Transcript 5.6 Exponential Functions as Mathematical Models

Exponential Functions as
Mathematical Models
BY
DR. JULIA ARNOLD AND MS. KAREN
OVERMAN
USING TAN’S 5TH EDITION APPLIED
CALCULUS FOR THE MANAGERIAL ,
LIFE, AND SOCIAL SCIENCES TEXT
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,
.
rt
This model is an example of exponential growth.
A  Pe
The general equation for exponential growth is
,
kt
where Q(t) is the amount present at any time t,
the
Q is
t 
Q oinitial
e
amount (or P, the amount invested in the interest example)
and k is
Qo
a positive constant,called the growth constant (or r, the interest
rate in the interest example).
Exponential Growth Model
Q t   Q o e
kt
Q(t) = the amount at any time, t
Q o = the initial amount
k = the growth constant
t = the time
You should note that at time t = 0, Q(0) = Q e k 0  Q e 0  Q  1  Q ,
o
o
o
o
which is the initial amount and as t increases without bound, Q(t) also
increases without bound.
Now let’s consider the rate of change of Q(t) otherwise known as
the derivative of Q(t).
Since, Q  t   Q o e
kt
The derivative will be: Q '  t   Q o e
kt
k  k Q oe
kt
 k  Q t 
If Q '  t   kQ  t  , 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.
It’s 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.
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’.
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
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
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.
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
k 
ln 2
20
 . 035
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.
1,000,000= 100 e.035t
10000 = e.035t
ln(10000) = ln (e.035t)
ln(10000) = .035t
ln 10000
. 035
 263 . 15 min
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 2000=1000e20k,
which would still yield 2=e20k, so k would remain the same value.
Part B would change to 1,000,000 = 1000e.035t
which would become ln1000/.035 ~ 197.36~3.28 hours
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.
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
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
Thus, ½ Q0 = Q0e-k5770
Now solve for k.
½ = e-k5770
ln (½) = ln(e-k5770 )
ln (½) = -5770k
Or
k = ln (½)/-5770=0.00012
Example 2 continued.
Now that we know k=0.00012, 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 Q0=Q0e-0.00012t
0.05 =e-0.00012t
ln0.05 =lne-0.00012t
ln 0.05 = -0.00012t
Or t = ln0.05/-0.00012
t = 24,964 years
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.
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
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
12.5 = 15ek4000
Now we substitute 12.5 for P and 4000 for h.
12.5 = 15e-k4000
12 . 5
15
ln
ln
5
6
5
6
ln
 e
 4000 k
 ln e
,
125
150

25  5
25  6

5
6
 4000 k
  4000 k
5
6
 k  . 000045
 4000
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
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
P  15 e
 . 00004 h
P   15 e
 . 00004 h
(  . 00004 )
 . 00004

P   . 0006 e
P    . 0006 e
h
 . 00004 ( 12000 )
  . 0004 lbin
The minus tells us the pressure is dropping.
2
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.