Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
1. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
Sec on 3.4
.
Exponen al Growth and Decay
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University
March 23, 2011
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Notes
Announcements
Quiz 3 next week in
recita on on 2.6, 2.8, 3.1,
3.2
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Notes
Objectives
Solve the ordinary
differen al equa on
y′ (t) = ky(t), y(0) = y0
Solve problems involving
exponen al growth and
decay
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. 1
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2. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
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
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Derivatives of exponential and Notes
logarithmic functions
y y′
ex ex
ax (ln a) · ax
1
ln x
x
1 1
loga x ·
ln a x
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Notes
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
.
.
. 2
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3. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
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.
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
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Notes
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
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Notes
The Equation y′ = 2
Example
Find a solu on to y′ (t) = 2.
Find the most general solu on to y′ (t) = 2.
Solu on
A solu on is y(t) = 2t.
The general solu on is y = 2t + C.
Remark
. If a func on has a constant rate of growth, it’s linear.
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4. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
The Equation y′ = 2t
Example
Find a solu on to y′ (t) = 2t.
Find the most general solu on to y′ (t) = 2t.
Solu on
A solu on is y(t) = t2 .
The general solu on is y = t2 + C.
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Notes
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).
Solu on
A solu on is y(t) = et .
The general solu on is y = Cet , not y = et + C.
(check this)
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Notes
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
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5. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
In general: y′ = ky
Example
Find a solu on to y′ = ky.
Find the general solu on to y′ = ky.
Solu on Remark
y=e kt 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.
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Constant Relative Growth =⇒ Notes
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
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Notes
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
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. 5
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6. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
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
.
.
Notes
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.
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Notes
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
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7. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
Bacteria Example Solution
Solu on (Con nued)
We have
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Notes
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
.
.
Notes
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.)
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8. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
Radioactive decay as a differential equation
The rela ve rate of decay is constant:
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.
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Notes
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?
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Notes
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
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9. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
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?
Solu on
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Notes
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
.
.
Notes
Newton’s Law of Cooling
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).
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10. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
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
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
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Notes
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.
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Notes
Finding k
Solu on (Con nued)
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11. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
Finding t
Solu on (Con nued)
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Notes
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.)
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Solu on Notes
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12. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
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
.
.
Notes
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.
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.
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Notes
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 = P((1.025)4 )t
The balance for the interest rate r compounded con nuously
account a er t years is
A2 (t) = A0 ert
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13. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
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.
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Computing doubling time with Notes
exponential growth
Example
How long does it take an ini al deposit of $100, compounded
con nuously, to double?
Solu on
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Notes
I-banking interview tip of the day
ln 2
The frac on can also
r
be 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.
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14. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
. March 23, 2011
Notes
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.
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Notes
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Notes
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