. . . . . .

Section3.4ExponentialGrowthandDecay

V63.0121.006/016, CalculusI

March10, 2010

Announcements

I Midtermis(almost)graded, willbereturnedinrecitationI ScoresandgradeswillbemassagedI WewilldropthefivelowestWebAssign-ments

. . . . . .

Announcements

I Midtermis(almost)graded, willbereturnedinrecitationI ScoresandgradeswillbemassagedI WewilldropthefivelowestWebAssign-ments

. . . . . .

Outline

Recall

Theequation y′ = ky

Modelingsimplepopulationgrowth

ModelingradioactivedecayCarbon-14Dating

Newton’sLawofCooling

ContinuouslyCompoundedInterest

. . . . . .

Derivativesofexponentialandlogarithmicfunctions

y y′

ex ex

ax (ln a)ax

ln x1x

loga x1ln a

· 1x

. . . . . .

Outline

Recall

Theequation y′ = ky

Modelingsimplepopulationgrowth

ModelingradioactivedecayCarbon-14Dating

Newton’sLawofCooling

ContinuouslyCompoundedInterest

. . . . . .

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(t) = mx′′(t) =⇒ x′′(t) +kmx(t) = 0.

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

√k/m.

. . . . . .

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(t) = mx′′(t) =⇒ x′′(t) +kmx(t) = 0.

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

√k/m.

. . . . . .

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(t) = mx′′(t) =⇒ x′′(t) +kmx(t) = 0.

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

√k/m.

. . . . . .

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(t) = mx′′(t) =⇒ x′′(t) +kmx(t) = 0.

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

√k/m.

. . . . . .

TheEquation y′ = 2t

Example

I Find a solutionto y′(t) = 2t.I Findthe mostgeneral solutionto y′(t) = 2t.

Solution

I A solutionis y(t) = t2.I Thegeneralsolutionis y = t2 + C.

(checkthis)

. . . . . .

TheEquation y′ = 2t

Example

I Find a solutionto y′(t) = 2t.I Findthe mostgeneral solutionto y′(t) = 2t.

Solution

I A solutionis y(t) = t2.

I Thegeneralsolutionis y = t2 + C.

(checkthis)

. . . . . .

TheEquation y′ = 2t

Example

I Find a solutionto y′(t) = 2t.I Findthe mostgeneral solutionto y′(t) = 2t.

Solution

I A solutionis y(t) = t2.I Thegeneralsolutionis y = t2 + C.

(checkthis)

. . . . . .

Theequation y′ = ky

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)

. . . . . .

Theequation y′ = ky

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)

. . . . . .

Theequation y′ = ky

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)

. . . . . .

Kickitupanotch

Example

I Findasolutionto y′ = 2y.I Findthegeneralsolutionto y′ = 2y.

Solution

I y = e2t

I y = Ce2t

. . . . . .

Kickitupanotch

Example

I Findasolutionto y′ = 2y.I Findthegeneralsolutionto y′ = 2y.

Solution

I y = e2t

I y = Ce2t

. . . . . .

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.

. . . . . .

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.

. . . . . .

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.

. . . . . .

ExponentialGrowth

I Itmeanstherateofchange(derivative)isproportionaltothecurrentvalue

I Examples: Naturalpopulationgrowth, compoundedinterest,socialnetworks

. . . . . .

Outline

Recall

Theequation y′ = ky

Modelingsimplepopulationgrowth

ModelingradioactivedecayCarbon-14Dating

Newton’sLawofCooling

ContinuouslyCompoundedInterest

. . . . . .

Bacteria

I Sinceyouneedbacteriatomakebacteria, theamountofnewbacteriaatanymomentisproportionaltothetotalamountofbacteria.

I Thismeansbacteriapopulationsgrowexponentially.

. . . . . .

BacteriaExample

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

SolutionSince y′ = ky forbacteria, wehave y = y0e

kt. Wehave

10, 000 = y0ek·3 40, 000 = y0e

k·5

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

10,000 = y0eln 2·3 = y0 · 8

So y0 =10,000

8= 1250.

. . . . . .

BacteriaExample

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

SolutionSince y′ = ky forbacteria, wehave y = y0e

kt. Wehave

10, 000 = y0ek·3 40, 000 = y0e

k·5

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

10,000 = y0eln 2·3 = y0 · 8

So y0 =10,000

8= 1250.

. . . . . .

BacteriaExample

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

SolutionSince y′ = ky forbacteria, wehave y = y0e

kt. Wehave

10, 000 = y0ek·3 40, 000 = y0e

k·5

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

10,000 = y0eln 2·3 = y0 · 8

So y0 =10,000

8= 1250.

. . . . . .

Couldyoudothatagainplease?

Wehave

10, 000 = y0ek·3

40, 000 = y0ek·5

Dividingthefirstintothesecondgives

40, 00010, 000

=y0e

5k

y0e3k

=⇒ 4 = e2k

=⇒ ln 4 = ln(e2k) = 2k

=⇒ k =ln 42

=ln 22

2=

2 ln 22

= ln 2

. . . . . .

Outline

Recall

Theequation y′ = ky

Modelingsimplepopulationgrowth

ModelingradioactivedecayCarbon-14Dating

Newton’sLawofCooling

ContinuouslyCompoundedInterest

. . . . . .

Modelingradioactivedecay

Radioactivedecayoccursbecausemanylargeatomsspontaneouslygiveoffparticles.

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

. . . . . .

Modelingradioactivedecay

Radioactivedecayoccursbecausemanylargeatomsspontaneouslygiveoffparticles.

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

. . . . . .

Thustherelativerateofdecayisconstant:

y′

y= k

where k is negative.

So

y′ = ky =⇒ y = y0ekt

again!It’scustomarytoexpresstherelativerateofdecayintheunitsofhalf-life: theamountoftimeittakesapuresampletodecaytoonewhichisonlyhalfpure.

. . . . . .

Thustherelativerateofdecayisconstant:

y′

y= k

where k is negative. So

y′ = ky =⇒ y = y0ekt

again!

It’scustomarytoexpresstherelativerateofdecayintheunitsofhalf-life: theamountoftimeittakesapuresampletodecaytoonewhichisonlyhalfpure.

. . . . . .

Thustherelativerateofdecayisconstant:

y′

y= k

where k is negative. So

y′ = ky =⇒ y = y0ekt

again!It’scustomarytoexpresstherelativerateofdecayintheunitsofhalf-life: theamountoftimeittakesapuresampletodecaytoonewhichisonlyhalfpure.

. . . . . .

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.

Notice y(t) = y0 · 2−t/t1/2 , where t1/2 isthehalf-life.

. . . . . .

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.

Notice y(t) = y0 · 2−t/t1/2 , where t1/2 isthehalf-life.

. . . . . .

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.

Notice y(t) = y0 · 2−t/t1/2 , where t1/2 isthehalf-life.

. . . . . .

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

. . . . . .

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

SolutionWearelookingforthevalueof t forwhich

p(t)p(0)

= 0.1

Fromtheequationwehave

2−t/5700 = 0.1

− t5700

ln 2 = ln 0.1

t =ln 0.1ln 2

· 5700 ≈ 18, 940

Sothefossilisalmost19,000yearsold.

. . . . . .

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

SolutionWearelookingforthevalueof t forwhich

p(t)p(0)

= 0.1

Fromtheequationwehave

2−t/5700 = 0.1

− t5700

ln 2 = ln 0.1

t =ln 0.1ln 2

· 5700 ≈ 18, 940

Sothefossilisalmost19,000yearsold.

. . . . . .

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

SolutionWearelookingforthevalueof t forwhich

p(t)p(0)

= 0.1

Fromtheequationwehave

2−t/5700 = 0.1

− t5700

ln 2 = ln 0.1

t =ln 0.1ln 2

· 5700 ≈ 18, 940

Sothefossilisalmost19,000yearsold.

. . . . . .

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

SolutionWearelookingforthevalueof t forwhich

p(t)p(0)

= 0.1

Fromtheequationwehave

2−t/5700 = 0.1

− t5700

ln 2 = ln 0.1

t =ln 0.1ln 2

· 5700 ≈ 18, 940

Sothefossilisalmost19,000yearsold.

. . . . . .

Outline

Recall

Theequation y′ = ky

Modelingsimplepopulationgrowth

ModelingradioactivedecayCarbon-14Dating

Newton’sLawofCooling

ContinuouslyCompoundedInterest

. . . . . .

Newton’sLawofCooling

I Newton’sLawofCooling statesthattherateofcoolingofanobjectisproportionaltothetemperaturedifferencebetweentheobjectanditssurroundings.

I Thisgivesusadifferentialequationoftheform

dTdt

= k(T− Ts)

(where k < 0 again).

. . . . . .

Newton’sLawofCooling

I Newton’sLawofCooling statesthattherateofcoolingofanobjectisproportionaltothetemperaturedifferencebetweentheobjectanditssurroundings.

I Thisgivesusadifferentialequationoftheform

dTdt

= k(T− Ts)

(where k < 0 again).

. . . . . .

GeneralSolutiontoNLC problems

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

and k(T− Ts) = ky. Theequationnowlookslike

dTdt

= k(T− Ts) ⇐⇒ dydt

= ky

Nowwecansolve!

y′ = ky =⇒ y = Cekt

=⇒ T− Ts = Cekt

=⇒ T = Cekt + Ts

Pluggingin t = 0, wesee C = y0 = T0 − Ts. So

T(t) = (T0 − Ts)ekt + Ts

. . . . . .

GeneralSolutiontoNLC problems

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

and k(T− Ts) = ky. Theequationnowlookslike

dTdt

= k(T− Ts) ⇐⇒ dydt

= ky

Nowwecansolve!

y′ = ky =⇒ y = Cekt

=⇒ T− Ts = Cekt

=⇒ T = Cekt + Ts

Pluggingin t = 0, wesee C = y0 = T0 − Ts. So

T(t) = (T0 − Ts)ekt + Ts

. . . . . .

GeneralSolutiontoNLC problems

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

and k(T− Ts) = ky. Theequationnowlookslike

dTdt

= k(T− Ts) ⇐⇒ dydt

= ky

Nowwecansolve!

y′ = ky =⇒ y = Cekt

=⇒ T− Ts = Cekt

=⇒ T = Cekt + Ts

Pluggingin t = 0, wesee C = y0 = T0 − Ts. So

T(t) = (T0 − Ts)ekt + Ts

. . . . . .

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.

. . . . . .

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.

. . . . . .

Finding k

38 = T(5) = 80e5k + 18

20 = 80e5k

14= e5k

ln(14

)= 5k

=⇒ k = −15ln 4.

Nowweneedtosolve

20 = T(t) = 80e−t5 ln 4 + 18

for t.

. . . . . .

Finding k

38 = T(5) = 80e5k + 18

20 = 80e5k

14= e5k

ln(14

)= 5k

=⇒ k = −15ln 4.

Nowweneedtosolve

20 = T(t) = 80e−t5 ln 4 + 18

for t.

. . . . . .

Finding t

20 = 80e−t5 ln 4 + 18

2 = 80e−t5 ln 4

140

= e−t5 ln 4

− ln 40 = − t5ln 4

=⇒ t =ln 4015 ln 4

=5 ln 40ln 4

≈ 13min

. . . . . .

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

. . . . . .

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.

. . . . . .

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.

. . . . . .

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.

. . . . . .

Outline

Recall

Theequation y′ = ky

Modelingsimplepopulationgrowth

ModelingradioactivedecayCarbon-14Dating

Newton’sLawofCooling

ContinuouslyCompoundedInterest

. . . . . .

Interest

I Ifanaccounthasancompoundinterestrateof r peryearcompounded n times, thenaninitialdepositof A0 dollarsbecomes

A0

(1+

rn

)ntafter t years.

I Fordifferentamountsofcompounding, thiswillchange. Asn → ∞, weget continouslycompoundedinterest

A(t) = limn→∞

A0

(1+

rn

)nt

= A0ert.

I Thusdollarsarelikebacteria.

. . . . . .

Interest

I Ifanaccounthasancompoundinterestrateof r peryearcompounded n times, thenaninitialdepositof A0 dollarsbecomes

A0

(1+

rn

)ntafter t years.

I Fordifferentamountsofcompounding, thiswillchange. Asn → ∞, weget continouslycompoundedinterest

A(t) = limn→∞

A0

(1+

rn

)nt= A0ert.

I Thusdollarsarelikebacteria.

. . . . . .

Interest

I Ifanaccounthasancompoundinterestrateof r peryearcompounded n times, thenaninitialdepositof A0 dollarsbecomes

A0

(1+

rn

)ntafter t years.

I Fordifferentamountsofcompounding, thiswillchange. Asn → ∞, weget continouslycompoundedinterest

A(t) = limn→∞

A0

(1+

rn

)nt= A0ert.

I Thusdollarsarelikebacteria.

. . . . . .

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

=696

= 11.5 years.

. . . . . .

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

=696

= 11.5 years.

. . . . . .

I-bankinginterviewtipoftheday

I Thefractionln 2r

can

alsobeapproximatedaseither70or72dividedbythepercentagerate(asanumberbetween0and100, notafractionbetween0and1.)

I Thisissometimescalledthe ruleof70 or ruleof72.

I 72haslotsoffactorssoit’susedmoreoften.

. . . . . .

Whathavewelearnedtoday?

I Whensomethinggrowsordecaysataconstant relative rate,thegrowthordecayisexponential.

I Equationswithunknownsinanexponentcanbesolvedwithlogarithms.

I Yourfriendlistislikecultureofbacteria(nooffense).