3.3

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Applications: Uninhibited and Limited Growth Models

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 . Then express in terms of .

Notice that

45 xf x e f x f x

44 5 xf x e

420 xf x e

4 45 and 4 5 . Thus 4 .x xf x e f x e f x f x

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THEOREM 8

A function y = f (x) satisfies the equation

if and only if

for some constant c.

dydx

ky or f (x) k f (x)

y cekx or f (x) cekx

3.3 Applications: Uninhibited and Limited Growth Models

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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:

The function is , or where is an arbitrary constant.

As a check, note that

.dN kNdt

ktN ce ktN t ce c

.ktN t ce k kN

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Uninhibited Population GrowthThe equation

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. So

where c is the initial population P0 , and t is time.

dPdt

kP or P (t) k P(t)

P t cekt or P t P0ekt

3.3 Applications: Uninhibited and Limited Growth Models

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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

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 ?

dPdt

0.07P.

3.3 Applications: Uninhibited and Limited Growth Models

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Example 2 (concluded):a)

b)

c)

P(t) P0e0.07t

3.3 Applications: Uninhibited and Limited Growth Models

(1)P

0.07(1)100e100 1.072508

$107.25

2002

ln2

0.07100 te0.07te

0.07ln te

ln2ln20.079.9

0.07t

t

t

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THEOREM 9

The growth rate k and the doubling time T are related by kT ln2 0.693147,

or

k ln2T

0.693147

T,

and

T ln2k

0.693147

k.

3.3 Applications: Uninhibited and Limited Growth Models

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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?

The exponential growth rate is approximately 0.69% per day.

3.3 Applications: Uninhibited and Limited Growth Models

ln 2kT

ln 2100

0.006931

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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,

where t is the time, in years, after 2000.

dPdt

0.016P,

3.3 Applications: Uninhibited and Limited Growth Models

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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?

3.3 Applications: Uninhibited and Limited Growth Models

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Example 3 (concluded):

a) P(t) 6.0400e0.016t

b) P(20) 6.0400e0.01620 6.0400e0.32 8.3179 billion

ln 2c) 43.3 years; 20430.016

T

3.3 Applications: Uninhibited and Limited Growth Models

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Models of Limited GrowthThe logistic equation

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.

P(t) L

1be kt , k 0,

3.3 Applications: Uninhibited and Limited Growth Models

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Models of Limited Growth

3.3 Applications: Uninhibited and Limited Growth Models

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Example 4: Spread by skin-to-skin contact or via shared towels or clothing, methicillin-resistantstaphylococcus 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 0 cases occurcan be modeled by

N(t) 568.803

1 62.200e 0.092t .

3.3 Applications: Uninhibited and Limited Growth Models

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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.

3.3 Applications: Uninhibited and Limited Growth Models

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Example 4 (continued):

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.

3.3 Applications: Uninhibited and Limited Growth Models

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Example 4 (continued): b) Find N (t) =

After 20 weeks, the disease is spreading through the campus at a rate of about 4 new cases per week.

1 62.200e 0.092t 0 568.803 62.200e 0.092t 0.092

1 62.200e 0.092t 2

N t 52.329876 62.200e 0.092t

1 62.200e 0.092t 2

N 20 4.368

3.3 Applications: Uninhibited and Limited Growth Models

N(t) 568.803

1 62.200e 0.092t .

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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.

3.3 Applications: Uninhibited and Limited Growth Models

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3.3 Applications: Uninhibited and Limited Growth Models

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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 type , whose solutions are .

• The rate of exponential growth and the doubling time are related by the equation , or .

• Certain kinds of limited growth can be modeled by equations such as and

dP kPdt

0ktP t P e

k Tln 2Tk

ln 2kT

1 kt

LP tbe

1 , for 0.ktP t L e k