1.
Section 3.4
Exponential Growth and Decay
V63.0121.002.2010Su, Calculus I
New York University
June 2, 2010
Announcements
Review in second half of class today
Oﬃce Hours after class today
Midterm tomorrow
2.
Announcements
Review in second half of
class today
Oﬃce Hours after class
today
Midterm tomorrow
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 2 / 37
3.
Objectives
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 3 / 37
4.
Outline
Recall
The diﬀerential equation y = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 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
5.
Derivatives of exponential and logarithmic functions
y y
ex ex
ax (ln a)ax
1
ln x
x
1 1
loga x ·
ln a x
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 5 / 37
6.
Outline
Recall
The diﬀerential equation y = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 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
7.
What is a diﬀerential equation?
Deﬁnition
A diﬀerential equation is an equation for an unknown function which
includes the function and its derivatives.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 7 / 37
8.
What is a diﬀerential equation?
Deﬁnition
A diﬀerential equation is an equation for an unknown function which
includes the function and its derivatives.
Example
Newton’s Second Law F = ma is a diﬀerential equation, where
a(t) = x (t).
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 7 / 37
9.
What is a diﬀerential equation?
Deﬁnition
A diﬀerential equation is an equation for an unknown function which
includes the function and its derivatives.
Example
Newton’s Second Law F = ma is a diﬀerential equation, where
a(t) = x (t).
In a spring, F (x) = −kx, where x is displacement from equilibrium
and k is a constant. So
k
−kx(t) = mx (t) =⇒ x (t) + x(t) = 0.
m
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 7 / 37
10.
What is a diﬀerential equation?
Deﬁnition
A diﬀerential equation is an equation for an unknown function which
includes the function and its derivatives.
Example
Newton’s Second Law F = ma is a diﬀerential equation, where
a(t) = x (t).
In a spring, F (x) = −kx, where x is displacement from equilibrium
and k is a constant. So
k
−kx(t) = mx (t) =⇒ x (t) + x(t) = 0.
m
The most general solution is x(t) = A sin ωt + B cos ωt, where
ω = k/m.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 7 / 37
11.
The Equation y = 2
Example
Find a solution to y (t) = 2.
Find the most general solution to y (t) = 2.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 8 / 37
12.
The Equation y = 2
Example
Find a solution to y (t) = 2.
Find the most general solution to y (t) = 2.
Solution
A solution is y (t) = 2t.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 8 / 37
13.
The Equation y = 2
Example
Find a solution to y (t) = 2.
Find the most general solution to y (t) = 2.
Solution
A solution is y (t) = 2t.
The general solution is y = 2t + C .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 8 / 37
14.
The Equation y = 2
Example
Find a solution to y (t) = 2.
Find the most general solution to y (t) = 2.
Solution
A solution is y (t) = 2t.
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
15.
The Equation y = 2t
Example
Find a solution to y (t) = 2t.
Find the most general solution to y (t) = 2t.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 9 / 37
16.
The Equation y = 2t
Example
Find a solution to y (t) = 2t.
Find the most general solution to y (t) = 2t.
Solution
A solution is y (t) = t 2 .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 9 / 37
17.
The Equation y = 2t
Example
Find a solution to y (t) = 2t.
Find the most general solution to y (t) = 2t.
Solution
A solution is y (t) = t 2 .
The general solution is y = t 2 + C .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 9 / 37
18.
The diﬀerential equation y = ky
Example
Find a solution to y (t) = y (t).
Find the most general solution to y (t) = y (t).
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 10 / 37
19.
The diﬀerential equation y = ky
Example
Find a solution to y (t) = y (t).
Find the most general solution to y (t) = y (t).
Solution
A solution is y (t) = e t .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 10 / 37
20.
The diﬀerential equation y = ky
Example
Find a solution to y (t) = y (t).
Find the most general solution to y (t) = y (t).
Solution
A solution is y (t) = e t .
The general solution is y = Ce t , not y = e t + C .
(check this)
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 10 / 37
21.
Kick it up a notch
Example
Find a solution to y = 2y .
Find the general solution to y = 2y .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 11 / 37
22.
Kick it up a notch
Example
Find a solution to y = 2y .
Find the general solution to y = 2y .
Solution
y = e 2t
y = Ce 2t
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 11 / 37
23.
In general
Example
Find a solution to y = ky .
Find the general solution to y = ky .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 12 / 37
24.
In general
Example
Find a solution to y = ky .
Find the general solution to y = ky .
Solution
y = e kt
y = Ce kt
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 12 / 37
25.
In general
Example
Find a solution to y = ky .
Find the general solution to y = ky .
Solution
y = e kt
y = Ce kt
Remark
What is C ? Plug in t = 0:
y (0) = Ce k·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
26.
Constant Relative Growth =⇒ Exponential Growth
Theorem
A function with constant relative growth rate k is an exponential function
with parameter k. Explicitly, the solution to the equation
y (t) = ky (t) y (0) = y0
is
y (t) = y0 e kt
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 13 / 37
27.
Exponential Growth is everywhere
Lots of situations have growth rates proportional to the current value
This is the same as saying the relative growth rate is constant.
Examples: Natural population growth, compounded interest, social
networks
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 14 / 37
28.
Outline
Recall
The diﬀerential equation y = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 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
29.
Bacteria
Since you need bacteria to
make bacteria, the amount
of new bacteria at any
moment is proportional to
the total amount of bacteria.
This means bacteria
populations grow
exponentially.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 16 / 37
30.
Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At the
end of 3 hours there are 10,000 bacteria. At the end of 5 hours there are
40,000. How many bacteria were present initially?
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 17 / 37
31.
Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At the
end of 3 hours there are 10,000 bacteria. At the end of 5 hours there are
40,000. How many bacteria were present initially?
Solution
Since y = ky for bacteria, we have y = y0 e kt . We have
10, 000 = y0 e k·3 40, 000 = y0 e k·5
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 17 / 37
32.
Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At the
end of 3 hours there are 10,000 bacteria. At the end of 5 hours there are
40,000. How many bacteria were present initially?
Solution
Since y = ky for bacteria, we have y = y0 e kt . We have
10, 000 = y0 e k·3 40, 000 = y0 e k·5
Dividing the ﬁrst into the second gives
4 = e 2k =⇒ 2k = ln 4 =⇒ k = ln 2. Now we have
10, 000 = y0 e ln 2·3 = y0 · 8
10, 000
So y0 = = 1250.
8
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 17 / 37
33.
Could you do that again please?
We have
10, 000 = y0 e k·3
40, 000 = y0 e k·5
Dividing the ﬁrst into the second gives
40, 000 y0 e 5k
=
10, 000 y0 e 3k
=⇒ 4 = e 2k
=⇒ ln 4 = ln(e 2k ) = 2k
ln 4 ln 22 2 ln 2
=⇒ k = = = = ln 2
2 2 2
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 18 / 37
34.
Outline
Recall
The diﬀerential equation y = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 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
35.
Modeling radioactive decay
Radioactive decay occurs because many large atoms spontaneously give oﬀ
particles.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 20 / 37
36.
Modeling radioactive decay
Radioactive decay occurs because many large atoms spontaneously give oﬀ
particles.
This means that in a sample of a
bunch of atoms, we can assume a
certain percentage of them will
“go oﬀ” at any point. (For
instance, if all atom of a certain
radioactive element have a 20%
chance of decaying at any point,
then we can expect in a sample
of 100 that 20 of them will be
decaying.)
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 20 / 37
37.
Radioactive decay as a diﬀerential equation
The relative rate of decay is constant:
y
=k
y
where k is negative.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 21 / 37
38.
Radioactive decay as a diﬀerential equation
The relative rate of decay is constant:
y
=k
y
where k is negative. So
y = ky =⇒ y = y0 e kt
again!
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 21 / 37
39.
Radioactive decay as a diﬀerential equation
The relative rate of decay is constant:
y
=k
y
where k is negative. So
y = ky =⇒ y = y0 e kt
again!
It’s customary to express the relative rate of decay in the units of half-life:
the amount of time it takes a pure sample to decay to one which is only
half pure.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 21 / 37
40.
Computing the amount remaining of a decaying
sample
Example
The half-life of polonium-210 is about 138 days. How much of a 100 g
sample remains after t years?
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 22 / 37
41.
Computing the amount remaining of a decaying
sample
Example
The half-life of polonium-210 is about 138 days. How much of a 100 g
sample remains after t years?
Solution
We have y = y0 e kt , where y0 = y (0) = 100 grams. Then
365 · ln 2
50 = 100e k·138/365 =⇒ k = − .
138
Therefore
365·ln 2
y (t) = 100e − 138
t
= 100 · 2−365t/138 .
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 22 / 37
42.
Computing the amount remaining of a decaying
sample
Example
The half-life of polonium-210 is about 138 days. How much of a 100 g
sample remains after t years?
Solution
We have y = y0 e kt , where y0 = y (0) = 100 grams. Then
365 · ln 2
50 = 100e k·138/365 =⇒ k = − .
138
Therefore
365·ln 2
y (t) = 100e − 138
t
= 100 · 2−365t/138 .
Notice y (t) = y0 · 2−t/t1/2 , where t1/2 is the half-life.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 22 / 37
43.
Carbon-14 Dating
The ratio of carbon-14 to
carbon-12 in an organism decays
exponentially:
p(t) = p0 e −kt .
The half-life of carbon-14 is
about 5700 years. So the
equation for p(t) is
ln2
p(t) = p0 e − 5700 t
Another way to write this would
be
p(t) = p0 2−t/5700
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 23 / 37
44.
Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is 10%
of that in a living organism. How old is the fossil?
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 24 / 37
45.
Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 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)
= 0.1
p(0)
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 24 / 37
46.
Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 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)
= 0.1
p(0)
From the equation we have
t ln 0.1
2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 ≈ 18, 940
5700 ln 2
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 24 / 37
47.
Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 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)
= 0.1
p(0)
From the equation we have
t ln 0.1
2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 ≈ 18, 940
5700 ln 2
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
48.
Outline
Recall
The diﬀerential equation y = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 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
49.
Newton’s Law of Cooling
Newton’s Law of Cooling
states that the rate of
cooling of an object is
proportional to the
temperature diﬀerence
between the object and its
surroundings.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 26 / 37
50.
Newton’s Law of Cooling
Newton’s Law of Cooling
states that the rate of
cooling of an object is
proportional to the
temperature diﬀerence
between the object and its
surroundings.
This gives us a diﬀerential
equation of the form
dT
= k(T − Ts )
dt
(where k < 0 again).
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 26 / 37
51.
General Solution to NLC problems
To solve this, change the variable y (t) = T (t) − Ts . Then y = T and
k(T − Ts ) = ky . The equation now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
52.
General Solution to NLC problems
To solve this, change the variable y (t) = T (t) − Ts . Then y = T and
k(T − Ts ) = ky . The equation now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
Now we can solve!
y = ky
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
53.
General Solution to NLC problems
To solve this, change the variable y (t) = T (t) − Ts . Then y = T and
k(T − Ts ) = ky . The equation now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
Now we can solve!
y = ky =⇒ y = Ce kt
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
54.
General Solution to NLC problems
To solve this, change the variable y (t) = T (t) − Ts . Then y = T and
k(T − Ts ) = ky . The equation now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
Now we can solve!
y = ky =⇒ y = Ce kt =⇒ T − Ts = Ce kt
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
55.
General Solution to NLC problems
To solve this, change the variable y (t) = T (t) − Ts . Then y = T and
k(T − Ts ) = ky . The equation now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
Now we can solve!
y = ky =⇒ y = Ce kt =⇒ T − Ts = Ce kt =⇒ T = Ce kt + Ts
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
56.
General Solution to NLC problems
To solve this, change the variable y (t) = T (t) − Ts . Then y = T and
k(T − Ts ) = ky . The equation now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
Now we can solve!
y = ky =⇒ y = Ce kt =⇒ T − Ts = Ce kt =⇒ T = Ce kt + 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 )e kt + Ts
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
57.
Computing cooling time with NLC
Example
A hard-boiled egg at 98 ◦ C is put in a sink of 18 ◦ C water. After 5
minutes, the egg’s temperature is 38 ◦ C. Assuming the water has not
warmed appreciably, how much longer will it take the egg to reach 20 ◦ C?
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 28 / 37
58.
Computing cooling time with NLC
Example
A hard-boiled egg at 98 ◦ C is put in a sink of 18 ◦ C water. After 5
minutes, the egg’s temperature is 38 ◦ C. Assuming the water has not
warmed 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 )e kt + Ts = 80e kt + 18
To ﬁnd k, plug in t = 5:
38 = T (5) = 80e 5k + 18
and solve for k.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 28 / 37
59.
Finding k
38 = T (5) = 80e 5k + 18
20 = 80e 5k
1
= e 5k
4
1
ln = 5k
4
1
=⇒ k = − ln 4.
5
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 29 / 37
60.
Finding k
38 = T (5) = 80e 5k + 18
20 = 80e 5k
1
= e 5k
4
1
ln = 5k
4
1
=⇒ k = − ln 4.
5
Now we need to solve
t
20 = T (t) = 80e − 5 ln 4 + 18
for t.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 29 / 37
61.
Finding t
t
20 = 80e − 5 ln 4 + 18
t
2 = 80e − 5 ln 4
1 t
= e − 5 ln 4
40
t
− ln 40 = − ln 4
5
ln 40 5 ln 40
=⇒ t = 1
= ≈ 13 min
5 ln 4 ln 4
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 30 / 37
62.
Computing time of death with NLC
Example
A murder victim is discovered at
midnight and the temperature of
the body is recorded as 31 ◦ C.
One hour later, the temperature
of the body is 29 ◦ C. Assume
that the surrounding air
temperature remains constant at
21 ◦ C. Calculate the victim’s
time of death. (The “normal”
temperature of a living human
being is approximately 37 ◦ C.)
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 31 / 37
63.
Solution
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 32 / 37
64.
Solution
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.
To ﬁnd k:
29 = 10e k·1 + 21 =⇒ k = ln 0.8
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 32 / 37
65.
Solution
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.
To ﬁnd k:
29 = 10e k·1 + 21 =⇒ k = ln 0.8
To ﬁnd t:
37 = 10e t·ln(0.8) + 21
1.6 = e t·ln(0.8)
ln(1.6)
t= ≈ −2.10 hr
ln(0.8)
So the time of death was just before 10:00 pm.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 32 / 37
66.
Outline
Recall
The diﬀerential equation y = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 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
67.
Interest
If an account has an compound interest rate of r per year
compounded n times, then an initial deposit of A0 dollars becomes
r nt
A0 1 +
n
after t years.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 34 / 37
68.
Interest
If an account has an compound interest rate of r per year
compounded n times, then an initial deposit of A0 dollars becomes
r nt
A0 1 +
n
after t years.
For diﬀerent amounts of compounding, this will change. As n → ∞,
we get continously compounded interest
r nt
A(t) = lim A0 1 + = A0 e rt .
n→∞ n
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 34 / 37
69.
Interest
If an account has an compound interest rate of r per year
compounded n times, then an initial deposit of A0 dollars becomes
r nt
A0 1 +
n
after t years.
For diﬀerent amounts of compounding, this will change. As n → ∞,
we get continously compounded interest
r nt
A(t) = lim A0 1 + = A0 e rt .
n→∞ n
Thus dollars are like bacteria.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 34 / 37
70.
Computing doubling time with exponential growth
Example
How long does it take an initial deposit of $100, compounded
continuously, to double?
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 35 / 37
71.
Computing doubling time with exponential growth
Example
How long does it take an initial deposit of $100, compounded
continuously, to double?
Solution
We need t such that A(t) = 200. In other words
ln 2
200 = 100e rt =⇒ 2 = e rt =⇒ ln 2 = rt =⇒ t = .
r
For instance, if r = 6% = 0.06, we have
ln 2 0.69 69
t= ≈ = = 11.5 years.
0.06 0.06 6
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 35 / 37
72.
I-banking interview tip of the day
ln 2
The fraction can also be
r
approximated as either 70 or
72 divided by the percentage
rate (as a number between 0
and 100, not a fraction
between 0 and 1.)
This is sometimes called the
rule of 70 or rule of 72.
72 has lots of factors so it’s
used more often.
V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 36 / 37
73.
Summary
When something grows or decays at a constant relative rate, the
growth or decay is exponential.
Equations with unknowns in an exponent can be solved with
logarithms.
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
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