Lesson 15: Exponential Growth and Decay (Section 021 slides)

Section 3.4Exponential Growth and Decay

V63.0121.021, Calculus I

New York University

October 28, 2010

Announcements

I Quiz 3 next week in recitation on 2.6, 2.8, 3.1, 3.2

. . . . . .

. . . . . .

Announcements

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

V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 2 / 40

. . . . . .

Objectives

I Solve the ordinarydifferential equationy′(t) = ky(t), y(0) = y0

I Solve problems involvingexponential growth anddecay


. . . . . .

Outline

Recall

The differential equation y′ = ky

Modeling simple population growth

Modeling radioactive decayCarbon-14 Dating

Newton’s Law of Cooling

Continuously Compounded Interest


. . . . . .

Derivatives of exponential and logarithmic functions

y y′

ex ex

ax (lna) · ax

ln x1x

loga x1ln a

· 1x


. . . . . .

Outline

Recall







. . . . . .

What is a differential equation?

DefinitionA differential equation is an equation for an unknown function whichincludes the function and its derivatives.

Example

I Newton’s Second Law F = ma is a differential equation, wherea(t) = x′′(t).

I In a spring, F(x) = −kx, where x is displacement from equilibriumand k is a constant. So

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

I The most general solution is x(t) = A sinωt+ B cosωt, whereω =

√k/m.


. . . . . .



Example



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


√k/m.


. . . . . .



Example



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


√k/m.


. . . . . .



Example



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


√k/m.


. . . . . .

Showing a function is a solution

Example (Continued)

Show that x(t) = A sinωt+ B cosωt satisfies the differential equation

x′′ +kmx = 0, where ω =

√k/m.

SolutionWe have

x(t) = A sinωt+ B cosωtx′(t) = Aω cosωt− Bω sinωt

x′′(t) = −Aω2 sinωt− Bω2 cosωt

Since ω2 = k/m, the last line plus k/m times the first line result in zero.


. . . . . .

Showing a function is a solution

Example (Continued)

Show that x(t) = A sinωt+ B cosωt satisfies the differential equation

x′′ +kmx = 0, where ω =

√k/m.

SolutionWe have

x(t) = A sinωt+ B cosωtx′(t) = Aω cosωt− Bω sinωt

x′′(t) = −Aω2 sinωt− Bω2 cosωt

Since ω2 = k/m, the last line plus k/m times the first line result in zero.


. . . . . .

The Equation y′ = 2

Example

I Find a solution to y′(t) = 2.I Find the most general solution to y′(t) = 2.

Solution

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

RemarkIf a function has a constant rate of growth, it’s linear.


. . . . . .


Example


Solution

I A solution is y(t) = 2t.

I The general solution is y = 2t+ C.



. . . . . .


Example


Solution




. . . . . .


Example


Solution




. . . . . .

The Equation y′ = 2t

Example

I Find a solution to y′(t) = 2t.I Find the most general solution to y′(t) = 2t.

Solution

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


. . . . . .


Example


Solution

I A solution is y(t) = t2.

I The general solution is y = t2 + C.


. . . . . .


Example


Solution

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


. . . . . .

The Equation y′ = y

Example

I Find a solution to y′(t) = y(t).I Find the most general solution to y′(t) = y(t).

Solution

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

(check this)


. . . . . .


Example


Solution

I A solution is y(t) = et.

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

(check this)


. . . . . .


Example


Solution

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

(check this)


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Kick it up a notch: y′ = 2y

Example

I Find a solution to y′ = 2y.I Find the general solution to y′ = 2y.

Solution

I y = e2t

I y = Ce2t


. . . . . .

Kick it up a notch: y′ = 2y

Example

I Find a solution to y′ = 2y.I Find the general solution to y′ = 2y.

Solution

I y = e2t

I y = Ce2t


. . . . . .

In general: y′ = ky

Example

I Find a solution to y′ = ky.I Find the general solution to y′ = ky.

Solution

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 initial value of y.


. . . . . .


Example


Solution

I y = ekt

I y = Cekt


y(0) = Cek·0 = C · 1 = C,

so y(0) = y0, the initial value of y.


. . . . . .


Example


Solution

I y = ekt

I y = Cekt


y(0) = Cek·0 = C · 1 = C,

so y(0) = y0, the initial value of y.V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 13 / 40

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Constant Relative Growth =⇒ Exponential Growth

TheoremA function with constant relative growth rate k is an exponentialfunction with parameter k. Explicitly, the solution to the equation

y′(t) = ky(t) y(0) = y0

isy(t) = y0e

kt


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Exponential Growth is everywhere

I Lots of situations have growth rates proportional to the currentvalue

I This is the same as saying the relative growth rate is constant.I Examples: Natural population growth, compounded interest,

social networks


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Outline

Recall







. . . . . .

Bacteria

I Since you need bacteria tomake bacteria, the amountof new bacteria at anymoment is proportional tothe total amount ofbacteria.

I This means bacteriapopulations growexponentially.


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Bacteria Example

Example

A colony of bacteria is grown under ideal conditions in a laboratory. Atthe end of 3 hours there are 10,000 bacteria. At the end of 5 hoursthere are 40,000. How many bacteria were present initially?

SolutionSince y′ = ky for bacteria, we have y = y0e

kt. We have

10,000 = y0ek·3 40,000 = y0e

k·5

Dividing the first into the second gives4 = e2k =⇒ 2k = ln 4 =⇒ k = ln 2. Now we have

10,000 = y0eln 2·3 = y0 · 8

So y0 =10,000

8= 1250.


. . . . . .

Bacteria Example

Example



kt. We have

10,000 = y0ek·3 40,000 = y0e

k·5


10,000 = y0eln 2·3 = y0 · 8

So y0 =10,000

8= 1250.


. . . . . .

Bacteria Example

Example



kt. We have

10,000 = y0ek·3 40,000 = y0e

k·5


10,000 = y0eln 2·3 = y0 · 8

So y0 =10,000

8= 1250.


. . . . . .

Bacteria Example

Example



kt. We have

10,000 = y0ek·3 40,000 = y0e

k·5


10,000 = y0eln 2·3 = y0 · 8

So y0 =10,000

8= 1250.


. . . . . .

Could you do that again please?

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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Outline

Recall







. . . . . .

Modeling radioactive decay

Radioactive decay occurs because many large atoms spontaneouslygive off particles.

This means that in a sample ofa bunch of atoms, we canassume a certain percentage ofthem will “go off” at any point.(For instance, if all atom of acertain radioactive elementhave a 20% chance of decayingat any point, then we canexpect in a sample of 100 that20 of them will be decaying.)


. . . . . .

Modeling radioactive decay

Radioactive decay occurs because many large atoms spontaneouslygive off particles.

This means that in a sample ofa bunch of atoms, we canassume a certain percentage ofthem will “go off” at any point.(For instance, if all atom of acertain radioactive elementhave a 20% chance of decayingat any point, then we canexpect in a sample of 100 that20 of them will be decaying.)


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Radioactive decay as a differential equation

The relative rate of decay is constant:

y′

y= k

where k is negative.

So

y′ = ky =⇒ y = y0ekt

again!It’s customary to express the relative rate of decay in the units ofhalf-life: the amount of time it takes a pure sample to decay to onewhich is only half pure.


. . . . . .



y′

y= k

where k is negative. So


again!

It’s customary to express the relative rate of decay in the units ofhalf-life: the amount of time it takes a pure sample to decay to onewhich is only half pure.


. . . . . .



y′

y= k

where k is negative. So


again!It’s customary to express the relative rate of decay in the units ofhalf-life: the amount of time it takes a pure sample to decay to onewhich is only half pure.


. . . . . .

Computing the amount remaining of a decaying

sample

Example

The half-life of polonium-210 is about 138 days. How much of a 100 gsample remains after t years?

SolutionWe have y = y0e

kt, 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

Notice y(t) = y0 · 2−t/t1/2 , where t1/2 is the half-life.


. . . . . .


sample

Example




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

.


365·ln 2138 t = 100 · 2−365t/138



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sample

Example




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

.


365·ln 2138 t = 100 · 2−365t/138



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Carbon-14 Dating

The ratio of carbon-14 tocarbon-12 in an organismdecays exponentially:

p(t) = p0e−kt.

The half-life of carbon-14 isabout 5700 years. So theequation for p(t) is

p(t) = p0e−ln25700 t

Another way to write this wouldbe

p(t) = p02−t/5700


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Computing age with Carbon-14 content

Example

Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is10% of that in a living organism. How old is the fossil?

Solution

We are looking for the value of t for whichp(t)p0

= 0.1

From the

equation 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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Example


Solution


= 0.1

From the

equation we have

2−t/5700 = 0.1

=⇒ − t5700

ln 2 = ln 0.1 =⇒ t =ln 0.1ln 2

· 5700 ≈ 18,940



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Example


Solution


= 0.1 From the

equation we have

2−t/5700 = 0.1

=⇒ − t5700

ln 2 = ln 0.1 =⇒ t =ln 0.1ln 2

· 5700 ≈ 18,940



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Example


Solution


= 0.1 From the

equation we have

2−t/5700 = 0.1 =⇒ − t5700

ln 2 = ln 0.1

=⇒ t =ln 0.1ln 2

· 5700 ≈ 18,940



. . . . . .


Example


Solution


= 0.1 From the

equation we have

2−t/5700 = 0.1 =⇒ − t5700

ln 2 = ln 0.1 =⇒ t =ln 0.1ln 2

· 5700

≈ 18,940



. . . . . .


Example


Solution


= 0.1 From the

equation we have

2−t/5700 = 0.1 =⇒ − t5700

ln 2 = ln 0.1 =⇒ t =ln 0.1ln 2

· 5700 ≈ 18,940



. . . . . .


Example


Solution


= 0.1 From the

equation we have

2−t/5700 = 0.1 =⇒ − t5700

ln 2 = ln 0.1 =⇒ t =ln 0.1ln 2

· 5700 ≈ 18,940



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Outline

Recall







. . . . . .

Newton's Law of Cooling

I Newton’s Law of Coolingstates that the rate ofcooling of an object isproportional to thetemperature differencebetween the object and itssurroundings.

I This gives us a differentialequation of the form

dTdt

= k(T− Ts)

(where k < 0 again).


. . . . . .

Newton's Law of Cooling

I Newton’s Law of Coolingstates that the rate ofcooling of an object isproportional to thetemperature differencebetween the object and itssurroundings.

I This gives us a differentialequation of the form

dTdt

= k(T− Ts)

(where k < 0 again).


. . . . . .

General Solution to NLC problems

To solve this, change the variable y(t) = T(t)− Ts. Then y′ = T′ andk(T− Ts) = ky. The equation now looks like

dTdt

= k(T− Ts) ⇐⇒ dydt

= 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. So

TheoremThe solution to the equation T′(t) = k(T(t)− Ts), T(0) = T0 is

T(t) = (T0 − Ts)ekt + Ts


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dTdt


= ky

Now we can solve!

y′ = ky

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





. . . . . .



dTdt


= ky

Now we can solve!

y′ = ky =⇒ y = Cekt

=⇒ T− Ts = Cekt =⇒ T = Cekt + Ts





. . . . . .



dTdt


= ky

Now we can solve!

y′ = ky =⇒ y = Cekt =⇒ T− Ts = Cekt

=⇒ T = Cekt + Ts





. . . . . .



dTdt


= ky

Now we can solve!






. . . . . .



dTdt


= ky

Now we can solve!






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Computing cooling time with NLC

Example

A hard-boiled egg at 98 ◦C is put in a sink of 18 ◦C water. After 5minutes, the egg’s temperature is 38 ◦C. Assuming the water has notwarmed appreciably, how much longer will it take the egg to reach20 ◦C?

SolutionWe know that the temperature function takes the form

T(t) = (T0 − Ts)ekt + Ts = 80ekt + 18

To find k, plug in t = 5:

38 = T(5) = 80e5k + 18

and solve for k.


. . . . . .

Computing cooling time with NLC

Example

A hard-boiled egg at 98 ◦C is put in a sink of 18 ◦C water. After 5minutes, the egg’s temperature is 38 ◦C. Assuming the water has notwarmed appreciably, how much longer will it take the egg to reach20 ◦C?

SolutionWe know that the temperature function takes the form

T(t) = (T0 − Ts)ekt + Ts = 80ekt + 18

To find k, plug in t = 5:

38 = T(5) = 80e5k + 18

and solve for k.


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Finding k

Solution (Continued)

38 = T(5) = 80e5k + 18

20 = 80e5k

14= e5k

ln(14

)= 5k

=⇒ k = −15ln 4.

Now we need to solve for t:

20 = T(t) = 80e−t5 ln 4 + 18


. . . . . .

Finding k


38 = T(5) = 80e5k + 18

20 = 80e5k

14= e5k

ln(14

)= 5k

=⇒ k = −15ln 4.


20 = T(t) = 80e−t5 ln 4 + 18


. . . . . .

Finding k


38 = T(5) = 80e5k + 18

20 = 80e5k

14= e5k

ln(14

)= 5k

=⇒ k = −15ln 4.


20 = T(t) = 80e−t5 ln 4 + 18


. . . . . .

Finding k


38 = T(5) = 80e5k + 18

20 = 80e5k

14= e5k

ln(14

)= 5k

=⇒ k = −15ln 4.


20 = T(t) = 80e−t5 ln 4 + 18


. . . . . .

Finding k


38 = T(5) = 80e5k + 18

20 = 80e5k

14= e5k

ln(14

)= 5k

=⇒ k = −15ln 4.


20 = T(t) = 80e−t5 ln 4 + 18


. . . . . .

Finding k


38 = T(5) = 80e5k + 18

20 = 80e5k

14= e5k

ln(14

)= 5k

=⇒ k = −15ln 4.


20 = T(t) = 80e−t5 ln 4 + 18


. . . . . .

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


. . . . . .

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


. . . . . .

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


. . . . . .

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


. . . . . .

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


. . . . . .

Computing time of death with NLC

Example

A murder victim is discovered atmidnight and the temperature ofthe body is recorded as 31 ◦C.One hour later, the temperatureof the body is 29 ◦C. Assumethat the surrounding airtemperature remains constantat 21 ◦C. Calculate the victim’stime of death. (The “normal”temperature of a living humanbeing is approximately 37 ◦C.)


. . . . . .

Solution

I Let time 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 time of death was just before 10:00pm.


. . . . . .

Solution


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



. . . . . .

Solution


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



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Outline

Recall







. . . . . .

Interest

I If an account has an compound interest rate of r per yearcompounded n times, then an initial deposit of A0 dollars becomes

A0

(1+

rn

)nt

after t years.

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

A(t) = limn→∞

A0

(1+

rn

)nt

= A0ert.

I Thus dollars are like bacteria.


. . . . . .

Interest


A0

(1+

rn

)nt

after t years.I For different amounts of compounding, this will change. As

n → ∞, we get continously compounded interest

A(t) = limn→∞

A0

(1+

rn

)nt= A0ert.



. . . . . .

Interest


A0

(1+

rn

)nt

after t years.I For different amounts of compounding, this will change. As

n → ∞, we get continously compounded interest

A(t) = limn→∞

A0

(1+

rn

)nt= A0ert.



. . . . . .

Continuous vs. Discrete Compounding of interest

Example

Consider two bank accounts: one with 10% annual interestedcompounded quarterly and one with annual interest rate r compundedcontinuously. If they produce the same balance after every year, whatis r?

SolutionThe balance for the 10% compounded quarterly account after t years is

A1(t) = A0(1.025)4t = P((1.025)4)t

The balance for the interest rate r compounded continuously accountafter t years is

A2(t) = A0ert


. . . . . .

Continuous vs. Discrete Compounding of interest

Example

Consider two bank accounts: one with 10% annual interestedcompounded quarterly and one with annual interest rate r compundedcontinuously. If they produce the same balance after every year, whatis r?

SolutionThe balance for the 10% compounded quarterly account after t years is

A1(t) = A0(1.025)4t = P((1.025)4)t

The balance for the interest rate r compounded continuously accountafter t years is

A2(t) = A0ert


. . . . . .

Solving


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 continuously.


. . . . . .

Computing doubling time with exponential growth

Example

How long does it take an initial deposit of $100, compoundedcontinuously, to double?

SolutionWe 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.


. . . . . .

Computing doubling time with exponential growth

Example

How long does it take an initial deposit of $100, compoundedcontinuously, to double?

SolutionWe 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 day

I The fractionln 2r

can alsobe approximated as either70 or 72 divided by thepercentage rate (as anumber between 0 and100, not a fraction between0 and 1.)

I This is sometimes calledthe rule of 70 or rule of 72.

I 72 has lots of factors so it’sused more often.


. . . . . .

Summary

I When something grows or decays at a constant relative rate, thegrowth or decay is exponential.

I Equations 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 (Section 021 slides)

Technology

t example

cos t b sin t x t

solution example

exponential growth

b cos t x t

general solution

differential equation

decay v63