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
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Objectives
I Solve the ordinarydifferential equationy′(t) = ky(t), y(0) = y0
I Solve problems involvingexponential growth anddecay
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 3 / 40
. . . . . .
Outline
Recall
The differential equation y′ = ky
Modeling simple population growth
Modeling radioactive decayCarbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 4 / 40
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Derivatives of exponential and logarithmic functions
y y′
ex ex
ax (lna) · ax
ln x1x
loga x1ln a
· 1x
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 5 / 40
. . . . . .
Outline
Recall
The differential equation y′ = ky
Modeling simple population growth
Modeling radioactive decayCarbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 6 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 8 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 8 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
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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)
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
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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)
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
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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)
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
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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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 12 / 40
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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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 12 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 13 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 13 / 40
. . . . . .
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.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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 14 / 40
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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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 15 / 40
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Outline
Recall
The differential equation y′ = ky
Modeling simple population growth
Modeling radioactive decayCarbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 16 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 17 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
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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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 19 / 40
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Outline
Recall
The differential equation y′ = ky
Modeling simple population growth
Modeling radioactive decayCarbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 20 / 40
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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.)
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 21 / 40
. . . . . .
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.)
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 21 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 40
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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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 24 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
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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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
. . . . . .
Outline
Recall
The differential equation y′ = ky
Modeling simple population growth
Modeling radioactive decayCarbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 26 / 40
. . . . . .
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).
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 27 / 40
. . . . . .
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).
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 27 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 29 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 29 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
. . . . . .
Finding t
Solution (Continued)
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
. . . . . .
Finding t
Solution (Continued)
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
. . . . . .
Finding t
Solution (Continued)
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
. . . . . .
Finding t
Solution (Continued)
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
. . . . . .
Finding t
Solution (Continued)
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
. . . . . .
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.)
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 32 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 33 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 33 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 33 / 40
. . . . . .
Outline
Recall
The differential equation y′ = ky
Modeling simple population growth
Modeling radioactive decayCarbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 34 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 35 / 40
. . . . . .
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. As
n → ∞, we get continously compounded interest
A(t) = limn→∞
A0
(1+
rn
)nt= A0ert.
I Thus dollars are like bacteria.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 35 / 40
. . . . . .
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. As
n → ∞, we get continously compounded interest
A(t) = limn→∞
A0
(1+
rn
)nt= A0ert.
I Thus dollars are like bacteria.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 35 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 36 / 40
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 36 / 40
. . . . . .
Solving
Solution (Continued)
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 37 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 38 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 38 / 40
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 39 / 40
. . . . . .
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).
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 40 / 40
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