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May 24th

May 24th

# Verbal, Operational, and Applied Mathematical Connections

1. Assuming that your students pronounce 3/7 as “three sevenths,” encourage them to pronounce the
difference between 0/7 (“ zero sevenths ”) and 0/7 (“seven zeroths?”) to help them understand why one of these
ex pressions is zero and the other is undefined.

2. The “three kinds” of polynomial multiplication ( Distributive Law , FOIL method, and vertical
multiplication) can be all be viewed as summing every term in the first polynomial times every term in the
second polynomial, once and only once (“everybody here times everybody there, once and only once”).

3. Combinatorial arithmetic can easily be viewed through the language and symbolism of ordered partitions .
Instead of just talking about choosing 5 of the 52 cards in a standard deck, think of the deck of 52 as
partitioned into the 5 cards you took and the 47 you didn’t take: Permutations can
also be done this way, giving students a new way to understand how the counting is different when order is
important. For example, if a club consisting of 30 members wants to select a President, a Vice-president, and
a Secretary-Treasurer, they will need to partition the membership so that one person will be President, one
will be Vice-president, one will be Secretary-Treasurer, and 27 will not hold office, so there would be
possible ways. If the same club decides that, instead of officers, it
only wants to select a 3-person steering committee, they only need to partition the membership into a group
of 3 who will serve and the group of 27 who will not, so different results are possible.
Students who are otherwise hesitant about deciding when order is important do rather well with this
approach.

Applying this insight to the Binomial Theorem, we view repetitive binomial multiplication as a
process
of choice. Ask (almost) any student the coefficient of and you’ll get the correct
answer: 1. Ask them “why” and you may get an explanation in which they mention multiplying the four a’s
together as in: Many of them also have the insight to expect terms in
If they obtained the term in by multiplying the four a’s together, then maybe a
term in is obtained by multiplying three of the a’s by one of the b’s. But which three a’s and which b
are being multiplied? There are a number of choices this time (as opposed to the one way of obtaining ).
Of course, with this example, they can count the ways on their fingers if you let them; but our goal is to
recognize the pattern. Since terms in occur when you choose to multiply 3 of the a’s and 1 of the b’s,
there are such choices. Likewise, for the term in , there are ways to
choose 2 of the a’s and thus 2 of the b’s. Note that the use of ordered partitions naturally addresses both of
the exponents. Don’t be too surprised if your students can tell you that when is expanded, the
coefficient of

4. The Order of Ope rations can be summarized as giving priority to higher-level arithmetic, resulting in an
order that makes it possible to write polynomials without using parentheses. More interestingly, we can
connect the Order of Operations with the single-base properties of exponents (or the corresponding
properties of logarithms), which can be summarized as bringing each operation “down a notch” algebraically.
That is, we multiply single-base expressions by adding their exponents, divide by subtracting their exponents,
and exponentiate by multiplying exponents. We can also find roots using division in the exponent, and better
understand why the multiplicative identity element and multiplicative inverses are shown with the additive
identity element and additive inverses in the exponent, respectively.

5. The ancient Babylonians noted that since 6 = 2* 3, the factor 2 is too small to be the square root of 6 and
the factor 3 is too large (of course, the square root is obtained when the two factors are equal). To get a
better estimate, they just averaged those two factors, and divided 6 by this new estimate to get another
factorization. In this case, they would compute (2 + 3) / 2 = 2.5 and 6 / 2.5 = 2.4 , to get the new factorization
6 = (2.4)(2.5). Of course, the next calculations were (2.4 + 2.5) / 2 = 2.45 and 6 / 2.45 = 2.448979592, giving
the factorization 6 = (2.448979592)(2.45) . The next estimate, (2.448979592 + 2.45) / 2 = 2.449489796, has
at least 7 significant digits , since 6 / 2.449489796 = 2.44948969. At the very least, we can make use of this
ancient insight to help our developmental students improve their lackluster number sense regarding radicals.
It’s quite an improvement when a student can estimate that simply from knowing that 90 = 9*10.
In a calculus setting, we can use standard algebra to prove that the Babylonian iteration formula is identical to
the iteration formula of Newton’s Method for finding as a zero of the polynomial function

6. Is really 2/3 of a factor of 8? Yes! Consider that 8 = 2* 2 * 2 and that More formally,
Similarly, should represent 3/4 of the factorization of 16: 16 = 2* 2 * 2* 2, so
and since 9 = 3* 3, the expression should just represent 3 (“half of a factor of 9”).

7. Make the connection between the tangent of an angle and the slope of the terminal side of that angle (in
standard position). When students know that the tangent button on their calculator is a “slope finder,” they
have a greater interest in this important function (and its inverse), and quickly develop insights regarding why
tanθ is positive in quadrants I and III, and negative in quadrants II and IV.

8. Inverse functions are often presented as what happens when the roles of x and y are reversed in a one-to-one
correspondence. For example, the inverse of y = x^3 can be studied by looking at the equation x = y^3 .
Many graph this reversal of roles by sketching a reflection of the original function across the line y = x .
Consider trying the following method, instead. Use a transparency (or a thin piece of paper with markers
that will bleed through to the back side) and draw the graph of the original function, carefully labeling the
(positive) x- and y-axes. Rotate the transparency (or paper) 90
ยบ counterclockwise and notice that the positive
x-axis is now where the positive y-axis used to be; unfortunately, the positive y-axis is where the negative x-axis
was. Now turn the transparency over (looking at the back side) and you will finally see the perfect
reversal of the roles of x and y. This transformation can also be performed with the transparent ViewScreen
designed for overhead projection with TI graphing calculators. Students who want to do this transformation
with their own graphing calculator will need a mirror (it’s hard to see through the back of a calculator)!
Graph y = x^3 and hold the calculator sideways in front of you, viewing the calculator’s back with the battery
chamber on the left as you look into the mirror. You’ll see the graph of x = y^3 (that is,

9. A couple of my favorite “overlooked” trig identities are:
(a nontrivial identity involving all six trig functions)
Both have proofs based on The second identity can be proven in several ways, but the
most revealing proof is based on simplifying

before performing the indicated multiplication on the left-hand side.

10. Once upon a time, there was a Community College Child Care Center Director (my wife) who was told
by the Board of Trustees that all billing for child care was to be done monthly, in advance. Of course, each
of the children had very different care schedules, with some attending only on Mondays, Wednesdays, and
Fridays; and others who were there each day, but with different hours. Added to that was the fact that no
two months had exactly the same number of Mondays, Tuesdays, etc., especially when holidays (like
Thanksgiving) were accounted for. So the billing system would need to allow for the individual schedules of
the children, the number of Mondays, Tuesdays, etc. in a given month; and, of course, the cost per hour for
day care, which was expected to rise over time.

Consider the following matrix product:

To obtain the final billing, the column matrix containing the total number of hours was multiplied by the
scalar cost of an hour of day care (in those days, \$3).

11. The Family Room Problem (a noncalculus approach featuring the discriminant).
A young couple wanted to build a rectangular family room addition of 450 square feet onto the back of their
existing house. How should the addition be designed in order to minimize the cost of construction?

Graphing and tracing the function to be minimized gives a rough idea of the solution.

Apparently, the function has a minimum value fairly close to 60 running feet of walls and foundations. Does
the function ever equal 60? Could it possibly get smaller than that? To be able to answer these questions in
an Intermediate Algebra class, set up the equation:

While it may be tempting to divide both sides by 2, let’s not. By keeping the equation in its present form,
we can see that the coefficient of x is the opposite of the function value we wish to investigate! Using the
quadratic formula, we can see that

So 60 feet is attainable and happens only when x =15. How about something smaller, say 59.98? Of
course, this will make the discriminant negative, so no real value of x gives a running length of 59.98 ft.
Students quickly realize that a positive discriminant (indicating two ways to achieve a specific value) and a
negative discriminant (indicating that there’s no way to achieve that value) do not correspond to the desired
minimum. This leaves the zero discriminant (one, unique way to achieve the value) as the desirable case.

We can then set up and solve the more general equation

Since we want the discriminant to equal zero,

Replacing q with 60 in the quadratic formula reveals the same x =15 we saw before.

Minimum when:

Design the addition to be 15 ft by 30 ft. It will then use the minimum length of walls and foundations,
namely 60 running feet.

Note also that if your students are studying quadratic inequalities, this problem can readily be posed as
finding the (positive) values of q that give real solutions for x. Thus the discriminant must be greater than or
equal to zero. So, solve the inequality

Ignoring the negative values, we see that 60 feet is the minimum attainable value for q (since all other
attainable values of q are larger than 60, according to the final inequality). Of course, once that is
established, x is found using the equations shown on the previous page.

12. The Geometry of a Can of Tennis Balls
Consider a can of tennis balls (a cylinder in which 3 spheres are tightly packed). Which is greater, the height
of the can or its circumference? The vast majority of students looking at the can will answer incorrectly! I
use this problem as an early example of modeling with a parameter. The parameter in this case is d, the
diameter of one of the tennis balls. The height of the can is 3 ball-diameters ( h = 3d ) and its circumference is
π d. Since π > 3, the circumference is greater than the height! By eliminating the parameter , we can model
the circumference in terms of the height (or vice-versa). Since h = 3d , we have that and thus C = π d
becomes which shows that C is about 4.72% longer than h.

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