Download Exp. Prob. 4 F95

Mathematical Investigations II
Name:
Functions - Teil Drei
The Exponential
Compounding Compound Interest and ex
1.
On Exp. 1, Compound Interest, we found out that a $10,000 CD invested at 6%,
compounded annually, is worth $10,600 at the end of the first year.
a.
Suppose that instead of compounding the money annually, the CD was
compounded quarterly. If the interest rate is still 6% annually, what will the
quarterly rate be?
Look back at your explicit function for compound interest. Write a new explicit
function A(t) to describe the CD's value after t-years, when the interest is
compounded quarterly. Use your new function to find the value of the CD at the
end of the first year.
A(t) =
b.
A(1) =
Now suppose that the CD is compounded monthly. What will be the monthly
rate?
Revise your function and determine the value of the CD at the end of the first
year, if it is compounded monthly.
c.
Suppose that the CD is compounded daily (assume 365 days per year). Write the
function and determine CD's value at the end of the first year.
d.
If we continue this process of compounding more and more often, what do you
think the maximum value of the CD will be at the end of the first year? What are
your reasons for your answer?
Exp. 4.1
Rev. S05
Mathematical Investigations II
Name:
2.
Consider the function
n
 0.06
V(n) = 10,000 1 


n 
which is the value of a CD at the end of the first year
if it is invested at 6% interested, compounded
n-times per year.
Graph the function for n  [1, 50], and
V  [10,600, 10,650].
Describe the graph.
3.
4.
A linear asymptote is a straight line that a function approaches. For example, if you
1
graphed y = , for "large" values of x, y gets close to, but never equal to, zero. Thus the
x
line y = 0 is an asymptote for this curve. [Graph the curve on your calculator if this is not
clear to you.] Similarly, for "large" values of y, x gets close to, but never equal to, zero.
Thus x = 0 is another asymptote for this graph.
a.
What is an asymptote for the exponential curve y = 2-x?
b.
What is an asymptote for the graph in problem 2?
Let's consider investing just ONE dollar, instead of $10 000, at 100% annual interest.
Now, our function looks like
n
 1.00
A(n) = 1 1 


n 
Graph the function on your calculator and complete this table:
A(100) =
A(2000) =
A(500) =
A(10,000) =
A(1000) =
A(30,000) =
What value is the function approaching; that is, as n , A(n) 
What is the approximate value of e1? On the TI-89, ex is found via green 'x'.
Exp. 4.2
Rev. S05
Mathematical Investigations II
Name:
Another look at Exponential Growth and Decay
Discrete vs Continuous Models
When we wrote the exponential functions modeling compound interest, bacteria growth and
radioactive decay, we assumed that at a certain time (like the end of each year or day), a certain
amount of growth or decay automatically took place. This is an acceptable assumption for
compound interest. At the end of each year, a predetermined amount of money is added to the
certificate. However, this is not an accurate assumption with respect to bacteria growth or
radioactive decay. At the end of each day, a predetermined amount of bacteria do not suddenly
come into existence, nor does a predetermined amount of radioactive material automatically
change at the end of each day. These changes occur throughout the day. Thus the discrete
model, or recursive function, is not a good model for these events. A better model involves the
number 'e', a number that will really become your friend as you move into calculus.
Let's review the functions you created on pages Exp. 2 and Exp. 3.
For bacteria growth, the explicit function was
A(t) =
t = days
For radioactive decay, the explicit function was
A(t) =
t = days
Since these events do not happen discretely, we need a better model.
As you might have noted on the last page, as you increase the frequency with which a certain
amount is compounded, there is a "ceiling" above which it cannot achieve. In fact, as n gets
larger and larger [We denote this with: n  ],
n
nt
 1.00
e
as n  11 
n 
r

or, in general, Ao  1    Ao ert

n
Thus rather than use a recursive function to model continuous changes, we can use
A(t) = Aoert,
where t is a unit time and r is determined by the type of event. If the CD on Exp. 1 was
compounded continually, its value at the end of the first year would be computed as
$10,000 e.06(1) = $10,618.37, which should be close to the value you found on problems 1d and
3b.
Rewrite the previous discrete functions from Exp. 2 and Exp. 3, using A(t) = Aoert.
For the bacteria model:
A(t) =
For the radioactive model:
t = days
A(t) =
t = days
Exp. 4.3
Rev. S05
Mathematical Investigations II
Name:
The Rule of 72
In Exp. 2, you discovered that it took about 29.22 days for the amount of bacteria to double at
2.4% regardless of the initial amount. How long will the bacteria take to double at each of the
following rates? Use the formula A(t) = Aoert, from the previous page. Leave your answers
rounded to the nearest tenth.
Rate
Time
Rate
a.
6%
b.
4%
c.
3%
d.
2%
e.
10%
f.
12%
g.
Graph the ordered pairs (rate, time).
h.
Describe the curve.
i.
What does rate·days = ?
Time
If you were careful with your arithmetic, you probably noticed that your graph is almost a curve.
Use your calculator to find a function that approximates the data. Use PowerReg.
The regression equation is:
You should have the equation time  69/rate. Which means that you can approximate the
amount of time it will take to double an amount under exponential growth by dividing the rate of
growth into 69.
So what's this about the Rule of 72? Well, 69 is not a nice number to divide into; whereas 72 is.
It has lots of factors. The rule states that you can approximate the amount of time it takes for a
CD (or anything else) to double in value by dividing the interest rate into 72. You might try this
on the previous data to see how accurate this rule is.
Exp. 4.4
Rev. S05
Mathematical Investigations II
Name:
Half-Life
In situations involving exponential decay, instead of being given the rate of decay of a substance,
you are often given the substance's half-life. A half-life is the amount of time that it takes for
half of the substance to decay into another substance. If Ao is a substance's initial amount, h is its
half-life, and r is its rate of decay, then the following must be true:
Ao
 Aoe rh
2
1
rh
e
2
2 1  erh
21/ h  er
2 t / h  e rt
Thus when modeling exponential decay, the function
A(t) = Aoert
is often replaced by the function
t/ h
A(t)  Ao 2
(1)
where Ao = the initial amount, h = the half-life and t = time. Note that h and t must be in the
same units.
Referring back to Exp. 3, what is the half-life of the isotope? (How long would it take until 500
mg had decayed?)
Use that figure to write an explicit function for the isotope using formula (1).
Exp. 4.5
Rev. S05