Lesson 15: Exponential Growth and Decay (slides)

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′

ax (ln a) · ax

ln x1x

loga x1ln a

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.

Example

−kx(t) = mx′′(t) =⇒ x′′(t) +kmx(t) = 0.

Example

−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

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

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

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.

Solu on

I y = ekt

I y = Cekt

y(0) = Cek·0 = C · 1 = C,

Solu on

I y = ekt

I y = Cekt

y(0) = Cek·0 = C · 1 = C,

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 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 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.)

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:

where k is nega ve.

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.

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.

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.

50 = 100ek·138/365 =⇒ k = −365 · ln 2138

365·ln 2138 t = 100 · 2−365t/138

50 = 100ek·138/365 =⇒ k = −365 · ln 2138

365·ln 2138 t = 100 · 2−365t/138

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

= 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

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

= k(T− Ts)

(where k < 0 again).

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

= 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

= k(T− Ts) ⇐⇒ dydt

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

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

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

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

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

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.

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

)= 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

)= 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

=⇒ k = −15ln 4.

20 = T(t) = 80e−t5 ln 4 + 18

38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k

14= e5k =⇒ ln

)= 5k =⇒ k = −1

5ln 4.

20 = T(t) = 80e−t5 ln 4 + 18

38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k

14= e5k =⇒ ln

)= 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 .

Solu on

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

Solu on

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

OutlineRecall

InterestI If an account has an compound interest rate of r per yearcompounded n mes, then an ini al deposit of A0 dollarsbecomes

a er t years.

I For different amounts of compounding, this will change. Asn → ∞, we get con nously compounded interest

A(t) = limn→∞

= A0ert.

I Thus dollars are like bacteria.

a er t years.I For different amounts of compounding, this will change. Asn → ∞, we get con nously compounded interest

A(t) = limn→∞

)nt= A0ert.

a er t years.I For different amounts of compounding, this will change. Asn → ∞, we get con nously compounded interest

A(t) = limn→∞

)nt= A0ert.

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

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

= 11.5 years.

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

= 11.5 years.

I-banking interview tip of the dayI The frac on

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).

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