Lesson 15: Exponential Growth and Decay (Section 021 slides)

Section 3.4
Exponential Growth and Decay

V63.0121.021, Calculus I

New York University

October 28, 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

. . . . . .

V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 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
1
ln x
x
1 1
loga x ·
ln a x

. . . . . .


Outline

Recall



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
k
−kx(t) = mx′′ (t) =⇒ x′′ (t) + x(t) = 0.
m

. . . . . .


Definition

Example

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

. . . . . .


Showing a function is a solution

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

Solution
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

Since ω 2 = 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.

. . . . . .



Example


Solution

A solution is y(t) = 2t.

. . . . . .



Example


Solution

The general solution is y = 2t + C.

. . . . . .



Example


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

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

. . . . . .



Example


Solution

A solution is y(t) = t2 .

. . . . . .



Example


Solution

A solution is y(t) = t2 .
The general solution is y = t2 + C.

. . . . . .


The Equation y′ = y

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)

. . . . . .


Kick it up a notch: y′ = 2y

Example

Find a solution to y′ = 2y.
Find the general solution to y′ = 2y.

. . . . . .


Kick it up a notch: y′ = 2y

Example

Find a solution to y′ = 2y.
Find the general solution to y′ = 2y.

Solution

y = e2t
y = Ce2t

. . . . . .


In general: y′ = ky
Example

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

. . . . . .


Example


Solution

y = ekt
y = Cekt

. . . . . .


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. . . . . . .


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 ekt

. . . . . .


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



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 = y0 ekt . We have

10, 000 = y0 ek·3 40, 000 = y0 ek·5

. . . . . .


Bacteria Example
Example

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



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′
=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.

. . . . . .


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 = y0 ekt , where y0 = y(0) = 100 grams. Then

365 · ln 2
50 = 100ek·138/365 =⇒ k = − .
138
Therefore
= 100 · 2−365t/138
365·ln 2
y(t) = 100e− 138
t

. . . . . .


sample

Example

Solution
We have y = y0 ekt , where y0 = y(0) = 100 grams. Then

365 · ln 2
50 = 100ek·138/365 =⇒ k = − .
138
Therefore
= 100 · 2−365t/138
365·ln 2
y(t) = 100e− 138
t

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) = 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
. . . . . .


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

. . . . . .



Example

Solution
p(t)
We are looking for the value of t for which = 0.1 From the
p0
equation we have

2−t/5700 = 0.1

. . . . . .



Example

Solution
p(t)
p0
equation we have

t
2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1
5700

. . . . . .



Example

Solution
p(t)
p0
equation we have

t ln 0.1
2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700
5700 ln 2

. . . . . .



Example

Solution
p(t)
p0
equation we have

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

. . . . . .



Example

Solution
p(t)
p0
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.

. . . . . .


Outline

Recall



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
= 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 equation 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 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

. . . . . .


Finding k

38 = T(5) = 80e5k + 18
20 = 80e5k

. . . . . .


Finding k

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

. . . . . .


Finding k

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

. . . . . .


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 for t:
t
20 = T(t) = 80e− 5 ln 4 + 18
. . . . . .


Finding t


t
20 = 80e− 5 ln 4 + 18

. . . . . .


Finding t


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

. . . . . .


Finding t


t
20 = 80e− 5 ln 4 + 18
t
2 = 80e− 5 ln 4
1 t
= e− 5 ln 4
40

. . . . . .


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

. . . . . .


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

. . . . . .


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)
ln(1.6)
t= ≈ −2.10 hr
ln(0.8)

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


Outline

Recall



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
( r )nt
A0 1 +
n
after t years.

. . . . . .


Interest

( 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

( 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.

. . . . . .


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

A1 (t) = A0 (1.025)4t = P((1.025)4 )t

The balance for the interest rate r compounded continuously account
after t years is
A2 (t) = A0 ert

. . . . . .


Solving


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 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

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 also
r
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 021 slides)

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Lesson 15: Exponential Growth and Decay (Section 021 slides)