2.1
A mathematical model is a description
of a real-world system using
mathematical language and ideas.
Exponential Growth
Differential equations are fundamental to modern science and engineering. Many of
the known laws of physics and chemistry are expressed as differential equations, and
differential equation models are used extensively in biology, economics, and other
social sciences.
The most common use of differential equations in science is to model dynamical
systems, i.e. systems that change in time according to some fixed rule. For such a
system, the variable under consideration (such as mass M, population P, or electric
charge Q) is a function of time t. Thus, a typical differential equation might be written
dM
t3 M2
dt
instead of the more familiar
y0 x 3 y 2 .
The Exponential Growth Equation
Perhaps the most common differential equation in science is the equation for exponential growth. It can be used to model a variety of phenomena in the natural and
social sciences, including unrestricted population growth, nuclear chain reactions,
continuously compounded interest, and macroeconomic growth.
The Exponential Growth Equation
The exponential growth equation is the differential equation
dy
k y,
dt
where k is a positive constant. Its solutions have the form
y y0 e kt ,
where y0 y (0) is the initial value of y.
The formula y y0 e kt represents exponential growth, as shown in Figure 1. The
constant k is called the growth constant, and has units of inverse time (e.g. number per
second).
a Figure 1: Exponential growth.
EXAMPLE 1
Solve the following initial value problem.
dy
3y,
dt
SOLUTION
y (0) 4.
According to the formulas above, the solution is y ( t ) 4 e 3t .
EXPONENTIAL GROWTH
2
EXAMPLE 2
Solve the following initial value problem.
dy
3y,
dt
y (2) 8.
SOLUTION This time the initial value is at t 2 instead of t 0, so we have some work to do.
We know that y has the form
y ( t ) y0 e 3t
for some constant y0 . Plugging in y (2) 8 gives
8 y0 e 6 ,
so y0 8 e −6 . Then
y ( t ) 8 e −6 e 3t 8 e 3t−6 .
Population Growth
Exponential growth is often used to model the growth of populations of organisms in a
resource-rich environment. Here “resource-rich” means that there is plenty of food
and other resources necessary for the population to grow indefinitely.
The relative growth rate of a population P ( t ) is the quantity
1 dP
.
P dt
This represents the rate of growth of P relative to the size of the population. For
example, if a bacteria culture has a relative growth rate of r 3/hour, it means
that each individual bacterium has an average of 3 offspring per hour (not counting
grandchildren).
Assuming a population P mantains a constant relative growth rate, its growth will
satisfy the differential equation
dP
rP.
dt
Such a population grows exponentially with time:
r P P0 e rt
a Figure 2: Exponential population growth.
where P0 is the population at time t 0 (see Figure 2).
EXAMPLE 3
During a biology experiment, a certain culture of cells has a constant relative growth rate of
0.04/minute. If there are 5,000 cells at the beginning of the experiment, how large will the
culture be one hour later?
SOLUTION
The population P ( t ) is given by the equation
P ( t ) 5,000 e 0.04t .
We are measuring t in minutes, so one hour later corresponds to t 60. Since
P (60) 5,000 e 0.04 (60) ≈ 55,115.9,
the population after one hour will be approximately 55,000 cells.
EXPONENTIAL GROWTH
3
EXAMPLE 4
A population of bacteria initially consists of 20,000 cells. Twenty minutes later, the population
has grown to 50,000 cells. How quickly is the population increasing at that time?
SOLUTION
Assuming constant relative growth rate, the population P ( t ) is given by the
equation
P ( t ) 20,000 e rt .
(1)
for some constant r. Plugging in P (20) 50,000 gives the equation
50,000 20,000 e r (20) .
To solve the equation
20,000 e 20r 50,000
for r, we first divide by 20,000 to get
e 20r 2.5,
and then we take the natural logarithm of
both sides, which yields
20r ln (2.5) .
We can now divide through by 20 to
obtain the value of r.
and solving for r yields
ln 2.5
≈ 0.045815.
20
To find the rate of increase at t 20 min, we must compute P 0 (20) . Taking the derivative of
equation (1) gives
P 0 ( t ) 20,000 r e rt
r so
P 0 (20) 20,000 (0.045815) e (0.045815)(20) ≈ 2290.73.
Thus the population is increasing at a rate of roughly 2,300 cells/min.
Describing the Growth Rate
For an exponential growing quantity y y0 e kt , there are a few alternatives to using
the growth constant k to describe the rate of exponential growth. The simplest is to use
the time constant τ, which is the reciprocal of the growth constant:
τ 1
.
k
Using the time constant, the formula for exponential growth can be written
y y0 e t/τ .
The time constant τ has units of time, and represents the amount of time that it takes
for the variable y to be multiplied by a factor of e. For this reason, τ is also known as
the e-folding time.
Another way to describe the rate of exponential growth is to use the doubling
time Td , which is the amount of time that it takes for y to double in size. The doubling
time satisfies the equation
e kTd 2
and hence
ln 2
(ln 2) τ.
k
Since ln 2 ≈ 0.6931, the doubling time is approximately 69% of the e-folding time.
Td EXAMPLE 5
A small tumor is growing exponentially with a doubling time of 15 days. If the tumor currently
has 1.3 million cells, how quickly is the number of cells increasing?
SOLUTION
Since the tumor is growing exponentially, the number of cells N is given by the
formula
N ( t ) N0 e kt ,
EXPONENTIAL GROWTH
4
where N0 1.3 million cells. Since the doubling time is 15 days, we know that
e 15k 2.
Taking the natural logarithm of both sides and then dividing through by 15 gives
k ≈ 0.046210/day.
To find N 0 (0) , we take the derivative of the formula for N to get
N 0 ( t ) N0 k e kt .
Plugging in t 0 gives
N 0 (0) N0 k (1.3 million cells)(0.046210/day) 60 thousand cells/day.
EXERCISES
1. Solve the following initial value problem:
dy
3y,
dt
y (ln 2) 40.
2. At the beginning of an experiment, a culture of E. coli contains 15,000 cells. Two
hours later, the population has grown to a size of 80,000 cells. Assuming exponential
growth, what is the time constant τ?
3. During a nuclear chain reaction, the number N of free neutrons in a sample of
fissile material obeys the differential equation
dN
kN,
dt
where k a constant known as the neutron multiplication factor.
(a) Suppose the sample initially contains 100 free neutrons. If the neutron multiplication factor is 7.0/µs, how many free neutrons will there be 2.0 µs later?
(b) How quickly will the number of free neutrons be increasing at that time?
4. The population of the United States is currently 320 million, and is increasing at a
rate of 2.4 million/year. Assuming exponential growth, what is the doubling time
of the population?
5. A cell culture is growing exponentially with a doubling time of 3.00 hours. If
there are 11,500 cells initially, how long will it take for the cell culture to grow to
30,000 cells?