Applications: Uninhibited
and Limited Growth Models
3.3
OBJECTIVE
• Find functions that satisfy dP/dt = kP.
• Convert between growth rate and doubling
time.
• Solve application problems using
exponential growth and limited growth
models.
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3.3 Applications: Uninhibited and Limited Growth
Models
Quick Check 1
Differentiate f  x   5e4 x. Then express f   x  in terms of f  x  .
f   x   4  5e4 x
f   x   20e4 x
Notice that f  x   5e 4 x and f   x   4  5e 4 x  . Thus f   x   4 f  x  .
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3.3 Applications: Uninhibited and Limited Growth
Models
THEOREM 8
A function y = f (x) satisfies the equation
dy
 ky or f ( x)  k  f ( x)
dx
if and only if
y  ce kx or f ( x)  ce kx
for some constant c.
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3.3 Applications: Uninhibited and Limited Growth
Models
Example 1: Find the general form of the function
that satisfies the equation
dA
 5 A.
dt
By Theorem 8, the function must be
A  ce5t
or
A(t )  ce5t .
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3.3 Applications: Uninhibited and Limited Growth
Models
Quick Check 2
Find the general form of the function that satisfies the equation:
dN
 kN .
dt
The function is N  ce kt, or N  t   ce kt where c is an arbitrary
constant.
As a check, note that N   t   ce kt  k  kN.
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3.3 Applications: Uninhibited and Limited Growth
Models
Uninhibited Population Growth
The equation
dP
 kP or P(t )  k  P (t )
dt
is the basic model of uninhibited (unrestrained)
population growth, whether the population is
comprised of humans, bacteria in a culture, or dollars
invested with interest compounded continuously. In the
absence of inhibiting or stimulation factors, a
population normally reproduces at a rate proportional
to its size, and this is exactly what dP/dt = kP says.
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3.3 Applications: Uninhibited and Limited Growth
Models
Uninhibited Population Growth
The only function that satisfies this differential
equation is given by
P (t ) = cekt or P (t ) = P0ekt
where t is time and k is the rate expressed in decimal
notation. Note that
k×0
0
P (0) = ce = ce = c ×1 = c,
so c represents the initial population, which we
denoted P0:
P (t ) = P0 ekt .
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3.3 Applications: Uninhibited and Limited Growth
Models
Example 2: Suppose that an amount P0, in dollars, is
invested in a savings account where the interest is
compounded continuously at 7% per year. That is, the
balance P grows at the rate given by dP  0.07 P.
dt
a) Find the function that satisfies the equation. Write it
in terms of P0 and 0.07.
b) Suppose that $100 is invested. What is the balance
after 1 yr?
c) In what period of time will an investment of $100
double itself?
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Slide
20123.3-9
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3.3 Applications: Uninhibited and Limited Growth
Models
Example 2 (concluded):
a)
P(t)  P0 e0.07t
b)
P(1) 


c)
100e0.07(1)
100  1.072508
$107.25
200  100e0.07t
e0.07t
2 
ln 2 
ln e0.07t
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ln 2
ln 2
0.07
9.9
 0.07t

t

t
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3.3 Applications: Uninhibited and Limited Growth
Models
THEOREM 9
The exponential growth rate k and the doubling time T
are related by
kT
 ln 2  0.693147,
or
𝑘
ln2
0.693147
,
=
≈
𝑇
𝑇
and
𝑇
ln2
0.693147
.
=
≈
𝑘
𝑘
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3.3 Applications: Uninhibited and Limited Growth
Models
Quick Check 3
Worldwide use of the Internet is increasing at an exponential rate,
with traffic doubling every 100 days. What is the exponential growth
rate?
ln 2
k
T

ln 2
100
 0.006931
The exponential growth rate is approximately 0.69% per day.
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3.3 Applications: Uninhibited and Limited Growth
Models
Example 3: The world population was
approximately 6.0400 billion at the beginning of 2000.
It has been estimated that the population is growing
exponentially at the rate of 0.016, or 1.6%, per year.
Thus,
dP
 0.016 P,
dt
where t is the time, in years, after 2000.
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3.3 Applications: Uninhibited and Limited Growth
Models
Example 3 (continued):
a) Find the function that satisfies the equation.
Assume that P0 = 6.0400 and k = 0.016.
b) Estimate the world population at the beginning of
2020 (t = 20).
c) After what period of time will the population be
double that in 2000?
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3.3 Applications: Uninhibited and Limited Growth
Models
Example 3 (concluded):
a) P (t )  6.0400e0.016t
b) P (20)  6.0400e
0.01620
 6.0400e
0.32
 8.3179 billion
ln 2
c) T 
 43.3 years
0.016
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3.3 Applications: Uninhibited and Limited Growth
Models
Models of Limited Growth
The logistic equation, or logistic function
L
P (t ) 
, k  0,
 kt
1  be
is one model for population growth, in which there are
factors preventing the population from exceeding some
limiting value L, perhaps a limitation on food, living
space, or other natural resources.
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3.3 Applications: Uninhibited and Limited Growth
Models
Models of Limited Growth
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3.3 Applications: Uninhibited and Limited Growth
Models
Example 4: Spread by skin-to-skin contact or via
shared towels or clothing, methicillin-resistant
staphylococcus aureus (MRSA) can easily spread a
staph infection throughout a university. Left
unchecked, the number of cases of MRSA on a
university campus t weeks after the first cases occur
can be modeled by
568.803
N (t ) 
.
0.092 t
1  62.200e
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3.3 Applications: Uninhibited and Limited Growth
Models
Example 4 (continued):
a) Find the number of infected students after 3 weeks;
40 weeks; 80 weeks.
b) Find the rate at which the disease is spreading after
20 weeks.
c) Explain why an uninhibited growth model is
inappropriate but a logistic equation is appropriate
for this situation. Then use a calculator to graph the
equation.
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3.3 Applications: Uninhibited and Limited Growth
Models
568.803
N (t ) 
Example 4 (continued):
1  62.200e
0.092 t
.
a) N(3) = 11.8. So, approximately 12 students are
infected after 3 weeks.
N(40) = 221.6. So, approximately 222 students are
infected after 40 weeks.
N(80) = 547.2. So, approximately 547 students are
infected after 80 weeks.
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3.3 Applications: Uninhibited and Limited Growth
Models
Example 4 (continued):
568.803
N (t ) 
.
0.092 t
1  62.200e
b) Find N(t) =
0.092 t
0.092 t
1  62.200e
 0  568.803  62.200e
 0.092 




0.092 t 2
1  62.200e 
N t  
52.329876  62.200e 0.092t 
1  62.200e

0.092 t 2
N   20   4.368
After 20 weeks, the disease is spreading about 4 new
cases per week.
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3.3 Applications: Uninhibited and Limited Growth
Models
Example 4 (continued):
c) Unrestricted growth is inappropriate for modeling
this situation because as more students become
infected, fewer are left to be newly infected. The
logistic equation displays the rapid spread of the
disease initially, as well as the slower growth in later
weeks when there are fewer students left to be newly
infected.
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3.3 Applications: Uninhibited and Limited Growth
Models
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3.3 Applications: Uninhibited and Limited Growth
Models
Models of Limited Growth
Another model of limited growth is provided by
 kt


P t  L 1  e  , for k  0.
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3.3 Applications: Uninhibited and Limited Growth
Models
Section Summary
• Uninhibited growth can be modeled by a differential equation of the
dP
form
 kP , whose solutions are P  t   P0ekt.
dt
• The exponential growth rate k and the doubling time T are related
ln 2
by the equation T 
, or k = ln 2 .
k
T
• Certain kinds of limited growth can be modeled by equations such
L
 kt
as P  t  
and
P
t

L
1

e
, for k  0.




 kt
1  be
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