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Chapter 5 Applications of Exponential and
Natural Logarithm Functions
We will rely significantly on the following fact
throughout chapter 5:
The function y = Cekt satisfies the differential
equation y’ = ky.
Conversely, if y = f(t) satisfies the differential
equation y’ = ky, then y = Cekt for some constant C.
It is important to note that if f(t) = Cekt, then by
setting t = 0, we have
f(0) = Ce0 = C
So, C is the value of f(t) at t = 0.
5.1 Exponential Growth and Decay
In many areas, including biology, chemistry and
economics, it is necessary to study the behavior of
a quantity that increases as time passes. Many of
the functions we will encounter are functions of
time t.
If, at every instant, the rate of increase of a quantity
is proportional to the quantity at that instant, then
we say that the quantity is growing exponentially
or exhibiting exponential growth.
Consider the growth of a bacteria culture.
Under ideal lab conditions, a bacteria culture grows
at a rate proportional to the number of bacteria
present.
Bacteria grow through division. The more bacteria
that is present at a given instant, the greater the rate
of growth will be.
Let P(t) denote the number of bacteria in a certain
culture at time t.
The rate of growth of the culture at time t is P’(t).
We assume that the rate of growth is proportional to
the size of the culture at time t so
P’(t) = kP(t)
k is called the positive constant of proportionality.
If we let y = P(t), then we can rewrite this
equation as
y’ = ky
So,
y = P(t) = P0ekt
P0 is the number of bacteria present at time t = 0
k is called the growth constant
Problem:
Suppose that a certain bacteria culture grows at a
rate proportional to its size. At time t = 0,
approximately 20,000 bacteria are present. In 5
hours there are 400,000 bacteria.
Determine a function that expresses the size of the
culture as a function of time, measured in hours.
Let P(t) be the number of bacteria present at time t.
We are given that the culture “grows at a rate
proportional to its size” in the problem.
We can then assume that P(t) satisfies a differential
equation of the form y’ = ky. So,
y = P(t) = P0ekt
P0 and k must be determined.
If we have data about the bacteria at two different
times, then we can discover P0 and k.
The problems gives us the bacteria population size
at times t= 0 and t = 5. So,
P(0) = 20,000
P(5) = 400,000
From P(0) = 20,000, we directly get P0 as 20,000.
Now we have
P(t) = 20,000ekt
By using the second condition P(5) = 400,000, we
have
P(5) = 20,000ek(5) = 400,000
so
20,000e5k = 400,000
After dividing both sides by 20,000, we have
e5k = 20 then
ln e5k = ln 20
5k = ln 20
k = (ln 20) / 5 or approximately .60
The graph of
P(t) = 20,000e.6t
appears as
Exponential Decay
It is known that, at any instant, the rate at which a
radioactive substance is decaying is proportional to
the amount of the substance that has not yet
disintegrated.
Let P(t) be the quantity of the substance present at
time t. Then, P’(t) is the rate of decay.
P’(t) must be negative since P(t) is decreasing, and
P’(t) = kP(t) for some negative constant k.
To emphasize that k is negative, k is often replaced
by –λ, where λ is a positive constant.
Now, P(t) satisfies the differential equation
P’(t) = –λP(t)
and P(t) has the form
P(t) = P0e-λt
P(t) is called an exponential decay function
λ is called the decay constant
Problem:
The decay constant for strontium 90 is λ = .0244,
where time is measured in years.
How long will it take for the quantity P0 of
strontium 90 to decay to half its original mass?
We have P(t) = P0e-.0244t
Now, we want to know when P(t) = .5 P0
We substitute for P(t)
P0e-.0244t = .5 P0
So,
e-.0244t = .5
ln e-.0244t = ln .5
-.0244t = ln .5
t = (ln.5) / -.0244 which is approximately 28 years
From our problem, the half-life of strontium 90 is
28 years.
We can infer that it will take an additional 28 years
for the original mass to decay to 1/4th of its
original mass (56 years total) and still another 28
years to decay to 1/8th of its original mass (74 total
years).