Lesson 15: Exponential Growth and Decay (slides)

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Sec on 3.4Exponen al Growth and Decay

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

March 23, 2011

Announcements

I Quiz 3 next week inrecita on on 2.6, 2.8, 3.1,3.2

Objectives

I Solve the ordinarydifferen al equa ony′(t) = ky(t), y(0) = y0

I Solve problems involvingexponen al growth anddecay

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

Derivatives of exponential andlogarithmic functions

y y′

ex ex

ax (ln a) · ax

ln x1x

loga x1ln a

· 1x

OutlineRecall






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.

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

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.



Solu on

I A solu on is y(t) = 2t.

I The general solu on is y = 2t+ C.




Solu on

I A solu on is y(t) = 2t.I The general solu on is y = 2t+ C.


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.



Solu on

I A solu on is y(t) = t2.

I The general solu on is y = t2 + C.



Solu on

I A solu on is y(t) = t2.I The general solu on is y = t2 + C.

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)



Solu on

I A solu on is y(t) = et.

I The general solu on is y = Cet, not y = et + C.

(check this)



Solu on

I A solu on is y(t) = et.I The general solu on is y = Cet, not y = et + C.

(check this)

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

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.

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

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

OutlineRecall






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.

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 onSince y′ = ky for bacteria, we have y = y0ekt. We have

10, 000 = y0ek·3 40, 000 = y0ek·5

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

Solu on (Con nued)

Since y = y0et ln 2, at me t = 3 we have

10, 000 = y0e3 ln 2 = y0 · 8 =⇒ y0 =10, 000

8= 1250

OutlineRecall






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.)

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.


y′

y= k

where k is nega ve. So


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.


y′

y= k

where k is nega ve. So


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.

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

.

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.

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

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 onWe are looking for the value of t for which

p(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.




p(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





p(t)p0


2−t/5700 = 0.1 =⇒ − t5700

ln 2 = ln 0.1

=⇒ t =ln 0.1ln 2

· 5700 ≈ 18, 940





p(t)p0


2−t/5700 = 0.1 =⇒ − t5700

ln 2 = ln 0.1 =⇒ t =ln 0.1ln 2

· 5700

≈ 18, 940





p(t)p0


2−t/5700 = 0.1 =⇒ − t5700

ln 2 = ln 0.1 =⇒ t =ln 0.1ln 2

· 5700 ≈ 18, 940


OutlineRecall






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).

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


dTdt


= 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



dTdt


= 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



dTdt


= 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



dTdt


= ky

Now we can solve!

y′ = ky =⇒ y = Cekt =⇒ T− Ts = Cekt =⇒ T = Cekt + Ts

Plugging 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



dTdt


= 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


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.

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


38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k

14= e5k =⇒ ln

(14

)= 5k =⇒ k = −1

5ln 4.


20 = T(t) = 80e−t5 ln 4 + 18


38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k

14= e5k

=⇒ ln(14

)= 5k =⇒ k = −1

5ln 4.


20 = T(t) = 80e−t5 ln 4 + 18


38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k

14= e5k =⇒ ln

(14

)= 5k

=⇒ k = −15ln 4.


20 = T(t) = 80e−t5 ln 4 + 18


38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k

14= e5k =⇒ ln

(14

)= 5k =⇒ k = −1

5ln 4.


20 = T(t) = 80e−t5 ln 4 + 18

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

Finding t

Solu on (Con nued)

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

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

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

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

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.)

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 .

OutlineRecall






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.

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.

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 = A0((1.025)4)t

The balance for the interest rate r compounded con nuouslyaccount a er t years is

A2(t) = A0ert

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.

Computing doubling timeExample

How long does it take an ini al deposit of $100, compoundedcon nuously, to double?

Solu onWe 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.

I-banking interview tip of the dayI The frac on

ln 2r

can also beapproximated as either 70 or 72divided by the percentage rate(as a number between 0 and100, not a frac on between 0and 1.)

I This is some mes called the ruleof 70 or rule of 72.

I 72 has lots of factors so it’s usedmore o en.

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).

Lesson 15: Exponential Growth and Decay (slides)

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