Lesson 15: Exponential Growth and Decay (handout)
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Transcript of Lesson 15: Exponential Growth and Decay (handout)
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..Sec on 3.4
Exponen al Growth and Decay
V63.0121.001: Calculus IProfessor Ma hew Leingang
New York University
March 23, 2011
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Announcements
I Quiz 3 next week inrecita on on 2.6, 2.8, 3.1,3.2
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Objectives
I Solve the ordinarydifferen al equa ony′(t) = ky(t), y(0) = y0
I Solve problems involvingexponen al growth anddecay
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Notes
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Notes
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Notes
. 1.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
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OutlineRecall
The differen al equa on y′ = ky
Modeling simple popula on growth
Modeling radioac ve decayCarbon-14 Da ng
Newton’s Law of Cooling
Con nuously Compounded Interest
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Derivatives of exponential andlogarithmic functions
y y′
ex ex
ax (ln a) · ax
ln x1x
loga x1ln a
· 1x
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OutlineRecall
The differen al equa on y′ = ky
Modeling simple popula on growth
Modeling radioac ve decayCarbon-14 Da ng
Newton’s Law of Cooling
Con nuously Compounded Interest
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Notes
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Notes
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Notes
. 2.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
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What is a differential equation?Defini onA differen al equa on is an equa on for an unknown func onwhich includes the func on and its deriva ves.
Example
I Newton’s Second Law F = ma is a differen al equa on, wherea(t) = x′′(t).
I In a spring, F(x) = −kx, where x is displacement fromequilibrium and k is a constant. So
−kx(t) = mx′′(t) =⇒ x′′(t) +kmx(t) = 0.
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Showing a function is a solutionExample (Con nued)
Show that x(t) = A sinωt+ B cosωt sa sfies the differen al
equa on x′′ +kmx = 0, where ω =
√k/m.
Solu onWe have
x(t) = A sinωt+ B cosωtx′(t) = Aω cosωt− Bω sinωtx′′(t) = −Aω2 sinωt− Bω2 cosωt
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The Equation y′ = 2Example
I Find a solu on to y′(t) = 2.I Find themost general solu on to y′(t) = 2.
Solu on
I A solu on is y(t) = 2t.I The general solu on is y = 2t+ C.
RemarkIf a func on has a constant rate of growth, it’s linear.
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Notes
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Notes
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Notes
. 3.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
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The Equation y′ = 2tExample
I Find a solu on to y′(t) = 2t.I Find themost general solu on to y′(t) = 2t.
Solu on
I A solu on is y(t) = t2.I The general solu on is y = t2 + C.
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The Equation y′ = yExample
I Find a solu on to y′(t) = y(t).I Find themost general solu on to y′(t) = y(t).
Solu on
I A solu on is y(t) = et.I The general solu on is y = Cet, not y = et + C.
(check this)
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Kick it up a notch: y′ = 2yExample
I Find a solu on to y′ = 2y.I Find the general solu on to y′ = 2y.
Solu on
I y = e2t
I y = Ce2t
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Notes
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Notes
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Notes
. 4.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
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In general: y′ = kyExample
I Find a solu on to y′ = ky.I Find the general solu on to y′ = ky.
Solu on
I y = ekt
I y = Cekt
RemarkWhat is C? Plug in t = 0:
y(0) = Cek·0 = C · 1 = C,
so y(0) = y0, the ini al valueof y.
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Constant Relative Growth =⇒Exponential Growth
TheoremA func on with constant rela ve growth rate k is an exponen alfunc on with parameter k. Explicitly, the solu on to the equa on
y′(t) = ky(t) y(0) = y0
isy(t) = y0ekt
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Exponential Growth is everywhereI Lots of situa ons have growth rates propor onal to the currentvalue
I This is the same as saying the rela ve growth rate is constant.I Examples: Natural popula on growth, compounded interest,social networks
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Notes
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Notes
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Notes
. 5.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
.
.
OutlineRecall
The differen al equa on y′ = ky
Modeling simple popula on growth
Modeling radioac ve decayCarbon-14 Da ng
Newton’s Law of Cooling
Con nuously Compounded Interest
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Bacteria
I Since you need bacteriato make bacteria, theamount of new bacteriaat any moment ispropor onal to the totalamount of bacteria.
I This means bacteriapopula ons growexponen ally.
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Bacteria ExampleExample
A colony of bacteria is grown under ideal condi ons in a laboratory.At the end of 3 hours there are 10,000 bacteria. At the end of 5hours there are 40,000. How many bacteria were present ini ally?
Solu on
Since y′ = ky for bacteria, we have y = y0ekt. We have
10, 000 = y0ek·3 40, 000 = y0ek·5
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Notes
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Notes
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Notes
. 6.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
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Bacteria Example SolutionSolu on (Con nued)
We have
10, 000 = y0ek·3 40, 000 = y0ek·5
Dividing the first into the second gives
40, 00010, 000
=y0e5k
y0e3k=⇒ 4 = e2k =⇒ ln 4 = ln(e2k) = 2k
=⇒ k =ln 42
=ln 22
2=
2 ln 22
= ln 2
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OutlineRecall
The differen al equa on y′ = ky
Modeling simple popula on growth
Modeling radioac ve decayCarbon-14 Da ng
Newton’s Law of Cooling
Con nuously Compounded Interest
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Modeling radioactive decayRadioac ve decay occurs because many large atoms spontaneouslygive off par cles.
This means that in a sample of abunch of atoms, we can assume acertain percentage of them will “gooff” at any point. (For instance, if allatom of a certain radioac ve elementhave a 20% chance of decaying at anypoint, then we can expect in asample of 100 that 20 of them will bedecaying.)
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Notes
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Notes
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Notes
. 7.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
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Radioactive decay as a differential equationThe rela ve rate of decay is constant:
y′
y= k
where k is nega ve. So
y′ = ky =⇒ y = y0ekt
again!It’s customary to express the rela ve rate of decay in the units ofhalf-life: the amount of me it takes a pure sample to decay to onewhich is only half pure.
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Computing the amount remainingExample
The half-life of polonium-210 is about 138 days. How much of a100 g sample remains a er t years?
Solu onWe have y = y0ekt, where y0 = y(0) = 100 grams. Then
50 = 100ek·138/365 =⇒ k = −365 · ln 2138
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Thereforey(t) = 100e−
365·ln 2138 t = 100 · 2−365t/138
No ce y(t) = y0 · 2−t/t1/2, where t1/2 is the half-life.
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Carbon-14 Dating
The ra o of carbon-14 to carbon-12 inan organism decays exponen ally:
p(t) = p0e−kt.
The half-life of carbon-14 is about 5700years. So the equa on for p(t) is
p(t) = p0e−ln25700 t = p02−t/5700
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Notes
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Notes
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Notes
. 8.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
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Computing age with Carbon-14Example
Suppose a fossil is found where the ra o of carbon-14 to carbon-12is 10% of that in a living organism. How old is the fossil?
Solu on
We are looking for the value of t for whichp(t)p0
= 0.1. From the equa on we have
2−t/5700 = 0.1 =⇒ − t5700
ln 2 = ln 0.1 =⇒ t =ln 0.1ln 2
· 5700 ≈ 18, 940
So the fossil is almost 19,000 years old.
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OutlineRecall
The differen al equa on y′ = ky
Modeling simple popula on growth
Modeling radioac ve decayCarbon-14 Da ng
Newton’s Law of Cooling
Con nuously Compounded Interest
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Newton’s Law of CoolingI Newton’s Law of Cooling statesthat the rate of cooling of anobject is propor onal to thetemperature difference betweenthe object and its surroundings.
I This gives us a differen alequa on of the form
dTdt
= k(T− Ts)
(where k < 0 again).
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Notes
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Notes
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Notes
. 9.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
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General Solution to NLC problemsTo solve this, change the variable y(t) = T(t)− Ts. Then y′ = T′ andk(T− Ts) = ky. The equa on now looks like
dTdt
= k(T− Ts) ⇐⇒ dydt
= ky
Now we can solve!
y′ = ky =⇒ y = Cekt =⇒ T− Ts = Cekt =⇒ T = Cekt + TsPlugging in t = 0, we see C = y0 = T0 − Ts. SoTheoremThe solu on to the equa on T′(t) = k(T(t)− Ts), T(0) = T0 is
T(t) = (T0 − Ts)ekt + Ts
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Computing cooling time with NLCExample
A hard-boiled egg at 98 ◦C is put in a sink of 18 ◦C water. A er 5minutes, the egg’s temperature is 38 ◦C. Assuming the water hasnot warmed appreciably, how much longer will it take the egg toreach 20 ◦C?
Solu onWe know that the temperature func on takes the form
T(t) = (T0 − Ts)ekt + Ts = 80ekt + 18
To find k, plug in t = 5 and solve for k.
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Finding kSolu on (Con nued)
38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k
14= e5k =⇒ ln
(14
)= 5k =⇒ k = −1
5ln 4.
Now we need to solve for t:
20 = T(t) = 80e−t5 ln 4 + 18
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Notes
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Notes
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Notes
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. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
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Finding t
Solu on (Con nued)
20 = 80e−t5 ln 4 + 18 =⇒ 2 = 80e−
t5 ln 4 =⇒ 1
40= e−
t5 ln 4
− ln 40 = − t5ln 4 =⇒ t =
ln 4015 ln 4
=5 ln 40ln 4
≈ 13min
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Computing time of death with NLCExample
A murder vic m is discovered atmidnight and the temperature of thebody is recorded as 31 ◦C. One hourlater, the temperature of the body is29 ◦C. Assume that the surroundingair temperature remains constant at21 ◦C. Calculate the vic m’s me ofdeath. (The “normal” temperature ofa living human being is approximately37 ◦C.)
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Solu on
I Let me 0 be midnight. We know T0 = 31, Ts = 21, andT(1) = 29. We want to know the t for which T(t) = 37.
I To find k:
29 = 10ek·1 + 21 =⇒ k = ln 0.8
I To find t:
37 = 10et·ln(0.8) + 21 =⇒ 1.6 = et·ln(0.8)
t =ln(1.6)ln(0.8)
≈ −2.10 hr
So the me of death was just before 10:00 .
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Notes
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Notes
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Notes
. 11.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
.
.
OutlineRecall
The differen al equa on y′ = ky
Modeling simple popula on growth
Modeling radioac ve decayCarbon-14 Da ng
Newton’s Law of Cooling
Con nuously Compounded Interest
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InterestI If an account has an compound interest rate of r per yearcompounded n mes, then an ini al deposit of A0 dollarsbecomes
A0
(1+
rn
)nt
a er t years.I For different amounts of compounding, this will change. Asn → ∞, we get con nously compounded interest
A(t) = limn→∞
A0
(1+
rn
)nt= A0ert.
I Thus dollars are like bacteria.
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Continuous vs. Discrete Compounding of interestExampleConsider two bank accounts: one with 10% annual interested compoundedquarterly and one with annual interest rate r compunded con nuously. If theyproduce the same balance a er every year, what is r?
Solu onThe balance for the 10% compounded quarterly account a er t yearsis
A1(t) = A0(1.025)4t = P((1.025)4)t
The balance for the interest rate r compounded con nuouslyaccount a er t years is
A2(t) = A0ert
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Notes
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Notes
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Notes
. 12.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
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SolvingSolu on (Con nued)
A1(t) = A0((1.025)4)t
A2(t) = A0(er)t
For those to be the same, er = (1.025)4, so
r = ln((1.025)4) = 4 ln 1.025 ≈ 0.0988
So 10% annual interest compounded quarterly is basically equivalentto 9.88% compounded con nuously.
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Computing doubling time withexponential growth
Example
How long does it take an ini al deposit of $100, compoundedcon nuously, to double?
Solu on
We need t such that A(t) = 200. In other words
200 = 100ert =⇒ 2 = ert =⇒ ln 2 = rt =⇒ t =ln 2r.
For instance, if r = 6% = 0.06, we have
t =ln 20.06
≈ 0.690.06
=696
= 11.5 years.
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I-banking interview tip of the dayI The frac on
ln 2r
can alsobe approximated aseither 70 or 72 divided bythe percentage rate (as anumber between 0 and100, not a frac onbetween 0 and 1.)
I This is some mes calledthe rule of 70 or rule of72.
I 72 has lots of factors soit’s used more o en.
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Notes
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Notes
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Notes
. 13.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011
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Summary
I When something grows or decays at a constant rela ve rate,the growth or decay is exponen al.
I Equa ons with unknowns in an exponent can be solved withlogarithms.
I Your friend list is like culture of bacteria (no offense).
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Notes
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Notes
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Notes
. 14.
. Sec on 3.4: Exponen al Growth and Decay. V63.0121.001: Calculus I . March 23, 2011