Exponential Growth and Radioactive Decay

Sec on 3.4
Exponen al Growth and Decay
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University

March 23, 2011

.

Announcements

Quiz 3 next week in
recita on on 2.6, 2.8, 3.1,
3.2

Objectives

Solve the ordinary
diﬀeren al equa on
y′ (t) = ky(t), y(0) = y0
Solve problems involving
exponen al growth and
decay

Outline
Recall
The diﬀeren al equa on y′ = ky
Modeling simple popula on growth
Modeling radioac ve decay
Carbon-14 Da ng
Newton’s Law of Cooling
Con nuously 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

What is a differential equation?
Defini on
A differen al equa on is an equa on for an unknown func on
which includes the func on and its deriva ves.

Deﬁni on
Example

Newton’s Second Law F = ma is a diﬀeren al equa on, where
a(t) = x′′ (t).

Deﬁni on
Example

Newton’s Second Law F = ma is a diﬀeren al equa on, 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

Showing a function is a solution
Example (Con nued)
Show that x(t) = A sin ωt + B cos ωt sa sﬁes the diﬀeren al
k √
equa on x′′ + x = 0, where ω = k/m.
m

Showing a function is a solution
Example (Con nued)
Show that x(t) = A sin ωt + B cos ωt sa sﬁes the diﬀeren al
k √
equa on x′′ + x = 0, where ω = k/m.
m
Solu on
We have

x(t) = A sin ωt + B cos ωt
x′ (t) = Aω cos ωt − Bω sin ωt
x′′ (t) = −Aω 2 sin ωt − Bω 2 cos ωt

The Equation y′ = 2
Example

Find a solu on to y′ (t) = 2.
Find the most general solu on to y′ (t) = 2.

Example


Solu on
A solu on is y(t) = 2t.

Example


Solu on
The general solu on is y = 2t + C.

Example


Solu on
The general solu on is y = 2t + C.

Remark
If a func on has a constant rate of growth, it’s linear.

The Equation y′ = 2t
Example

Find a solu on to y′ (t) = 2t.
Find the most general solu on to y′ (t) = 2t.

Example


Solu on
A solu on is y(t) = t2 .

Example


Solu on
A solu on is y(t) = t2 .
The general solu on is y = t2 + C.

The Equation y′ = y
Example

Find a solu on to y′ (t) = y(t).
Find the most general solu on to y′ (t) = y(t).

Example


Solu on
A solu on is y(t) = et .

Example


Solu on
A solu on is y(t) = et .
The general solu on is y = Cet , not y = et + C.
(check this)

Kick it up a notch: y′ = 2y
Example

Find a solu on to y′ = 2y.
Find the general solu on to y′ = 2y.

Kick it up a notch: y′ = 2y
Example

Find a solu on to y′ = 2y.
Find the general solu on to y′ = 2y.

Solu on
y = e2t
y = Ce2t

In general: y′ = ky
Example

Find a solu on to y′ = ky.
Find the general solu on to y′ = ky.

Example


Solu on
y = ekt
y = Cekt

Example


Solu on Remark
y = ekt What is C? Plug in t = 0:
y = Cekt y(0) = Cek·0 = C · 1 = C,

so y(0) = y0 , the ini al value
of y.

Constant Relative Growth =⇒
Exponential Growth
Theorem
A func on with constant rela ve growth rate k is an exponen al
func on with parameter k. Explicitly, the solu on to the equa on

y′ (t) = ky(t) y(0) = y0

is
y(t) = y0 ekt

Exponential Growth is everywhere
Lots of situa ons have growth rates propor onal to the current
value
This is the same as saying the rela ve growth rate is constant.
Examples: Natural popula on growth, compounded interest,
social networks

Bacteria
Since you need bacteria
to make bacteria, the
amount of new bacteria
at any moment is
propor onal to the total
amount of bacteria.
This means bacteria
popula ons grow
exponen ally.

Bacteria Example
Example
A colony of bacteria is grown under ideal condi ons 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 ini ally?

Bacteria Example
Example
A colony of bacteria is grown under ideal condi ons 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 ini ally?

Solu on
Since y′ = ky for bacteria, we have y = y0 ekt . We have

10, 000 = y0 ek·3 40, 000 = y0 ek·5

Bacteria Example Solution
Solu on (Con nued)
We have

10, 000 = y0 ek·3 40, 000 = y0 ek·5

Dividing the ﬁrst into the second gives

40, 000 y0 e5k
= 3k
=⇒ 4 = e2k =⇒ ln 4 = ln(e2k ) = 2k
10, 000 y0 e
ln 4 ln 22 2 ln 2
=⇒ k = = = = ln 2
2 2 2

Solu on (Con nued)
Since y = y0 et ln 2 , at me t = 3 we have
10, 000
10, 000 = y0 e3 ln 2 = y0 · 8 =⇒ y0 = = 1250
8

Modeling radioactive decay
Radioac ve decay occurs because many large atoms spontaneously
give oﬀ par cles.

Modeling radioactive decay
Radioac ve decay occurs because many large atoms spontaneously
give oﬀ par cles.
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 radioac ve 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 rela ve rate of decay is constant:
y′
=k
y
where k is nega ve.

y′
=k
y
where k is nega ve. So

y′ = ky =⇒ y = y0 ekt

again!

y′
=k
y
where k is nega ve. So

y′ = ky =⇒ y = y0 ekt

again!
It’s customary to express the rela ve rate of decay in the units of
half-life: the amount of me it takes a pure sample to decay to one
which is only half pure.

Computing the amount remaining
Example
The half-life of polonium-210 is about 138 days. How much of a
100 g sample remains a er t years?

Computing the amount remaining
Example
The half-life of polonium-210 is about 138 days. How much of a
100 g sample remains a er t years?

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

Carbon-14 Dating
The ra o of carbon-14 to carbon-12 in
an organism decays exponen ally:

p(t) = p0 e−kt .

The half-life of carbon-14 is about 5700
years. So the equa on for p(t) is

p(t) = p0 e− 5700 t = p0 2−t/5700
ln2

Computing age with Carbon-14
Example
Suppose a fossil is found where the ra o of carbon-14 to carbon-12
is 10% of that in a living organism. How old is the fossil?

Example

Solu on
p(t)
We are looking for the value of t for which = 0.1.
p0

Example

Solu on
p(t)
We are looking for the value of t for which = 0.1. From the equa on we have
p0

2−t/5700 = 0.1

Example

Solu on
p(t)
p0
t
2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1
5700

Example

Solu on
p(t)
p0
t ln 0.1
2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700
5700 ln 2

Example

Solu on
p(t)
p0
t ln 0.1
2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 ≈ 18, 940
5700 ln 2

Example

Solu on
p(t)
p0
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.

Newton’s Law of Cooling states
that the rate of cooling of an
object is propor onal 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 propor onal to the
temperature diﬀerence between
the object and its surroundings.
This gives us a diﬀeren al
equa on of the form
dT
= k(T − Ts )
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 equa on now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt

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

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

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

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

′ ′
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
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 solu on to the equa on 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. A er 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. A er 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?
Solu on
We know that the temperature func on takes the form

T(t) = (T0 − Ts )ekt + Ts = 80ekt + 18

To ﬁnd k, plug in t = 5 and solve for k.

Finding k
Solu on (Con nued)

38 = T(5) = 80e5k + 18

Finding k
Solu on (Con nued)

38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k

Finding k
Solu on (Con nued)

38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k
1
= e5k
4

Finding k
Solu on (Con nued)

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

Finding k
Solu on (Con nued)

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

Finding k
Solu on (Con nued)

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

20 = T(t) = 80e− 5 ln 4 + 18
t

Finding t
Solu on (Con nued)

20 = 80e− 5 ln 4 + 18
t

Finding t
Solu on (Con nued)

20 = 80e− 5 ln 4 + 18 =⇒ 2 = 80e− 5 ln 4
t t

Finding t
Solu on (Con nued)

1
20 = 80e− 5 ln 4 + 18 =⇒ 2 = 80e− 5 ln 4 =⇒ = e− 5 ln 4
t t t

40

Finding t
Solu on (Con nued)

1
20 = 80e− 5 ln 4 + 18 =⇒ 2 = 80e− 5 ln 4 =⇒ = e− 5 ln 4
t t t

40
t
− ln 40 = − ln 4
5

Finding t
Solu on (Con nued)

1
20 = 80e− 5 ln 4 + 18 =⇒ 2 = 80e− 5 ln 4 =⇒ = e− 5 ln 4
t t t

40
t ln 40 5 ln 40
− ln 40 = − ln 4 =⇒ t = 1 = ≈ 13 min
5 5 ln 4 ln 4

Computing time of death with NLC
Example
A murder vic m 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 vic m’s me of
death. (The “normal” temperature of
a living human being is approximately
37 ◦ C.)

Solu on
Let me 0 be midnight. We know T0 = 31, Ts = 21, and
T(1) = 29. We want to know the t for which T(t) = 37.

Solu on
To ﬁnd k:

29 = 10ek·1 + 21 =⇒ k = ln 0.8

Solu on
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 me of death was just before 10:00 .

Interest
If an account has an compound interest rate of r per year
compounded n mes, then an ini al deposit of A0 dollars
becomes ( r )nt
A0 1 +
n
a er t years.

Interest
becomes ( r )nt
A0 1 +
n
a er t years.
For diﬀerent amounts of compounding, this will change. As
n → ∞, we get con nously compounded interest
( r )nt
A(t) = lim A0 1 + = A0 ert .
n→∞ n

Interest
becomes ( r )nt
A0 1 +
n
a er t years.
For diﬀerent amounts of compounding, this will change. As
n → ∞, we get con nously compounded interest
( r )nt
A(t) = lim A0 1 + = A0 ert .
n→∞ n

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 con nuously. If they
produce the same balance a er 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 con nuously. If they
produce the same balance a er every year, what is r?

Solu on
The balance for the 10% compounded quarterly account a er t years
is
A1 (t) = A0 (1.025)4t = A0 ((1.025)4 )t
The balance for the interest rate r compounded con nuously
account a er t years is
A2 (t) = A0 ert

Solving
Solu on (Con nued)

A1 (t) = A0 ((1.025)4 )t
A2 (t) = A0 (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 con nuously.

Computing doubling time
Example
How long does it take an ini al deposit of $100, compounded
con nuously, to double?

Computing doubling time
Example
How long does it take an ini al deposit of $100, compounded
con nuously, to double?

Solu on
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 frac on can also be
r
approximated as either 70 or 72
divided by the percentage rate
(as a number between 0 and
100, not a frac on between 0
and 1.)
This is some mes called the rule
of 70 or rule of 72.
72 has lots of factors so it’s used
more o en.

Summary

When something grows or decays at a constant rela ve rate,
the growth or decay is exponen al.
Equa ons with unknowns in an exponent can be solved with
logarithms.
Your friend list is like culture of bacteria (no oﬀense).

Exponential Growth and Radioactive Decay

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