Section 3.4
Exponential Growth and Decay
V63.0121.027, Calculus I
October 27, 2009
Announcements
Quiz 3 this week in recitation
. . . . . .

Outline
Recall
The 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
1
ln x
x
1 1
loga x ·
ln a x
. . . . . .

Outline
Recall
The equation y′ = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
. . . . . .

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

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).
. . . . . .

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 = mx′′ =⇒ x′′ + = 0.
m
. . . . . .

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 = mx′′ =⇒ x′′ + = 0.
m
The most general solution is x(t) = A sin ω t + B cos ω t, where
√
ω = k/m.
. . . . . .

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

The 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) = et .
. . . . . .

The 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) = et .
The general solution is y = Cet , not y = et + C.
(check this)
. . . . . .

In general
Example
Find a solution to y′ = ky.
Find the general solution to y′ = ky.
. . . . . .

In general
Example
Find a solution to y′ = ky.
Find the general solution to y′ = ky.
Solution
y = ekt
y = Cekt
. . . . . .

In general
Example
Find a solution to y′ = ky.
Find the general solution to y′ = 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.
. . . . . .

Exponential Growth
It means the rate of change (derivative) is proportional to the
current value
Examples: Natural population growth, compounded interest,
social networks
. . . . . .

Outline
Recall
The equation y′ = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
. . . . . .

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
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 ekt . We have
10, 000 = y0 ek·3 40, 000 = y0 ek·5
. . . . . .

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 ekt . We have
10, 000 = y0 ek·3 40, 000 = y0 ek·5
Dividing the first into the second gives
4 = e2k =⇒ 2k = ln 4 =⇒ k = ln 2. Now we have
10, 000 = y0 eln 2·3 = y0 · 8
10, 000
So y0 = = 1250.
8
. . . . . .

Could you do that again please?
We have
10, 000 = y0 ek·3
40, 000 = y0 ek·5
Dividing the first into the second gives
40, 000 y e5k
= 0 3k
10, 000 y0 e
4 = e2k
ln 4 = ln(e2k ) = 2k
ln 4 ln 22 2 ln 2
k= = = = ln 2
2 2 2
. . . . . .

Outline
Recall
The equation y′ = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
. . . . . .

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

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.)
. . . . . .

Thus the relative rate of decay is constant:
y′
=k
y
where k is negative.
. . . . . .

Thus the relative rate of decay is constant:
y′
=k
y
where k is negative. So
y′ = ky =⇒ y = y0 ekt
again!
. . . . . .

Thus the relative rate of decay is constant:
y′
=k
y
where k is negative. So
y′ = ky =⇒ y = y0 ekt
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.
. . . . . .

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

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 ekt , where y0 = y(0) = 100 grams. Then
365 · ln 2
50 = 100ek·138/365 =⇒ k = − .
138
Therefore
365·ln 2
y(t) = 100e− 138
t
= 100 · 2−365t/138 .
. . . . . .

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
. . . . . .

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
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)
. . . . . .

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
2−t/5700 = 0.1
t
− ln 2 = ln 0.1
5700
ln 0.1
t= · 5700 ≈ 18, 940
ln 2
. . . . . .

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
2−t/5700 = 0.1
t
− ln 2 = ln 0.1
5700
ln 0.1
t= · 5700 ≈ 18, 940
ln 2
So the fossil is almost 19,000 years old.
. . . . . .

Outline
Recall
The equation y′ = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
. . . . . .

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.
. . . . . .

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 − T s )
dt
(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
dy
= ky
dt
. . . . . .

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
dy
= ky
dt
which we can solve:
y = Cekt
T − Ts = Cekt
=⇒ T = Cekt + Ts
. . . . . .

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
dy
= ky
dt
which we can solve:
y = Cekt
T − Ts = Cekt
=⇒ T = Cekt + Ts
Here C = y0 = T0 − Ts .
. . . . . .

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?
. . . . . .

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

Finding k
38 = T(5) = 80e5k + 18
20 = 80e5k
1
= e5k
( )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.
. . . . . .

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
. . . . . .

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
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 = 10ek·1 + 21 =⇒ k = ln 0.8
. . . . . .

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 = 10ek·1 + 21 =⇒ k = ln 0.8
To find t:
37 = 10et·ln(0.8) + 21
1.6 = et·ln(0.8)
ln(1.6)
t= ≈ −2.10 hr
ln(0.8)
So the time of death was just before 10:00pm.
. . . . . .

Outline
Recall
The equation y′ = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
. . . . . .

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.
. . . . . .

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 ert .
n→∞ n
. . . . . .

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 ert .
n→∞ n
Thus dollars are like bacteria.
. . . . . .

Example
How long does it take an initial deposit of $100, compounded
continuously, to double?
. . . . . .

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 = 100ert =⇒ 2 = ert =⇒ 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
. . . . . .

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