Lesson 15: Exponential Growth and Decay (Section 041 handout)

Section 3.4
Exponential Growth and Decay
V63.0121.041, Calculus I
New York University
October 27, 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
V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 2 / 40
Objectives
Solve the ordinary
diﬀerential equation
y (t) = ky(t), y(0) = y0
Solve problems involving
exponential growth and
decay
Notes
Notes
Notes
1
Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010

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
ln x
1
x
loga x
1
ln a
·
1
x
Outline
Recall
Carbon-14 Dating
Notes
Notes
Notes
2

What is a differential equation?
Definition
A differential equation is an equation for an unknown function which
includes the function and its derivatives.
Example
Newton’s Second Law F = ma is a differential equation, where
a(t) = x (t).
In a spring, F(x) = −kx, where x is displacement from equilibrium
and k is a constant. So
−kx(t) = mx (t) =⇒ x (t) +
k
m
x(t) = 0.
The most general solution is x(t) = A sin ωt + B cos ωt, where
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
x +
k
m
x = 0, where ω = k/m.
Solution
We have
x(t) = A sin ωt + B cos ωt
x (t) = Aω cos ωt − Bω sin ωt
x (t) = −Aω2
sin ωt − Bω2
sin ωt
Since ω2
= 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.
Solution
A solution is y(t) = 2t.
The general solution is y = 2t + C.
Remark
If a function has a constant rate of growth, it’s linear.
Notes
Notes
Notes
3

The Equation y = 2t
Example
Find a solution to y (t) = 2t.
Find the most general solution to y (t) = 2t.
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).
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.
Solution
y = e2t
y = Ce2t
Notes
Notes
Notes
4

In general: y = ky
Example
Find a solution to y = ky.
Find the general solution to y = ky.
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) = y0ekt
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
Notes
Notes
Notes
5

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?
Solution
Notes
Notes
Notes
6

Outline
Recall
Carbon-14 Dating
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
y
= k
where k is negative. So
y = ky =⇒ y = y0ekt
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.
Notes
Notes
Notes
7

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?
Carbon-14 Dating
The ratio of carbon-14 to
carbon-12 in an organism decays
exponentially:
p(t) = p0e−kt
.
The half-life of carbon-14 is
about 5700 years. So the
equation for p(t) is
p(t) = p0e− ln2
5700
t
Another way to write this would
be
p(t) = p02−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?
Solution
Notes
Notes
Notes
8

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.
This gives us a diﬀerential
equation of the form
dT
dt
= k(T − Ts)
(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
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky
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
Notes
Notes
Notes
9

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)
Finding t
Notes
Notes
Notes
10

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
Outline
Recall
Carbon-14 Dating
Notes
Notes
Notes
11

Interest
If an account has an compound interest rate of r per year
compounded n times, then an initial deposit of A0 dollars becomes
A0 1 +
r
n
nt
after t years.
For diﬀerent amounts of compounding, this will change. As n → ∞,
we get continously compounded interest
A(t) = lim
n→∞
A0 1 +
r
n
nt
= A0ert
.
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?
Solution
The balance for the 10% compounded quarterly account after t years is
B1(t) = P(1.025)4t
= P((1.025)4
)t
The balance for the interest rate r compounded continuously account after
t years is
B2(t) = Pert
Solving
B1(t) = P((1.025)4
)t
B2(t) = P(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.
Notes
Notes
Notes
12

Computing doubling time with exponential growth
Example
How long does it take an initial deposit of $100, compounded
continuously, to double?
Solution
I-banking interview tip of the day
The fraction
ln 2
r
can 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.
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.
Notes
Notes
Notes
13

Lesson 15: Exponential Growth and Decay (Section 041 handout)

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