Lesson 15: Exponential Growth and Decay

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Transcript of Lesson 15: Exponential Growth and Decay
Section 3.4Exponential Growth and Decay
V63.0121.002.2010Su, Calculus I
New York University
June 2, 2010
Announcements
I Review in second half of class today
I Office Hours after class today
I Midterm tomorrow
Announcements
I Review in second half ofclass today
I Office Hours after classtoday
I Midterm tomorrow
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 2 / 37
Objectives
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 3 / 37
Outline
Recall
The differential equation y ′ = ky
Modeling simple population growth
Modeling radioactive decayCarbon14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 4 / 37
Derivatives of exponential and logarithmic functions
y y ′
ex ex
ax (ln a)ax
ln x1
x
loga x1
ln a· 1
x
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 5 / 37
Outline
Recall
The differential equation y ′ = ky
Modeling simple population growth
Modeling radioactive decayCarbon14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 6 / 37
What is a differential equation?
Definition
A 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) +k
mx(t) = 0.
I The most general solution is x(t) = A sinωt + B cosωt, whereω =
√k/m.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 7 / 37
What is a differential equation?
Definition
A 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) +k
mx(t) = 0.
I The most general solution is x(t) = A sinωt + B cosωt, whereω =
√k/m.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 7 / 37
What is a differential equation?
Definition
A 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) +k
mx(t) = 0.
I The most general solution is x(t) = A sinωt + B cosωt, whereω =
√k/m.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 7 / 37
What is a differential equation?
Definition
A 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) +k
mx(t) = 0.
I The most general solution is x(t) = A sinωt + B cosωt, whereω =
√k/m.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 7 / 37
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 .
Remark
If a function has a constant rate of growth, it’s linear.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 8 / 37
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 .
Remark
If a function has a constant rate of growth, it’s linear.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 8 / 37
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 .
Remark
If a function has a constant rate of growth, it’s linear.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 8 / 37
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 .
Remark
If a function has a constant rate of growth, it’s linear.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 8 / 37
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 .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 9 / 37
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 .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 9 / 37
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 .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 9 / 37
The differential equation y ′ = ky
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)
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 10 / 37
The differential equation y ′ = ky
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)
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 10 / 37
The differential equation y ′ = ky
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)
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 10 / 37
Kick it up a notch
Example
I Find a solution to y ′ = 2y .
I Find the general solution to y ′ = 2y .
Solution
I y = e2t
I y = Ce2t
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 11 / 37
Kick it up a notch
Example
I Find a solution to y ′ = 2y .
I Find the general solution to y ′ = 2y .
Solution
I y = e2t
I y = Ce2t
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 11 / 37
In general
Example
I Find a solution to y ′ = ky .
I Find the general solution to y ′ = ky .
Solution
I y = ekt
I y = Cekt
Remark
What is C ? Plug in t = 0:
y(0) = Cek·0 = C · 1 = C ,
so y(0) = y0, the initial value of y .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 12 / 37
In general
Example
I Find a solution to y ′ = ky .
I Find the general solution to y ′ = ky .
Solution
I y = ekt
I y = Cekt
Remark
What is C ? Plug in t = 0:
y(0) = Cek·0 = C · 1 = C ,
so y(0) = y0, the initial value of y .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 12 / 37
In general
Example
I Find a solution to y ′ = ky .
I Find the general solution to y ′ = ky .
Solution
I y = ekt
I y = Cekt
Remark
What is C ? Plug in t = 0:
y(0) = Cek·0 = C · 1 = C ,
so y(0) = y0, the initial value of y .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 12 / 37
Constant Relative Growth =⇒ Exponential Growth
Theorem
A function with constant relative growth rate k is an exponential functionwith parameter k. Explicitly, the solution to the equation
y ′(t) = ky(t) y(0) = y0
isy(t) = y0ekt
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 13 / 37
Exponential Growth is everywhere
I Lots of situations have growth rates proportional to the current value
I This is the same as saying the relative growth rate is constant.
I Examples: Natural population growth, compounded interest, socialnetworks
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 14 / 37
Outline
Recall
The differential equation y ′ = ky
Modeling simple population growth
Modeling radioactive decayCarbon14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 15 / 37
Bacteria
I Since you need bacteria tomake bacteria, the amountof new bacteria at anymoment is proportional tothe total amount of bacteria.
I This means bacteriapopulations growexponentially.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 16 / 37
Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At theend of 3 hours there are 10,000 bacteria. At the end of 5 hours there are40,000. How many bacteria were present initially?
Solution
Since y ′ = ky for bacteria, we have y = y0ekt . We have
10, 000 = y0ek·3 40, 000 = y0ek·5
Dividing the first into the second gives4 = e2k =⇒ 2k = ln 4 =⇒ k = ln 2. Now we have
10, 000 = y0e ln 2·3 = y0 · 8
So y0 =10, 000
8= 1250.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 17 / 37
Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At theend of 3 hours there are 10,000 bacteria. At the end of 5 hours there are40,000. How many bacteria were present initially?
Solution
Since y ′ = ky for bacteria, we have y = y0ekt . We have
10, 000 = y0ek·3 40, 000 = y0ek·5
Dividing the first into the second gives4 = e2k =⇒ 2k = ln 4 =⇒ k = ln 2. Now we have
10, 000 = y0e ln 2·3 = y0 · 8
So y0 =10, 000
8= 1250.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 17 / 37
Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At theend of 3 hours there are 10,000 bacteria. At the end of 5 hours there are40,000. How many bacteria were present initially?
Solution
Since y ′ = ky for bacteria, we have y = y0ekt . We have
10, 000 = y0ek·3 40, 000 = y0ek·5
Dividing the first into the second gives4 = e2k =⇒ 2k = ln 4 =⇒ k = ln 2. Now we have
10, 000 = y0e ln 2·3 = y0 · 8
So y0 =10, 000
8= 1250.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 17 / 37
Could you do that again please?
We have
10, 000 = y0ek·3
40, 000 = y0ek·5
Dividing the first into the second gives
40, 000
10, 000=
y0e5k
y0e3k
=⇒ 4 = e2k
=⇒ ln 4 = ln(e2k) = 2k
=⇒ k =ln 4
2=
ln 22
2=
2 ln 2
2= ln 2
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 18 / 37
Outline
Recall
The differential equation y ′ = ky
Modeling simple population growth
Modeling radioactive decayCarbon14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 19 / 37
Modeling radioactive decay
Radioactive decay occurs because many large atoms spontaneously give offparticles.
This means that in a sample of abunch of atoms, we can assume acertain percentage of them will“go off” at any point. (Forinstance, if all atom of a certainradioactive element have a 20%chance of decaying at any point,then we can expect in a sampleof 100 that 20 of them will bedecaying.)
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 20 / 37
Modeling radioactive decay
Radioactive decay occurs because many large atoms spontaneously give offparticles.
This means that in a sample of abunch of atoms, we can assume acertain percentage of them will“go off” at any point. (Forinstance, if all atom of a certainradioactive element have a 20%chance of decaying at any point,then we can expect in a sampleof 100 that 20 of them will bedecaying.)
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 20 / 37
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 of halflife:the amount of time it takes a pure sample to decay to one which is onlyhalf pure.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 21 / 37
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 of halflife:the amount of time it takes a pure sample to decay to one which is onlyhalf pure.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 21 / 37
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 of halflife:the amount of time it takes a pure sample to decay to one which is onlyhalf pure.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 21 / 37
Computing the amount remaining of a decayingsample
Example
The halflife of polonium210 is about 138 days. How much of a 100 gsample remains after t years?
Solution
We have y = y0ekt , where y0 = y(0) = 100 grams. Then
50 = 100ek·138/365 =⇒ k = −365 · ln 2
138.
Thereforey(t) = 100e−
365·ln 2138
t = 100 · 2−365t/138.
Notice y(t) = y0 · 2−t/t1/2 , where t1/2 is the halflife.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 22 / 37
Computing the amount remaining of a decayingsample
Example
The halflife of polonium210 is about 138 days. How much of a 100 gsample remains after t years?
Solution
We have y = y0ekt , where y0 = y(0) = 100 grams. Then
50 = 100ek·138/365 =⇒ k = −365 · ln 2
138.
Thereforey(t) = 100e−
365·ln 2138
t = 100 · 2−365t/138.
Notice y(t) = y0 · 2−t/t1/2 , where t1/2 is the halflife.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 22 / 37
Computing the amount remaining of a decayingsample
Example
The halflife of polonium210 is about 138 days. How much of a 100 gsample remains after t years?
Solution
We have y = y0ekt , where y0 = y(0) = 100 grams. Then
50 = 100ek·138/365 =⇒ k = −365 · ln 2
138.
Thereforey(t) = 100e−
365·ln 2138
t = 100 · 2−365t/138.
Notice y(t) = y0 · 2−t/t1/2 , where t1/2 is the halflife.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 22 / 37
Carbon14 Dating
The ratio of carbon14 tocarbon12 in an organism decaysexponentially:
p(t) = p0e−kt .
The halflife of carbon14 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
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 23 / 37
Computing age with Carbon14 content
Example
Suppose a fossil is found where the ratio of carbon14 to carbon12 is 10%of that in a living organism. How old is the fossil?
Solution
We are looking for the value of t for which
p(t)
p(0)= 0.1
From the equation we have
2−t/5700 = 0.1 =⇒ − t
5700ln 2 = ln 0.1 =⇒ t =
ln 0.1
ln 2· 5700 ≈ 18, 940
So the fossil is almost 19,000 years old.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 24 / 37
Computing age with Carbon14 content
Example
Suppose a fossil is found where the ratio of carbon14 to carbon12 is 10%of that in a living organism. How old is the fossil?
Solution
We are looking for the value of t for which
p(t)
p(0)= 0.1
From the equation we have
2−t/5700 = 0.1 =⇒ − t
5700ln 2 = ln 0.1 =⇒ t =
ln 0.1
ln 2· 5700 ≈ 18, 940
So the fossil is almost 19,000 years old.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 24 / 37
Computing age with Carbon14 content
Example
Suppose a fossil is found where the ratio of carbon14 to carbon12 is 10%of that in a living organism. How old is the fossil?
Solution
We are looking for the value of t for which
p(t)
p(0)= 0.1
From the equation we have
2−t/5700 = 0.1 =⇒ − t
5700ln 2 = ln 0.1 =⇒ t =
ln 0.1
ln 2· 5700 ≈ 18, 940
So the fossil is almost 19,000 years old.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 24 / 37
Computing age with Carbon14 content
Example
Suppose a fossil is found where the ratio of carbon14 to carbon12 is 10%of that in a living organism. How old is the fossil?
Solution
We are looking for the value of t for which
p(t)
p(0)= 0.1
From the equation we have
2−t/5700 = 0.1 =⇒ − t
5700ln 2 = ln 0.1 =⇒ t =
ln 0.1
ln 2· 5700 ≈ 18, 940
So the fossil is almost 19,000 years old.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 24 / 37
Outline
Recall
The differential equation y ′ = ky
Modeling simple population growth
Modeling radioactive decayCarbon14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 25 / 37
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
dT
dt= k(T − Ts)
(where k < 0 again).
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 26 / 37
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
dT
dt= k(T − Ts)
(where k < 0 again).
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 26 / 37
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
dT
dt= k(T − Ts) ⇐⇒ dy
dt= 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
Theorem
The solution to the equation T ′(t) = k(T (t)− Ts), T (0) = T0 is
T (t) = (T0 − Ts)ekt + Ts
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
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
dT
dt= k(T − Ts) ⇐⇒ dy
dt= 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
Theorem
The solution to the equation T ′(t) = k(T (t)− Ts), T (0) = T0 is
T (t) = (T0 − Ts)ekt + Ts
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
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
dT
dt= k(T − Ts) ⇐⇒ dy
dt= 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
Theorem
The solution to the equation T ′(t) = k(T (t)− Ts), T (0) = T0 is
T (t) = (T0 − Ts)ekt + Ts
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
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
dT
dt= k(T − Ts) ⇐⇒ dy
dt= 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
Theorem
The solution to the equation T ′(t) = k(T (t)− Ts), T (0) = T0 is
T (t) = (T0 − Ts)ekt + Ts
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
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
dT
dt= k(T − Ts) ⇐⇒ dy
dt= 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
Theorem
The solution to the equation T ′(t) = k(T (t)− Ts), T (0) = T0 is
T (t) = (T0 − Ts)ekt + Ts
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
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
dT
dt= k(T − Ts) ⇐⇒ dy
dt= 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
Theorem
The solution to the equation T ′(t) = k(T (t)− Ts), T (0) = T0 is
T (t) = (T0 − Ts)ekt + Ts
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
Computing cooling time with NLC
Example
A hardboiled 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 reach 20 ◦C?
Solution
We 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 28 / 37
Computing cooling time with NLC
Example
A hardboiled 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 reach 20 ◦C?
Solution
We 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 28 / 37
Finding k
38 = T (5) = 80e5k + 18
20 = 80e5k
1
4= e5k
ln
(1
4
)= 5k
=⇒ k = −1
5ln 4.
Now we need to solve
20 = T (t) = 80e−t5ln 4 + 18
for t.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 29 / 37
Finding k
38 = T (5) = 80e5k + 18
20 = 80e5k
1
4= e5k
ln
(1
4
)= 5k
=⇒ k = −1
5ln 4.
Now we need to solve
20 = T (t) = 80e−t5ln 4 + 18
for t.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 29 / 37
Finding t
20 = 80e−t5ln 4 + 18
2 = 80e−t5ln 4
1
40= e−
t5ln 4
− ln 40 = − t
5ln 4
=⇒ t =ln 4015 ln 4
=5 ln 40
ln 4≈ 13min
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 30 / 37
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 constant at21 ◦C. Calculate the victim’stime of death. (The “normal”temperature of a living humanbeing is approximately 37 ◦C.)
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 31 / 37
Solution
I Let time 0 be midnight. We know T0 = 31, Ts = 21, and T (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)≈ −2.10hr
So the time of death was just before 10:00pm.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 32 / 37
Solution
I Let time 0 be midnight. We know T0 = 31, Ts = 21, and T (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)≈ −2.10hr
So the time of death was just before 10:00pm.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 32 / 37
Solution
I Let time 0 be midnight. We know T0 = 31, Ts = 21, and T (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)≈ −2.10hr
So the time of death was just before 10:00pm.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 32 / 37
Outline
Recall
The differential equation y ′ = ky
Modeling simple population growth
Modeling radioactive decayCarbon14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 33 / 37
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 +
r
n
)ntafter t years.
I For different amounts of compounding, this will change. As n→∞,we get continously compounded interest
A(t) = limn→∞
A0
(1 +
r
n
)nt
= A0ert .
I Thus dollars are like bacteria.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 34 / 37
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 +
r
n
)ntafter t years.
I For different amounts of compounding, this will change. As n→∞,we get continously compounded interest
A(t) = limn→∞
A0
(1 +
r
n
)nt= A0ert .
I Thus dollars are like bacteria.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 34 / 37
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 +
r
n
)ntafter t years.
I For different amounts of compounding, this will change. As n→∞,we get continously compounded interest
A(t) = limn→∞
A0
(1 +
r
n
)nt= A0ert .
I Thus dollars are like bacteria.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 34 / 37
Computing doubling time with exponential growth
Example
How long does it take an initial deposit of $100, compoundedcontinuously, to double?
Solution
We need t such that A(t) = 200. In other words
200 = 100ert =⇒ 2 = ert =⇒ ln 2 = rt =⇒ t =ln 2
r.
For instance, if r = 6% = 0.06, we have
t =ln 2
0.06≈ 0.69
0.06=
69
6= 11.5 years.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 35 / 37
Computing doubling time with exponential growth
Example
How long does it take an initial deposit of $100, compoundedcontinuously, to double?
Solution
We need t such that A(t) = 200. In other words
200 = 100ert =⇒ 2 = ert =⇒ ln 2 = rt =⇒ t =ln 2
r.
For instance, if r = 6% = 0.06, we have
t =ln 2
0.06≈ 0.69
0.06=
69
6= 11.5 years.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 35 / 37
Ibanking interview tip of the day
I The fractionln 2
rcan also be
approximated as either 70 or72 divided by the percentagerate (as a number between 0and 100, not a fractionbetween 0 and 1.)
I This is sometimes called therule of 70 or rule of 72.
I 72 has lots of factors so it’sused more often.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 36 / 37
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).
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 37 / 37