Lesson 15: Exponential Growth and Decay

Section 3.4
Exponential Growth and Decay

V63.0121.002.2010Su, Calculus I

New York University

June 2, 2010

Announcements

Review in second half of class today
Oﬃce Hours after class today
Midterm tomorrow

Announcements

Review in second half of
class today
Oﬃce Hours after class
today
Midterm tomorrow

V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 2 / 37

Objectives


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

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


Solution

A solution is y (t) = 2t.


The Equation y = 2

Example


Solution

The general solution is y = 2t + C .


The Equation y = 2

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.


The Equation y = 2t

Example


Solution

A solution is y (t) = t 2 .


The Equation y = 2t

Example


Solution

A solution is y (t) = t 2 .
The general solution is y = t 2 + C .



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) = e t .



Example


Solution

A solution is y (t) = e t .
The general solution is y = Ce t , not y = e t + C .
(check this)


Kick it up a notch

Example

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


Kick it up a notch

Example

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

Solution

y = e 2t
y = Ce 2t


In general
Example

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


In general
Example


Solution

y = e kt
y = Ce kt


In general
Example


Solution

y = e kt
y = Ce kt

Remark
What is C ? Plug in t = 0:

y (0) = Ce k·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 e kt


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 e kt . We have

10, 000 = y0 e k·3 40, 000 = y0 e k·5


Bacteria Example
Example

Solution
Since y = ky for bacteria, we have y = y0 e kt . We have

10, 000 = y0 e k·3 40, 000 = y0 e k·5

Dividing the ﬁrst into the second gives
4 = e 2k =⇒ 2k = ln 4 =⇒ k = ln 2. Now we have

10, 000 = y0 e ln 2·3 = y0 · 8
10, 000
So y0 = = 1250.
8

Could you do that again please?

We have

10, 000 = y0 e k·3
40, 000 = y0 e k·5

Dividing the ﬁrst into the second gives

40, 000 y0 e 5k
=
10, 000 y0 e 3k
=⇒ 4 = e 2k
=⇒ ln 4 = ln(e 2k ) = 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 e kt

again!




y
=k
y
where k is negative. So

y = ky =⇒ y = y0 e kt

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

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


sample

Example

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

365 · ln 2
50 = 100e k·138/365 =⇒ k = − .
138
Therefore
365·ln 2
y (t) = 100e − 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) = 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
We are looking for the value of t for which

p(t)
= 0.1
p(0)



Example

Solution

p(t)
= 0.1
p(0)

From the 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)
= 0.1
p(0)

From the 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





states that the rate of
cooling of an object is
proportional to the
temperature diﬀerence
between the object and its
surroundings.



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 = Ce kt




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

y = ky =⇒ y = Ce kt =⇒ T − Ts = Ce kt




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

y = ky =⇒ y = Ce kt =⇒ T − Ts = Ce kt =⇒ T = Ce kt + Ts




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

y = ky =⇒ y = Ce kt =⇒ T − Ts = Ce kt =⇒ T = Ce kt + 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 )e kt + 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 )e kt + Ts = 80e kt + 18

To ﬁnd k, plug in t = 5:

38 = T (5) = 80e 5k + 18

and solve for k.


Finding k

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


Finding k

38 = T (5) = 80e 5k + 18
20 = 80e 5k
1
= e 5k
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


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 = 10e k·1 + 21 =⇒ k = ln 0.8


Solution

To ﬁnd k:

29 = 10e k·1 + 21 =⇒ k = ln 0.8

To ﬁnd t:

37 = 10e t·ln(0.8) + 21
1.6 = e t·ln(0.8)
ln(1.6)
t= ≈ −2.10 hr
ln(0.8)

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


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 e rt .
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 e rt .
n→∞ n

Thus dollars are like bacteria.


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 = 100e rt =⇒ 2 = e rt =⇒ 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 be
r
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

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