Dr. Julia Arnold and Ms. Karen Overman

1
Exponential Functions as Mathematical Models

• By
• Dr. Julia Arnold and Ms. Karen Overman
• using Tans 5th edition Applied Calculus for the
  managerial , life, and social sciences text

2
Exponential Growth

• There are some real life situations that can be
  modeled by the exponential functions previously
  studied. Recall in Section 5.3 on compound
  interest we looked at the formula for interest
  compounded continuously, .
• This model is an example of exponential growth.
• The general equation for exponential growth is
  ,
• where Q(t) is the amount present at any time t,
  is the initial amount (or P, the amount
  invested in the interest example) and k is a
  positive constant,called the growth constant (or
  r, the interest rate in the interest example).

3
Exponential Growth Model
Q(t) the amount at any time, t
the initial amount
k the growth constant
t the time
You should note that at time t 0, Q(0)
, which is the initial amount and as t
increases without bound, Q(t) also increases
without bound.
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Now lets consider the rate of change of Q(t)
otherwise known as the derivative of Q(t).
Since,
The derivative will be
If , this says that the rate of growth at any
time, t is directly proportional to the amount,
Q(t).
Thus when a function has this quality, that the
rate of growth is directly proportional to the
amount, it is said to exhibit exponential
growth.
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Its amazing how mathematics can attach formulas
to real life examples.
Example 1 Exponential Growth The growth rate
of the bacterium E-Coli, is proportional to its
size. Under ideal laboratory conditions, when
this bacterium is grown in a nutrient broth
medium, the number of cells in a culture doubles
approximately every 20 min. If the initial cell
population is 100, determine the function Q(t)
that expresses the exponential growth of the
number of cells of this bacterium as a function
of time t in minutes.
6
Example 1 Exponential Growth The growth rate
of the bacterium E-Coli, is proportional to its
size. Under ideal laboratory conditions, when
this bacterium is grown in a nutrient broth
medium, the number of cells in a culture doubles
approximately every 20 min. If the initial cell
population is 100, determine the function Q(t)
that expresses the exponential growth of the
number of cells of this bacterium as a function
of time t in minutes.
Key words growth rate would refer to the
derivative of Q or Q.
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Example 1 Exponential Growth The growth rate
of the bacterium E-Coli, is proportional to its
size. Under ideal laboratory conditions, when
this bacterium is grown in a nutrient broth
medium, the number of cells in a culture doubles
approximately every 20 min. If the initial cell
population is 100, determine the function Q(t)
that expresses the exponential growth of the
number of cells of this bacterium as a function
of time t in minutes.
Key words growth rate would refer to the
derivative of Q or Q.
The growth rate of the bacterium E-Coli, is
proportional to its size implies the equation Q
(t) k Q(t) which in turn implies the
exponential growth equation Q(t) Q0ekt
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Example 1 Exponential Growth The growth rate
of the bacterium E-Coli, is proportional to its
size. Under ideal laboratory conditions, when
this bacterium is grown in a nutrient broth
medium, the number of cells in a culture doubles
approximately every 20 min. If the initial cell
population is 100, determine the function Q(t)
that expresses the exponential growth of the
number of cells of this bacterium as a function
of time t in minutes.
Q(t) Q0ekt is our equation where Q0 represents
the amount present at time t 0, also known as
the initial amount.
Q(t) 100ekt
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Example 1 Exponential Growth The growth rate
of the bacterium E-Coli, is proportional to its
size. Under ideal laboratory conditions, when
this bacterium is grown in a nutrient broth
medium, the number of cells in a culture doubles
approximately every 20 min. If the initial cell
population is 100, determine the function Q(t)
that expresses the exponential growth of the
number of cells of this bacterium as a function
of time t in minutes.
Since the number of cells doubles every 20 min,
this means Q(t) 200 when t 20.
200 100ek20
Now we can solve for k.
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For Example 1, we are trying to determine the
function Q(t) that expresses the exponential
growth of the number of cells of this bacterium
as a function of time t in minutes.
200 100ek20 2 e20k ln2 lne20k ln2 20k
Now that we know k, we have an equation that
describes the growth of the bacteria for this
problem. Q(t) 100 e.035t
Part B How long will it take for a colony of
100 cells to increase to a population of 1
million?
1,000,000 100 e.035t and we want to find the
time t. This is another exponential equation
which requires logs to solve.
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1,000,000 100 e.035t10000 e.035tln(10000)
ln (e.035t)ln(10000) .035t
263.15 min is 4.38 hours
Part C If the initial cell population were
1000, how would this alter our model?
This equation, 200 100ek20, would become
20001000e20k, which would still yield 2e20k,
so k would remain the same value.
Part B would change to 1,000,000
1000e.035t which would become ln1000/.035
197.363.28 hours
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Exponential Decay

• Radioactive substances decay exponentially.
• These substances obey the rule Q(t) Q0e-kt
  where Q0 is the initial amount present and k is a
  suitable positive constant.
• With the exponential decay model, when t 0,
  Q(t) Q0 just like the previous model, but as t
  increases without bound Q(t) approaches 0. This
  is a decreasing function since the exponent on e
  is negative.

The half-life of a radioactive substance is the
time required for a given amount to be reduced
by ½.
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Example 2 Fossils discovered in South America
contain 5 of the Carbon 14 they originally
contained. If you know that the half-life of
Carbon 14 is 5770 years, calculate the age of the
fossils.
Solution If you know the half-life of Carbon 14
is 5770 years, then when t 5770 years the
amount of Carbon 14 is ½ the original amount.
Say the original amount is Q0
Then ½ the original amount is ½ Q0
14
Example 2 Fossils discovered in South America
contain 5 of the Carbon 14 they original
contained. If you know that the half-life of
Carbon 14 is 5770 years, calculate the age of the
fossils.
Solution If you know the half-life of Carbon 14
is 5770 years, then when t 5770 years the
amount of Carbon 14 is ½ the original amount.
Say the original amount is Q0
Then ½ the original amount is ½ Q0
Now solve for k.
Thus, ½ Q0
Q0e-k5770
½ e-k5770
ln (½) ln(e-k5770 )
Or k ln (½)/-57700.00012
ln (½) -5770k
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Example 2 continued. Now that we know k0.00012,
we can substitute it back into the exponential
decay model and use that model to answer the
question.
Q(t) Q0e-0.00012t
The question was how old are the fossils if they
contain 5 of the original amount of Carbon
14. Here we know the amount is 5 of the
original amount or 0.05Q0 . Substitute this in
for Q(t) and solve for t.
0.05 Q0Q0e-0.00012t
Or t ln0.05/-0.00012 t 24,964 years
0.05 e-0.00012t
ln0.05 lne-0.00012t
ln 0.05 -0.00012t
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Example 3 Atmospheric Pressure If the
temperature is constant, then the atmospheric
pressure P ( in pounds per square inch) varies
with the altitude above sea level h in accordance
with the law, P p0e-kh where p0 is the
atmospheric pressure at sea level and k is a
constant. If the atmospheric pressure is 15
lb/in2 at sea level and 12.5 lb/in2 at 4000 ft,
find the atmospheric pressure at an altitude of
12,000 ft. How fast is the atmospheric pressure
changing with respect to altitude at an altitude
of 12,000 ft?
Solution Begin with the generic equation and
substitute what we know.
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Example 3 Atmospheric Pressure If the
temperature is constant, then the atmospheric
pressure P ( in pounds per square inch) varies
with the altitude above sea level h in accordance
with the law, P p0e-kh where p0 is the
atmospheric pressure at sea level and k is a
constant. If the atmospheric pressure is 15
lb/in2 at sea level and 12.5 lb/in2 at 4000 ft,
find the atmospheric pressure at an altitude of
12,000 ft. How fast is the atmospheric pressure
changing with respect to altitude at an altitude
of 12,000 ft?
Solution Begin with the generic equation and
substitute what we know.
This tells us p0 15 lb/in2
P 15e-kh
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Example 3 Atmospheric Pressure If the
temperature is constant, then the atmospheric
pressure P ( in pounds per square inch) varies
with the altitude above sea level h in accordance
with the law, P p0e-kh where p0 is the
atmospheric pressure at sea level and k is a
constant. If the atmospheric pressure is 15
lb/in2 at sea level and 12.5 lb/in2 at 4000 ft,
find the atmospheric pressure at an altitude of
12,000 ft. How fast is the atmospheric pressure
changing with respect to altitude at an altitude
of 12,000 ft?
Solution
P 15e-kh
Now we substitute 12.5 for P and 4000 for h.
12.5 15e-k4000
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12.5 15e-k4000
Now the equation is P 15 e-.000045h
Find the atmospheric pressure at an altitude of
12,000 ft
P 15 e-.000045(12000)8.7 lb/in2
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The second part of the question asks How fast is
the atmospheric pressure changing with respect to
altitude at an altitude of 12,000 ft?
This is asking for the rate of change, P at h
12,000
The minus tells us the pressure is dropping.
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You should use these examples as a guide when
working the practice problems. Below is some
additional information you may need for the
practice exercises. Use this information as
appropriate. The half-life of radium is know to
be approximately 1600 years. Carbon 14 has a
half-life of 5770 years.