Download Exercise 27: Applications of the Exponential Function

Pre Calculus 40S
Exercise 27: Applications of the Exponential Function
Objectives: Model and apply exponential and logarithmic functions.
Lesson:
The graph of f(x) = eax where a > 0 is called an exponential growth function.
The graph of f(x) = eax where a < 0 is called an exponential decay function.
Exponential Growth Function:
Exponential Decay Function:
These formulas can be used to model many real-world situations such as compound
interest, radioactive decay and concentration of solutions.
Compound Interest:
Consider an investment of P dollars.
Suppose it earns interest at a yearly rate of r percent and that the interest is compounded
(calculated) annually.
After the first year, the investment will be worth the original amount plus the interest
earned on that amount:
P + Pr
We could simplify this as:
P(1 + r)
At the end of the second year we would the amount from the first year plus the interest on
the amount from the first year:
P(1 + r) + P(1 + r)r
We could simplify this as:
P(1 + r)(1 + r) = P(1 + r)2
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At the end of the third year we would the amount from the second year plus the interest
on the amount from the second year:
P(1 + r)2 + P(1 + r)2r
We could simplify this as:
P(1 + r)2(1 + r) = P(1 + r)3
In general, we could say that the total value A, after t years of P dollars invested at a
yearly rate of r percent, compounded annually is:
A = P(1 + r)t
The interest periods for compound interest are usually less than a year. If the interest is
compounded n times a year and the annual or nominal interest rate is r percent, then the
r
interest rate per period is n .
So, after one year we have:
r
A = P(1 + n )n
After investing P dollars for t years at a nominal rate of r percent compounded n times a
year, we have:
r
A = P(1 + n )nt
Example 1:
A $5000 investment earns interest at an annual rate of 8.4% compounded monthly.
a) What is the investment worth after one year?
0.084
A = 5000(1 + 12 )12(1) = $5436.55
b) What is the investment worth after ten years?
0.084
A = 5000(1 + 12 )12(10) = $11 547.99
c) How much interest was earned in ten years?
$11 547.99  $5000 = $6547.99
Continuous Compound Interest:
Notice that as the number of compounding periods becomes larger, the expression
r
(1 + n )n
approaches er.
This leads to the following formula for continuous compound interest:
A = Pert
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Example 2:
A $5000 investment earns interest at an annual rate of 8.4% compounded continuously.
What is the investment worth after 10 years?
A = Pert = 5000e(0.084)(10) = $11 581.84
Notice that this is $33.85 more than the result obtained when the same principle is
earning 8.4% interest compounded monthly.
Exponential Growth and Decay:
We can generalize the formula for continuous compound interest to any situation
involving continuous exponential growth or decay. In general
A = A0ekt
Where A0 is the original amount, k is the rate of growth (<0 if decay) and t is the time.
Example 3:
At the present time, there are 1000 type A bacteria. If the rate of increase per hour is
0.025, how many bacteria can you expect in 24 hours?
A0 = 1000
k = 0.025
t = 24
A = A0ekt
= 1000e(0.025)24
= 1000e6
= 1000(1.8221188)
= 1822 bacteria
Example 4:
A radioactive substance decays at a daily rate of 0.13. How long does it take for this
substance to decompose to half its size?
There are two ways to approach this question – one is as a decay question and the other is
as a growth question. If we think of this as a decay question, we consider that we start
with 2x of the substance and are left with x amount. Because it decays, the rate will be
negative:
A=x
A0 = 2x
k = 0.13
A = A0ekt
x = (2x) e0.13t
0.5 = e0.13t
ln(0.5) = 0.13t
ln(0.5)
t=
= 5.332
0.13
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If we think of this as a growth question, we consider that we start with x of the substance
and are left with 2x amount. Because it grows, the rate will be positive:
A = 2x
A0 = x
k = 0.13
A = A0ekt
2x = x e0.13t
2 = e0.13t
ln(2) = 0.13t
ln(2)
t = 0.13 = 5.332
In both cases, it takes about five and one-third days to decompose to half its amount.
Homework: Exercise 27
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