Chapter 5 Applications of the Exponential and Natural Logarithm Functions

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Chapter 5Applications of the Exponential and
Natural Logarithm Functions
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Chapter Outline

• Exponential Growth and Decay
• Compound Interest
• Applications of the Natural Logarithm Function to
  Economics
• Further Exponential Models

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5.1
Exponential Growth and Decay
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Section Outline

• Exponential Growth
• The Exponential Growth and Decay Model
• Exponential Growth in Application
• Exponential Decay
• Exponential Decay in Application

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Exponential Growth
Definition Example
Exponential Growth A quantity, such that, at every instant the rate of increase of the quantity is proportional to the amount of the quantity at that instant
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Exponential Growth Decay Model
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Exponential Growth in Application
EXAMPLE
(Worlds Population) The worlds population was
5.51 billion on January 1, 1993 and 5.88 billion
on January 1, 1998. Assume that at any time the
population grows at a rate proportional to the
population at that time. In what year will the
worlds population reach 7 billion?
SOLUTION
Since the oldest information were given
corresponds to 1993, that will serve as our
initial time. Therefore the year 1993 will be
the year t 0 and the population at time t 0
is 5.51 (measured in billions). Therefore, the
year 1998 will be year t 5 and the population
at time t 5 is 5.88 (measured in billions).
Since the population grows at a rate proportional
to the size of the population, we can use the
exponential growth model P(t) P0ekt to describe
the population of the world.
Since P0 is the initial quantity, P0 5.51.
Therefore, our formula becomes
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Exponential Growth in Application
CONTINUED
Now we use the other given information (5.88
billion in 1998) to determine k.
This is our function so far.
When t 5, the population is 5.88 billion people.
Divide.
Rewrite in logarithmic form.
Solve for k.
Therefore, our formula to model this situation is
Now we can determine when the worlds population
will be 7 billion.
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Exponential Growth in Application
CONTINUED
This the derived function.
Replace P(t) with 7.
Divide.
Rewrite in logarithmic form.
Solve for t.
Therefore, the worlds population will be 7
billion people about 18.36 years after our
initial year, 1993. This would be the year 1993
18.36 2011.36. That is, around the year
2011. The graph is given below.
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Exponential Growth in Application
CONTINUED
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Exponential Decay
Definition Example
Exponential Decay A quantity, such that, at every instant the rate of decrease of the quantity is proportional to the amount of the quantity at that instant
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Exponential Decay in Application
EXAMPLE
(Radioactive Decay) Radium-226 is used in cancer
radiotherapy, as a neutron source for some
research purposes, and as a constituent of
luminescent paints. Let P(t) be the number of
grams of radium-226 in a sample remaining after t
years, and suppose that P(t) satisfies the
differential equation
(a) Find the formula for P(t).
(b) What was the initial amount?
(c) What is the decay constant?
(d) Approximately how much of the radium will
remain after 943 years?
(e) How fast is the sample disintegrating when
just one gram remains? Use the differential
equation.
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Exponential Decay in Application
CONTINUED
(f) What is the weight of the sample when it is
disintegrating at the rate of 0.004 grams per
year?
(g) The radioactive material has a half-life of
about 1612 years. How much will remain after
1612 years? 3224 years?
SOLUTION
(a) Since the function y Cekt satisfies the
differential equation y? ky, the function P(t)
Cekt Ce-0.00043t (where k -0.00043).
Since for the function y Cekt, C is always the
initial quantity (at time t 0), C 12 (since
P(0) 12). Therefore, our function is
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Exponential Decay in Application
CONTINUED
(b) We were given P(0) 12. Therefore the
initial amount is 12 grams.
(c) Since our exponential decay function is
, the decay constant, being
the coefficient of t, is -0.00043.
(d) To determine approximately how much of the
radium will remain after 943 years, we will
evaluate the function at t 943.
This is the decay function.
Evaluate the function at t 943.
Simplify.
Therefore, after 943 years, there will be
approximately 8 grams remaining.
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Exponential Decay in Application
CONTINUED
(e) To determine how fast the sample is
disintegrating when just one gram remains, we
must first recognize that this is a situation
concerning the rate of change of a quantity,
namely the rate at which the radium is
disintegrating. This of course involves the
derivative function. This function was given to
us and is . Now
we will determine the value of the derivative
function at P(t) 1 (when one gram remains).
This is the derivative function.
Replace P(t) with 1.
So, when there is just one gram remaining, the
radium is disintegrating at a rate of 0.00043
grams/year.
(f) To determine the weight of the sample when it
is disintegrating at the rate of 0.004 grams per
year, we must determine P(t) when P?(t) -0.004.
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Exponential Decay in Application
CONTINUED
This is the derivative function.
Replace P?(t) with -0.004.
Solve for P(t).
So, the weight of the sample when it is
disintegrating at the rate of 0.004 grams per
year, is 9.3 grams.
(g) To determine how much of the radium will
remain after 1612 years, that is one half-life,
we will simply recognize that after one
half-life, half of the original amount of radium
will be disintegrated. That is, 12/2 6 grams
will be disintegrated and therefore 6 grams will
remain. After 3224 years, two half-lives, half
of what was remaining at the end of the first
1612 years (6 grams) will remain. That is, 6/2
3 grams. These results can be verified using the
formula for P(t).
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5.2
Compound Interest
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Section Outline

• Compound Interest Non-Continuous
• Compound Interest Continuous
• Applications of Interest Compounded Continuously

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Compound Interest Non-Continuous
Compound Interest Non-Continuous

• P principal amount invested
• m the number of times per year interest is
  compounded
• r the interest rate
• t the number of years interest is being
  compounded
• A the compound amount, the balance after t
  years

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Compound Interest
Notice that as m increases, so does A.
Therefore, the maximum amount of interest can be
acquired when m is being compounded all the time
- continuously.
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Compound Interest Continuous
Compound Interest Continuous

• P principal amount invested
• r the interest rate
• t the number of years interest is being
  compounded
• A the compound amount, the balance after t
  years

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Compound Interest Continuous
EXAMPLE
(Continuous Compound) Ten thousand dollars is
invested at 6.5 interest compounded
continuously. When will the investment be worth
41,787?
SOLUTION
We must first determine the formula for A(t).
Since interest is being compounded continuously,
the basic formula to be used is
Since the interest rate is 6.5, r 0.065.
Since ten thousand dollars is being invested, P
10,000. And since the investment is to grow to
become 41,787, A 41,787. We will make the
appropriate substitutions and then solve for t.
This is the formula to use.
P 10,000, r 0.065, and A 41,787.
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Compound Interest Continuous
CONTINUED
Divide by 10,000.
Rewrite the equation in logarithmic form.
Divide by 0.065 and solve for t.
Therefore, the 10,000 investment will grow to
41,787, via 6.5 interest compounded
continuously, in 22 years.
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Compound Interest Present Value
Variables are defined the same as in Slide 21.
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Compound Interest Present Value
EXAMPLE
(Investment Analysis) An investment earns 5.1
interest compounded continuously and is currently
growing at the rate of 765 per year. What is
the current value of the investment?
SOLUTION
Since the problem involves a rate of change, we
will use the formula for the derivative of
That is, A? rA. Since the
investment is growing at a rate of 765 per year,
A? 765. Since the interest rate is 5.1, r
0.051.
This is the given function.
A? 765 and r 0.051.
Solve for A.
Therefore, the value of A for this situation is
15,000. We can now use this, and the present
value formula, to determine P.
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Compound Interest Present Value
CONTINUED
This is the present value formula.
A 15,000, r 0.051 and t 1 (since we were
given the rate of growth per year).
Simplify.
Therefore, the current value is 14,254.18.
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5.3
Applications of the Natural Logarithm Function
to Economics
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Section Outline

• Relative Rates of Change
• Elasticity of Demand

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Relative Rate of Change
Definition Example
Relative Rate of Change The quantity on either side of the equation is often called the relative rate of change of f (t) per unit change of t (a way of comparing rates of change for two different situations). An example will be given immediately hereafter.
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Relative Rate of Change
EXAMPLE
(Percentage Rate of Change) Suppose that the
price of wheat per bushel at time t (in months)
is approximated by What is the percentage rate
of change of f (t) at t 0? t 1? t 2?
SOLUTION
Since
we see that
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Relative Rate of Change
CONTINUED
So at t 0 months, the price of wheat per bushel
contracts at a relative rate of 0.22 per month
1 month later, the price of wheat per bushel is
still contracting, but more so, at a relative
rate of 0.65. One month after that (t
2), the price of wheat per bushel is contracting,
but much less so, at a relative rate of 0.0087.
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Elasticity of Demand
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Elasticity of Demand
EXAMPLE
(Elasticity of Demand) A subway charges 65 cents
per person and has 10,000 riders each day. The
demand function for the subway is
(a) Is demand elastic or inelastic at p 65?
(b) Should the price of a ride be raised or
lowered in order to increase the amount of money
taken in by the subway?
SOLUTION
(a) We must first determine E(p).
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Elasticity of Demand
CONTINUED
Now we will determine for what value of p E(p)
1.
Set E(p) 1.
Multiply by 180 2p.
Add 2p to both sides.
Divide both sides by 3.
So, p 60 is the point at which E(p) changes
from elastic to inelastic, or visa versa.
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Elasticity of Demand
CONTINUED
Through simple inspection, which we could have
done in the first place, we can determine whether
the value of the function E(p) is greater than 1
(elastic) or less than 1 (inelastic) at p 65.
So, demand is elastic at p 65.
(b) Since demand is elastic when p 65, this
means that for revenue to increase, price should
decrease.
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5.4
Further Exponential Models
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Section Outline

• Population Growth Equations
• Exponential Models in Application

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Population Growth Equations
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Exponential Models in Application
EXAMPLE
(Spread of News) A news item is spread by word of
mouth to a potential audience of 10,000 people.
After t days, people will have heard the news.
The graph of this function is found below.
(a) Approximately how many people will have heard
the news after 7 days?
(b) At approximately what rate will the news be
spreading after 14 days?
(c) Approximately when will 7000 people have
heard the news?
(d) Approximately when will the news be spreading
at the rate of 600 people per day?
(e) When will the news be spreading at the
greatest rate?
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Exponential Models in Application
CONTINUED
(f) Use and
to determine the differential
equation satisfied by f (t).
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Exponential Models in Application
CONTINUED
SOLUTION
(a) To determine approximately how many people
will have heard the news after 7 days, we will
evaluate f (7).
So, after 7 days, we would expect about 2475
people to have heard the news.
(b) To determine at approximately what rate the
news will be spreading after 14 days, we will
evaluate f ?(14) (we use the derivative of the
function since we seek a rate of change). We
first use the quotient rule.
This is the given function.
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Exponential Models in Application
CONTINUED
Use the quotient rule.
Simplify.
Evaluate f ?(14).
So, after 14 days, the news will be spreading at
approximately 526.76 people per day.
(c) To determine approximately when 7000 people
will have heard the news, we replace f (t) with
7000 and then solve for t.
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Exponential Models in Application
CONTINUED
This is the given function.
Replace f (t) with 7000.
Multiply by the denominator.
Distribute.
Subtract.
Divide.
Rewrite in logarithm form.
Divide.
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Exponential Models in Application
CONTINUED
So, 7000 people will have heard the news after
approximately 11.89 days.
(d) To determine approximately when the news will
be spreading at the rate of 600 people per day,
we need to replace f ?(t) with 600. However,
since this will be a long, messy process (a good
algebraic exercise for you), we will just look at
the given graph where f ?(t) 600.
As can be seen on the graph, the derivative has a
value of 600 when t 6 or t 13.5.
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Exponential Models in Application
CONTINUED
(e) The news will be spreading at the greatest
rate when the rate (derivative) is greatest.
That is,
at t 10 (or when f ??(t) 0).
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Exponential Models in Application
CONTINUED
(f) Since we know
satisfies the differential equation
then we can rewrite f (t) in the form
. Upon doing this, we will
have defined M, k, and B for our function. We
can then use to create
a differential equation satisfied by f (t).
This is the given function.
Since the number 10,000 is by itself in the
numerator, it must be that M 10,000. And since
the number 50 is the only coefficient of e-0.4t,
it must be that B 50. So we must now rewrite
-0.4 in the form Mk to determine k. That is
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Exponential Models in Application
CONTINUED
Therefore,
So, f (t) satisfies the differential equation