lesson 16: exponential growth and decay
DESCRIPTION
When a quantity changes in proportion to itself (for instance, bacteria reproduction or radioactive decay), the growth or decay is exponential in nature. There are many many examples of this to be found.TRANSCRIPT
. . . . . .
Section3.4ExponentialGrowthandDecay
V63.0121.027, CalculusI
October27, 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.
. . . . . .
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.
. . . . . .
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.