehave y = y0e

kt. Wehave

10, 000 = y0ek·3 40, 000 = y0e

Dividingthefirstintothesecondgives4 = e2k =⇒ 2k = ln 4 =⇒ k = ln 2. Nowwehave

10,000 = y0eln 2·3 = y0 · 8

So y0 =10,000

8= 1250.

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BacteriaExample

kt. Wehave

10, 000 = y0ek·3 40, 000 = y0e

10,000 = y0eln 2·3 = y0 · 8

So y0 =10,000

8= 1250.

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BacteriaExample

kt. Wehave

10, 000 = y0ek·3 40, 000 = y0e

10,000 = y0eln 2·3 = y0 · 8

So y0 =10,000

8= 1250.

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Couldyoudothatagainplease?

Wehave

10, 000 = y0ek·3

40, 000 = y0ek·5

Dividingthefirstintothesecondgives

40, 00010, 000

4 = e2k

ln 4 = ln(e2k) = 2k

k =ln 42

=ln 22

2 ln 22

= ln 2

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Outline

Recall

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Modelingradioactivedecay

Radioactivedecayoccursbecausemanylargeatomsspontaneouslygiveoffparticles.

Thismeansthatinasampleofabunchofatoms, wecanassumeacertainpercentageofthemwill“gooff”atanypoint. (Forinstance, ifallatomofacertainradioactiveelementhavea20%chanceofdecayingatanypoint,thenwecanexpectinasampleof100that20ofthemwillbedecaying.)

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Modelingradioactivedecay

Radioactivedecayoccursbecausemanylargeatomsspontaneouslygiveoffparticles.

Thismeansthatinasampleofabunchofatoms, wecanassumeacertainpercentageofthemwill“gooff”atanypoint. (Forinstance, ifallatomofacertainradioactiveelementhavea20%chanceofdecayingatanypoint,thenwecanexpectinasampleof100that20ofthemwillbedecaying.)

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Thustherelativerateofdecayisconstant:

where k is negative.

y′ = ky =⇒ y = y0ekt

again!It’scustomarytoexpresstherelativerateofdecayintheunitsofhalf-life: theamountoftimeittakesapuresampletodecaytoonewhichisonlyhalfpure.

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where k is negative. So

again!

It’scustomarytoexpresstherelativerateofdecayintheunitsofhalf-life: theamountoftimeittakesapuresampletodecaytoonewhichisonlyhalfpure.

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where k is negative. So

again!It’scustomarytoexpresstherelativerateofdecayintheunitsofhalf-life: theamountoftimeittakesapuresampletodecaytoonewhichisonlyhalfpure.

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ExampleThehalf-lifeofpolonium-210isabout138days. Howmuchofa100gsampleremainsafter t years?

SolutionWehave y = y0e

kt, where y0 = y(0) = 100 grams. Then

50 = 100ek·138/365 =⇒ k = −365 · ln 2138

Therefore

y(t) = 100e−365·ln 2138 t = 100 · 2−365t/138.

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ExampleThehalf-lifeofpolonium-210isabout138days. Howmuchofa100gsampleremainsafter t years?

SolutionWehave y = y0e

kt, where y0 = y(0) = 100 grams. Then

50 = 100ek·138/365 =⇒ k = −365 · ln 2138

Therefore

y(t) = 100e−365·ln 2138 t = 100 · 2−365t/138.

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Carbon-14Dating

Theratioofcarbon-14tocarbon-12inanorganismdecaysexponentially:

p(t) = p0e−kt.

Thehalf-lifeofcarbon-14isabout5700years. Sotheequationfor p(t) is

p(t) = p0e− ln2

5700 t

Anotherwaytowritethiswouldbe

p(t) = p02−t/5700

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ExampleSupposeafossilisfoundwheretheratioofcarbon-14tocarbon-12is10%ofthatinalivingorganism. Howoldisthefossil?

SolutionWearelookingforthevalueof t forwhich

p(t)p(0)

Fromtheequationwehave

2−t/5700 = 0.1

− t5700

ln 2 = ln 0.1

t =ln 0.1ln 2

· 5700 ≈ 18, 940

Sothefossilisalmost19,000yearsold.

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p(t)p(0)

2−t/5700 = 0.1

− t5700

ln 2 = ln 0.1

t =ln 0.1ln 2

· 5700 ≈ 18, 940

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p(t)p(0)

2−t/5700 = 0.1

− t5700

ln 2 = ln 0.1

t =ln 0.1ln 2

· 5700 ≈ 18, 940

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p(t)p(0)

2−t/5700 = 0.1

− t5700

ln 2 = ln 0.1

t =ln 0.1ln 2

· 5700 ≈ 18, 940

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Outline

Recall

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I Newton’sLawofCooling statesthattherateofcoolingofanobjectisproportionaltothetemperaturedifferencebetweentheobjectanditssurroundings.

I Thisgivesusadifferentialequationoftheform

= k(T− Ts)

(where k < 0 again).

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I Newton’sLawofCooling statesthattherateofcoolingofanobjectisproportionaltothetemperaturedifferencebetweentheobjectanditssurroundings.

I Thisgivesusadifferentialequationoftheform

= k(T− Ts)

(where k < 0 again).

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GeneralSolutiontoNLC problems

Tosolvethis, changethevariable y(t) = T(t) − Ts. Then y′ = T′

and k(T− Ts) = ky. Theequationnowlookslike

whichwecansolve:

y = Cekt

T− Ts = Cekt

=⇒ T = Cekt + Ts

Here C = y0 = T0 − Ts.

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whichwecansolve:

y = Cekt

T− Ts = Cekt

=⇒ T = Cekt + Ts

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whichwecansolve:

y = Cekt

T− Ts = Cekt

=⇒ T = Cekt + Ts

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ExampleA hard-boiledeggat 98◦C isputinasinkof 18◦C water. After5minutes, theegg’stemperatureis 38◦C. Assumingthewaterhasnotwarmedappreciably, howmuchlongerwillittaketheeggtoreach 20◦C?

SolutionWeknowthatthetemperaturefunctiontakestheform

T(t) = (T0 − Ts)ekt + Ts = 80ekt + 18

Tofind k, plugin t = 5:

38 = T(5) = 80e5k + 18

andsolvefor k.

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ExampleA hard-boiledeggat 98◦C isputinasinkof 18◦C water. After5minutes, theegg’stemperatureis 38◦C. Assumingthewaterhasnotwarmedappreciably, howmuchlongerwillittaketheeggtoreach 20◦C?

SolutionWeknowthatthetemperaturefunctiontakestheform

T(t) = (T0 − Ts)ekt + Ts = 80ekt + 18

Tofind k, plugin t = 5:

38 = T(5) = 80e5k + 18

andsolvefor k.

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Finding k

38 = T(5) = 80e5k + 18

20 = 80e5k

=⇒ k = −15ln 4.

Nowweneedtosolve

20 = T(t) = 80e−t5 ln 4 + 18

for t.

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Finding k

38 = T(5) = 80e5k + 18

20 = 80e5k

=⇒ k = −15ln 4.

Nowweneedtosolve

20 = T(t) = 80e−t5 ln 4 + 18

for t.

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Finding t

20 = 80e−t5 ln 4 + 18

2 = 80e−t5 ln 4

= e−t5 ln 4

− ln 40 = − t5ln 4

=⇒ t =ln 4015 ln 4

=5 ln 40ln 4

≈ 13min

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ExampleA murdervictimisdiscoveredatmidnightandthetemperatureofthebodyisrecordedas 31 ◦C. Onehourlater, thetemperatureofthebodyis 29 ◦C. Assumethatthesurroundingairtemperatureremainsconstantat 21 ◦C. Calculatethevictim’stimeofdeath.(The“normal”temperatureofalivinghumanbeingisapproximately 37 ◦C.)

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Solution

I Lettime 0 bemidnight. Weknow T0 = 31, Ts = 21, andT(1) = 29. Wewanttoknowthe t forwhich T(t) = 37.

I Tofind k:

29 = 10ek·1 + 21 =⇒ k = ln 0.8

I Tofind t:

37 = 10et·ln(0.8) + 21

1.6 = et·ln(0.8)

t =ln(1.6)

ln(0.8)≈ −2.10 hr

Sothetimeofdeathwasjustbefore10:00pm.

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Solution

I Tofind k:

29 = 10ek·1 + 21 =⇒ k = ln 0.8

I Tofind t:

37 = 10et·ln(0.8) + 21

1.6 = et·ln(0.8)

t =ln(1.6)

ln(0.8)≈ −2.10 hr

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Solution

I Tofind k:

29 = 10ek·1 + 21 =⇒ k = ln 0.8

I Tofind t:

37 = 10et·ln(0.8) + 21

1.6 = et·ln(0.8)

t =ln(1.6)

ln(0.8)≈ −2.10 hr

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Outline

Recall

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Interest

I Ifanaccounthasancompoundinterestrateof r peryearcompounded n times, thenaninitialdepositof A0 dollarsbecomes

after t years.

I Fordifferentamountsofcompounding, thiswillchange. Asn → ∞, weget continouslycompoundedinterest

A(t) = limn→∞

= A0ert.

I Thusdollarsarelikebacteria.

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Interest

after t years.I Fordifferentamountsofcompounding, thiswillchange. Asn → ∞, weget continouslycompoundedinterest

A(t) = limn→∞

)nt= A0ert.

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Interest

after t years.I Fordifferentamountsofcompounding, thiswillchange. Asn → ∞, weget continouslycompoundedinterest

A(t) = limn→∞

)nt= A0ert.

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ExampleHowlongdoesittakeaninitialdepositof$100, compoundedcontinuously, todouble?

SolutionWeneed t suchthat A(t) = 200. Inotherwords

200 = 100ert =⇒ 2 = ert =⇒ ln 2 = rt =⇒ t =ln 2r

Forinstance, if r = 6% = 0.06, wehave

t =ln 20.06

≈ 0.690.06

= 11.5 years.

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ExampleHowlongdoesittakeaninitialdepositof$100, compoundedcontinuously, todouble?

SolutionWeneed t suchthat A(t) = 200. Inotherwords

200 = 100ert =⇒ 2 = ert =⇒ ln 2 = rt =⇒ t =ln 2r

Forinstance, if r = 6% = 0.06, wehave

t =ln 20.06

≈ 0.690.06

= 11.5 years.

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I-bankinginterviewtipoftheday

I Thefractionln 2r

alsobeapproximatedaseither70or72dividedbythepercentagerate(asanumberbetween0and100, notafractionbetween0and1.)

I Thisissometimescalledthe ruleof70 or ruleof72.

I 72haslotsoffactorssoit’susedmoreoften.

Lesson 16: Exponential Growth and Decay

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Transcript of Lesson 16: Exponential Growth and Decay