Lesson 16: Exponential Growth and Decay

Section 3.4
Exponential Growth and Decay

V63.0121.034, Calculus I

October 26, 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

. . . . . .

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
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

Example
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.

. . . . . .


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

. . . . . .


Example

Solution
A solution is y(t) = et .

. . . . . .


Example

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

. . . . . .

In general

. . . . . .

In general

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

. . . . . .

In general

Example

Solution
y = ekt
y = Cekt

. . . . . .

In general

Example

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

again,

. . . . . .

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
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
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

. . . . . .


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.)

. . . . . .

Thus the relative rate of decay is constant:

y′
=k
y

where k is negative.

. . . . . .


y′
=k
y

where k is negative. So

y′ = ky =⇒ y = y0 ekt

again!

. . . . . .


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
fossil?

Solution
We are looking for the value of t for which

p(t)
= 0.1
p(0)

. . . . . .

Example
fossil?

Solution

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
fossil?

Solution

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.
. . . . . .


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
= 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

. . . . . .



dy
= ky
dt
which we can solve:

y = Cekt
T − Ts = Cekt
=⇒ T = Cekt + Ts

. . . . . .



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
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)
ln(1.6)
t= ≈ −2.10 hr
ln(0.8)

So the time of death was just before 10:00pm.

. . . . . .

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

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

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.

. . . . . .

Lesson 16: Exponential Growth and Decay

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Lesson 16: Exponential Growth and Decay