Lesson 15: Exponential Growth and Decay (Section 041 slides)

Section 3.4
Exponential Growth and Decay
V63.0121.041, Calculus I
New York University
October 27, 2010
Announcements
Quiz 3 next week in recitation on 2.6, 2.8, 3.1, 3.2
. . . . . .

. . . . . .
Announcements
Quiz 3 next week in
recitation on 2.6, 2.8, 3.1,
3.2
V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 2 / 40

. . . . . .
Objectives
Solve the ordinary
differential equation
y′
(t) = ky(t), y(0) = y0
Solve problems involving
exponential growth and
decay

. . . . . .
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

. . . . . .
Derivatives of exponential and logarithmic functions
y y′
ex
ex
ax
(ln a) · ax
ln x
1
x
loga x
1
ln a
·
1
x

. . . . . .
Outline
Recall
= ky
Carbon-14 Dating

. . . . . .
What is a differential equation?
Definition
A differential equation is an equation for an unknown function which
includes the function and its derivatives.

. . . . . .
Definition
Example
Newton’s Second Law F = ma is a differential equation, where
a(t) = x′′
(t).

. . . . . .
Definition
Example
a(t) = x′′
(t).
In a spring, F(x) = −kx, where x is displacement from equilibrium
and k is a constant. So
−kx(t) = mx′′
(t) =⇒ x′′
(t) +
k
m
x(t) = 0.

. . . . . .
Definition
Example
a(t) = x′′
(t).
In a spring, F(x) = −kx, where x is displacement from equilibrium
and k is a constant. So
−kx(t) = mx′′
(t) =⇒ x′′
(t) +
k
m
x(t) = 0.
The most general solution is x(t) = A sin ωt + B cos ωt, where
ω =
√
k/m.

. . . . . .
Showing a function is a solution
Example (Continued)
Show that x(t) = A sin ωt + B cos ωt satisfies the differential equation
x′′
+
k
m
x = 0, where ω =
√
k/m.

. . . . . .
Showing a function is a solution
Example (Continued)
Show that x(t) = A sin ωt + B cos ωt satisfies the differential equation
x′′
+
k
m
x = 0, where ω =
√
k/m.
Solution
We have
x(t) = A sin ωt + B cos ωt
x′
(t) = Aω cos ωt − Bω sin ωt
x′′
(t) = −Aω2
sin ωt − Bω2
sin ωt
Since ω2
= k/m, the last line plus k/m times the first line result in zero.

. . . . . .
The Equation y′
= 2
Example
Find a solution to y′
(t) = 2.
Find the most general solution to y′
(t) = 2.

. . . . . .
The Equation y′
= 2
Example
(t) = 2.
(t) = 2.
Solution
A solution is y(t) = 2t.

. . . . . .
The Equation y′
= 2
Example
(t) = 2.
(t) = 2.
Solution
The general solution is y = 2t + C.

. . . . . .
The Equation y′
= 2
Example
(t) = 2.
(t) = 2.
Solution
The general solution is y = 2t + C.
Remark
If a function has a constant rate of growth, it’s linear.

. . . . . .
The Equation y′
= 2t
Example
(t) = 2t.
(t) = 2t.

. . . . . .
The Equation y′
= 2t
Example
(t) = 2t.
(t) = 2t.
Solution
A solution is y(t) = t2
.

. . . . . .
The Equation y′
= 2t
Example
(t) = 2t.
(t) = 2t.
Solution
A solution is y(t) = t2
.
The general solution is y = t2
+ C.

. . . . . .
The Equation y′
= y
Example
(t) = y(t).
(t) = y(t).

. . . . . .
The Equation y′
= y
Example
(t) = y(t).
(t) = y(t).
Solution
A solution is y(t) = et
.

. . . . . .
The Equation y′
= y
Example
(t) = y(t).
(t) = y(t).
Solution
A solution is y(t) = et
.
The general solution is y = Cet
, not y = et
+ C.
(check this)

. . . . . .
Kick it up a notch: y′
= 2y
Example
= 2y.
Find the general solution to y′
= 2y.

. . . . . .
Kick it up a notch: y′
= 2y
Example
= 2y.
= 2y.
Solution
y = e2t
y = Ce2t

. . . . . .
In general: y′
= ky
Example
= ky.
= ky.

. . . . . .
In general: y′
= ky
Example
= ky.
= ky.
Solution
y = ekt
y = Cekt

. . . . . .
In general: y′
= ky
Example
= ky.
= ky.
Solution
y = ekt
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.

. . . . . .
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) = y0ekt

. . . . . .
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

. . . . . .
Outline
Recall
= ky
Carbon-14 Dating

. . . . . .
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.

. . . . . .
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?

. . . . . .
Bacteria Example
Example
Solution
Since y′
= ky for bacteria, we have y = y0ekt
. We have
10, 000 = y0ek·3
40, 000 = y0ek·5

. . . . . .
Bacteria Example
Example
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 gives
4 = e2k
=⇒ 2k = ln 4 =⇒ k = ln 2. Now we have
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, 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

. . . . . .
Outline
Recall
= ky
Carbon-14 Dating

. . . . . .
Radioactive decay occurs because many large atoms spontaneously
give off particles.

. . . . . .
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.)

. . . . . .
Radioactive decay as a differential equation
The relative rate of decay is constant:
y′
y
= k
where k is negative.

. . . . . .
y′
y
= k
where k is negative. So
y′
= ky =⇒ y = y0ekt
again!

. . . . . .
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
half-life: the amount of time it takes a pure sample to decay to one
which 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 g
sample remains after t years?

. . . . . .
sample
Example
Solution
We have y = y0ekt
, where y0 = y(0) = 100 grams. Then
50 = 100ek·138/365
=⇒ k = −
365 · ln 2
138
.
Therefore
y(t) = 100e−365·ln 2
138
t
= 100 · 2−365t/138
.

. . . . . .
sample
Example
Solution
We have y = y0ekt
, where y0 = y(0) = 100 grams. Then
50 = 100ek·138/365
=⇒ k = −
365 · ln 2
138
.
Therefore
y(t) = 100e−365·ln 2
138
t
= 100 · 2−365t/138
.
Notice y(t) = y0 · 2−t/t1/2 , where t1/2 is the half-life.

. . . . . .
Carbon-14 Dating
The ratio of carbon-14 to
carbon-12 in an organism
decays exponentially:
p(t) = p0e−kt
.
The half-life of carbon-14 is
about 5700 years. So the
equation for p(t) is
p(t) = p0e− ln2
5700
t
Another way to write this would
be
p(t) = p02−t/5700

. . . . . .
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?

. . . . . .
Example
Solution
We are looking for the value of t for which
p(t)
p(0)
= 0.1

. . . . . .
Example
Solution
p(t)
p(0)
= 0.1
From the equation we have
2−t/5700
= 0.1 =⇒ −
t
5700
ln 2 = ln 0.1 =⇒ t =
ln 0.1
ln 2
· 5700 ≈ 18, 940

. . . . . .
Example
Solution
p(t)
p(0)
= 0.1
From the equation we have
2−t/5700
= 0.1 =⇒ −
t
5700
ln 2 = ln 0.1 =⇒ t =
ln 0.1
ln 2
· 5700 ≈ 18, 940
So the fossil is almost 19,000 years old.

. . . . . .
Outline
Recall
= ky
Carbon-14 Dating

. . . . . .
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.

. . . . . .
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
dt
= 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′
and
k(T − Ts) = ky. The equation now looks like
dT
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky

. . . . . .
= T′
and
dT
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky
Now we can solve!
y′
= ky

. . . . . .
= T′
and
dT
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky
Now we can solve!
y′
= ky =⇒ y = Cekt

. . . . . .
= T′
and
dT
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky
Now we can solve!
y′
= ky =⇒ y = Cekt
=⇒ T − Ts = Cekt

. . . . . .
= T′
and
dT
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky
Now we can solve!
y′
= ky =⇒ y = Cekt
=⇒ T = Cekt
+ Ts

. . . . . .
= T′
and
dT
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky
Now we can solve!
y′
= ky =⇒ y = 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

. . . . . .
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?

. . . . . .
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)ekt
+ Ts = 80ekt
+ 18
To find k, plug in t = 5:
38 = T(5) = 80e5k
+ 18
and solve for k.

. . . . . .
Finding k
Solution (Continued)
38 = T(5) = 80e5k
+ 18
20 = 80e5k
1
4
= e5k
ln
(
1
4
)
= 5k
=⇒ k = −
1
5
ln 4.

. . . . . .
Finding k
38 = T(5) = 80e5k
+ 18
20 = 80e5k
1
4
= e5k
ln
(
1
4
)
= 5k
=⇒ k = −
1
5
ln 4.
Now we need to solve for t:
20 = T(t) = 80e− t
5
ln 4
+ 18

. . . . . .
Finding t
20 = 80e− t
5
ln 4
+ 18
2 = 80e− t
5
ln 4
1
40
= e− t
5
ln 4
− ln 40 = −
t
5
ln 4
=⇒ t =
ln 40
1
5 ln 4
=
5 ln 40
ln 4
≈ 13 min

. . . . . .
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.)

. . . . . .
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.

. . . . . .
Solution
To find k:
29 = 10ek·1
+ 21 =⇒ k = ln 0.8

. . . . . .
Solution
To find k:
29 = 10ek·1
+ 21 =⇒ k = ln 0.8
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.

. . . . . .
Outline
Recall
= ky
Carbon-14 Dating

. . . . . .
Interest
If an account has an compound interest rate of r per year
compounded n times, then an initial deposit of A0 dollars becomes
A0
(
1 +
r
n
)nt
after t years.

. . . . . .
Interest
A0
(
1 +
r
n
)nt
after t years.
For different amounts of compounding, this will change. As
n → ∞, we get continously compounded interest
A(t) = lim
n→∞
A0
(
1 +
r
n
)nt
= A0ert
.

. . . . . .
Interest
A0
(
1 +
r
n
)nt
after t years.
For different amounts of compounding, this will change. As
n → ∞, we get continously compounded interest
A(t) = lim
n→∞
A0
(
1 +
r
n
)nt
= A0ert
.
Thus dollars are like bacteria.

. . . . . .
Continuous vs. Discrete Compounding of interest
Example
Consider two bank accounts: one with 10% annual interested
compounded quarterly and one with annual interest rate r compunded
continuously. If they produce the same balance after every year, what
is r?

. . . . . .
Continuous vs. Discrete Compounding of interest
Example
Consider two bank accounts: one with 10% annual interested
compounded quarterly and one with annual interest rate r compunded
continuously. If they produce the same balance after every year, what
is r?
Solution
The balance for the 10% compounded quarterly account after t years is
B1(t) = P(1.025)4t
= P((1.025)4
)t
The balance for the interest rate r compounded continuously account
after t years is
B2(t) = Pert

. . . . . .
Solving
B1(t) = P((1.025)4
)t
B2(t) = P(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 equivalent
to 9.88% compounded continuously.

. . . . . .
Computing doubling time with exponential growth
Example
How long does it take an initial deposit of $100, compounded
continuously, to double?

. . . . . .
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
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.

. . . . . .
I-banking interview tip of the day
The fraction
ln 2
r
can also
be 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.

. . . . . .
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).

Lesson 15: Exponential Growth and Decay (Section 041 slides)

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