Lesson 16: Exponential Growth and Decay
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Transcript of Lesson 16: Exponential Growth and Decay
. . . . . .
Section3.4ExponentialGrowthandDecay
V63.0121.034, CalculusI
October26, 2009
Announcements
I Quiz3thisweekinrecitation
. . . . . .
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 = mx′′ =⇒ x′′ +km
= 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 = mx′′ =⇒ x′′ +km
= 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 = mx′′ =⇒ x′′ +km
= 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 = mx′′ =⇒ x′′ +km
= 0.
I Themostgeneralsolutionis x(t) = A sinωt+ B cosωt, whereω =
√k/m.
. . . . . .
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)
. . . . . .
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.
. . . . . .
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.
. . . . . .
ExponentialGrowthI Itmeanstherateofchange(derivative)isproportionaltothecurrentvalue
I Examples: Naturalpopulationgrowth, compoundedinterest,socialnetworks
again,
. . . . . .
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.
. . . . . .
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.
. . . . . .
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
dydt
= ky
whichwecansolve:
y = Cekt
T− Ts = Cekt
=⇒ T = Cekt + Ts
Here C = y0 = T0 − Ts.
. . . . . .
GeneralSolutiontoNLC problems
Tosolvethis, changethevariable y(t) = T(t) − Ts. Then y′ = T′
and k(T− Ts) = ky. Theequationnowlookslike
dydt
= ky
whichwecansolve:
y = Cekt
T− Ts = Cekt
=⇒ T = Cekt + Ts
Here C = y0 = T0 − Ts.
. . . . . .
GeneralSolutiontoNLC problems
Tosolvethis, changethevariable y(t) = T(t) − Ts. Then y′ = T′
and k(T− Ts) = ky. Theequationnowlookslike
dydt
= ky
whichwecansolve:
y = Cekt
T− Ts = Cekt
=⇒ T = Cekt + Ts
Here C = y0 = T0 − 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
)nt
after 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
)nt
after 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
)nt
after 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.