Download Project: Take two points x and y1 distance 1 apart

Project: Take two points x and y1 distance 1 apart. Construct a point y2 so that the
triangle xy1y2 has area 1, and so that xy1⊥ y1y2. Now find a point y3 such that xy2y3 has
area 1 and xy2⊥ y2y3. Keep going; what sort of figure do you get? What happens if you
make the area smaller and smaller?
Construction
What does this figure look like?
The answer to the first question posed in the problem statement involves doing
some basic geometric constructions using the basic tools of a straight edge and a compass
and by applying the Pythagorean Theorem and the formula for finding the area of a
triangle. Through class lectures and demonstrations, we proved that a point described as
the intersection of two circles, a line and a circle, or two lines could be constructed using
only straight edge and compass. Since our points can be described as the intersections of
lines and circles, we can construct this figure by hand.
To construct our first triangle, we applied that formula ½ bh=A in which A
(area)=1 and b (base)=1 also. As a result, we found the height to be 2. Knowing the two
legs of the triangle, we could find the hypotenuse using the Pythagorean Theorem and
thus obtaining a length of 5 . Then, applying the same process using b= 5 and A=1
again, we found the next height to be 2/ 5 . And similarly, using the Pythagorean
Theorem we found the next hypotenuse to be 29 . Now, we were able to construct
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this new triangle but could intuit that the next triangle would be significantly more
difficult to construct since we would obtain larger and larger fractions inside the square
root. For example, we found the side lengths of the third triangle to be 29 , 2/ 29 ,
5
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and 941 . We quickly abandoned the idea of trying to calculate the side lengths
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using exact numbers and choose to use decimal approximations of these irrational
numbers. (We know that the square root of a rational number is rational if and only if it
is of the form a2/b2 with a and b both integers.) First, we simply used a TI-85 calculator
to get a twelve-digit approximation and proceeded to construct more triangles. As we
performed the appropriate calculations, we began to notice a pattern. The most obvious
pattern was the relation between the two legs of the triangle. If we let ak be the length of
the hypotenuse of one triangle, then this becomes the base of triangle k+1. Then the
height of triangle k+1 is 2/ak if we keep the area 1. Naturally, then, we tried to find a way
of expressing the length of the hypotenuse of triangle k+1 in terms of the other two legs
through the Pythagorean Theorem. The following equation is what we derived:
a k 1 =
ak2  (2 / ak ) 2
Back to answering the first question as to the shape of the figure: as we began to
construct the first few triangles, it became apparent that the figure would resemble a
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spiral or a nautilus shell. We were reminded of an activity involving the golden ratio and
of an exponential function that was a close approximation of the spiral generated by an
actual nautilus shell. See figures 1 and 2.
How many triangles are needed to complete a full 3600 revolution?
As we began calculating triangle side lengths, we were curious as to the number
of triangles that would need to be created in order to traverse a full 3600, or 2π radians.
Since we now have a way to find the side lengths of any number of triangles, we could
find what we referred to as the central angle, meaning the angle at the point x using
trigonometry functions. The sum of the central angles of each triangle will then tell us
when we reach and or surpass 3600. So, back to the TI-85, we found that 3600 was
between the sum of the first twenty-seven central angles and the sum of the first twentyeight central angles. Each sum was about 40 different from exactly 3600.
A natural question then to ask would be, will the central angles of the triangles
ever measure to exactly some multiple of 3600? When we began to explore this question,
we realized that all of our calculations were not exact, but were some decimal
approximation. Our calculations of central angles would be especially susceptible to
large error because the trigonometry functions do not have exact values except at a very
few angle measures. Since tanΘ= opposite/adjacent, and our side lengths are also
approximations, the possible error could be quite significant. Since our triangles are all
dependent on the previous triangle, we knew the error could possibly get quite large since
we are not only approximating the rational numbers inside the square root, but the square
root itself. This error would then carry over to the next triangle since we use the
hypotenuse of one triangle as the base of the next in the sequence. Thus, the error will
just keep accumulating.
In an attempt to find better approximations, or better yet, exact values, we tried
creating a program on a TI-92 calculator. On this more advanced calculator, were had
the option of getting an output with an exact value. This program worked quite well for
finding exact values for the first two triangles but failed when attempting to calculate
exact values for the third triangle. In fact, the program did not yield any number, exact or
approximate. The program just kept running without end. We reasoned that the
calculator was trying to rationalize the denominator and couldn’t. At this point, we knew
trying to find exact answers was going to be almost impossible.
So, we then tried to find better approximations using Excel, a computer spread
sheet program, which could hold sixteen digits. At this point, we decided that the error in
our calculations would not be significant enough to really affect the work we were doing.
Using the spreadsheet, calculating a large number of triangles would be very
simple since we could enter our specific equations and then click and drag the formula to
any number of cells on the spread sheet. See Data Sheets. Within a matter of minutes,
we were able to find enough triangles to complete four full revolutions around point x.
Upon examining the multiples of 3600, we found that the second revolution came the
closest to exactly a multiple of 3600. In fact, we were within one degree above and below
7200. We then came back to our problem of error. Was a sixteen-digit approximation
accurate enough? Was that less than one-degree error caused by the use of
approximations?
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The answer to these questions is still unknown to our group. We reasoned that
even if we had used exact values for our side lengths, we would still have to use decimal
approximations when employing the trigonometric functions to find the central angle
measure. So, there is essentially no way to find the central angle measures without error.
Most tables that give values for the trigonometric functions only carry four or five
decimal places. We were using thirteen or fourteen decimal places, which is quite a bit
more accurate than four or five decimal places. It seemed highly unlikely that after any
number of revolutions, the central angles would measure exactly some multiple of 3600.
Although, as we construct more and more triangles, we would at least come closer and
closer to getting the sum of the central angels to add up to some multiple of 3600 since
the central angles get smaller and smaller with each successive triangle.
Does this sequence of triangles have a limit?
Due to the nature of the equation we derived to find the hypotenuse of the next
triangle given the previous, we were reminded of sequences we encountered in classes
such as analysis and calculus. Usually, when such objects appeared in class, we were to
examine the sequence and try to find a limit as n goes to infinity. So, naturally, we tried
to find out if this sequence of triangles, or equivalently this spiral, has a limit.
Before actually trying to do the traditional epsilon-delta proof, we simply tried to
reason our way to an answer, or at least strengthen our convictions one way or the other.
The first situation that we examined involved the Pythagorean Theorem and the formula
for finding the area of a triangle. We reasoned that if this sequence was bounded, then
the hypotenuses of the triangles could only get as big as the bound. In order to find the
hypotenuse of a triangle, we need only substitute the appropriate values into the derived
equation. Due to the nature of the derivation, we find that any number we choose as the
hypotenuse will yield a larger number for the hypotenuse of the next triangle.
For example, lets use a hypotenuse of length 10. Then this becomes the base of
the next triangle and we can find the length of the other leg by substituting into the
formula for finding the area of a triangle. We then find the other leg to be 1/5. Now we
can substitute into the equation we derived and obtain a hypotenuse of length 2501 25
which is 10.0019998 as a decimal which is slightly larger than 10.
Going along with this idea, we tried to find a function that could approximate the
graph of hypotenuse length verse triangle number on a Cartesian Plane. Using a
computer to find a regression line, we obtained a polynomial function, y=1.7682x0.2426.
See Graphs. We were expecting to find an exponential function such as h=eΘ. As we
know from calculus, though, polynomial functions are not bounded. This fact also
strengthened our resolve that the sequence did not have a bound, but again we are faced
with the idea of error when finding a regression line from a finite set of points.
The next attempt to find out whether or not the sequence had a limit lead us to
polar coordinates and hopefully a way to approximate the graph of the spiral generated by
the sequence of yi’s where i goes to infinity. By imposing a set of axes such that the
origin is at the point x. Then we can find the x and y coordinates of all the yi’s using sine
and cosine. What we found was that each yi corresponds to the ordered pair
(h cosΘ, h sinΘ) where Θ is the cumulative central angle measure and h is a calculated
hypotenuse created by drawing a perpendicular to the x axis from some yi. See figure 3.
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These ordered pairs correspond to the points on a circle of radius h. But, in our situation,
h, the length of the hypotenuse is always getting larger, and thus would not be bounded
by a circle of any radius.
Now, although each of these methods led us to believe that this sequence is not
bounded, we still had to give a concrete proof. That is to say, we really needed to give an
epsilon-delta proof. The proof we will give does not involve an epsilon per se, but the
idea is the same. The idea behind the proof is to find the difference between the y
coordinates, the hypotenuse length, of two successive triangles and show that if the first
triangle is within δ of the bound n, then the next triangle will have hypotenuse length
greater than n. So, we want to show:
(n   ) 2  (2 /( n   )) 2 > n
(n   ) 2  (2 /( n   )) 2 > n 2
(n   ) 2 > n 2  (2 /( n   )) 2
n   >  n 2  (2 /( n   )) 2
-  >  n 2  (2 /( n   )) 2  n >  n 2  (2 / n) 2  n
 <   n 2  ( 2 / n) 2  n
Concerning the plus/minus sign in front of the square root, we are only concerned with
the positive value and then the extra negative sign will cause the square root to be
negative which will yield a smaller value than if we took the positive value, and we want
delta to be small. Thus, we want:
 <  n 2  ( 2 / n) 2  n
This means that given a triangle with hypotenuse of length n-δ, the next triangle will
have a hypotenuse greater than n if  <  n 2  (2 / n) 2  n . And, given any length n,
we can find a triangle of area one with n as the length of the base, 2/n as length of the
second leg and the hypotenuse will have length greater than n, or more accurately the
hypotenuse will have length
n 2  (2 / n) 2 which is larger than n.
What happens to the spiral if we change the area?
We began to explore this question in much the same way as we did the original
question using area 1. First, we considered the case in which we kept the area of the
triangle ½ and started with two points distance one apart. As a result, we found the
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first triangle to have side lengths of 1, 1, and 2 , a 45-45-90 triangle. In order to
find the side lengths of the second triangle we can again apply the Pythagorean
Theorem and the formula for finding the area of a triangle. The calculations are fairly
simple, so we will not describe them step by step. We found the second triangle to
have side lengths 2 , 1/ 2 , and 3 / 2 . And again, we noticed a pattern as we
began to calculate more and more triangles. This pattern was almost the same as the
pattern with triangles of area one. We found that the sequence of hypotenuses were
related by a similar equation:
ak2  (1 / ak ) 2
a k 1 =
The only difference between the two equations is the numerator in the second term
under the radical. Similarly, we found the equations for areas of 1/3, 1/4, and 1/5.
Area 1/3
a k 1 =
ak2  (2 / 3ak ) 2
Area 1/4
a k 1 =
ak2  (1 / 2ak ) 2
Area 1/5
a k 1 =
ak2  (2 / 5ak ) 2
The only thing that changes is the constant of the second term under the radical. In
each case, the epsilon-delta proof would hold true. Thus, no matter the area, the
sequence of triangles will not have a limit. However, the δ value will get extremely small
as the area gets smaller.
Unanswered question
In our explorations, we found that the sequences of triangles were not bounded.
This led us to ask the question, “What needs to change in order to find a sequence of
triangles that is bounded?”
One situation that we did not have the time to examine would be when the area of
the triangle gets smaller with each consecutive triangle. For example, if we start with a
triangle of area one, and the second triangle has area one half, the third triangle has area
one third…
Although we did not get to explore this question, we conjectured that this
sequence would be bounded by a circle since we now not only have the height and the
angle going to zero, but the area as well.
High School
How can this project be applicable for the High School Curriculum?
The project of the Nautilus Shell has numerous applications and implications for a
high school curriculum, ranging from applications in Algebra, Geometry, Statistics, Pre-
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Calculus, Discrete Math, Statistics, and even Calculus. The first thing that came into our
minds when attacking this problem was the idea of spirals since what we first drew
looked very similar to what a spiral looks like. However, we did not explore this idea
very far, it could be a great way to introduce students at the high school level to spirals.
More influential in this problem was the ability to construct the picture, which
would not be easily done without some practice in the construction of numbers using only
straight edge and compass. Also, we needed ideas like Pythagorean Theorem, and
formula of the Area of a triangle to find the sides of each successive triangle, which are
concepts often introduced in Algebra or earlier. Since our recursion had 1 square root in
it, we were forced to understand how to construct a side length that is a square root. This
could have many uses in a high school geometry class for this project would be an
opportunity for students to apply the construction tools they have learned to a real
problem that may not have an answer (We certainly found no concrete all ending answer
and we are college seniors). The idea that there is still math to be proven, even at the
level that high school students can understand, is not an idea that is stressed enough in
high schools. Most high school students think that math is just memorizing what has
already been discovered and that there really is no more math to discover, a
misconception we wish we could wipe away. Math is supposed to be a creative
endeavor, in which students can explore the world around them. The idea that students
with some simple construction tools could at least begin to explore this topic, is an
interesting topic for teachers to explore.
Now, this topic problem has many more implications for the high school
curriculum, including the entire idea of a recursion and how they work. If students were
for instance in a Algebra II or Pre-Calculus class they would most definitely see the idea
of a recursion formula, and would probably be familiar with the Fibonacci numbers. We
used this recursion very early in order to try to get a grasp on what an equation would
look like. We however were not successful in finding an equation for the nth hypotenuse
that was not dependent on the previous one. This could be an excellent opportunity to
talk to students about the question whether all things can be written as an explicit
equation related merely to n, the triangle in which you are trying to find. We had no luck
in finding one and are pretty convinced we could not find one if we’d had more time.
This is where we used our calculators to write a program in which we could be
told what the hypotenuse length and angle measure (central angle) could be given by
merely plugging in the area you wanted to keep constant and the first side length. This
program was pretty much our recursion plugged into a calculator program, which would
do the calculations of our recursion for us. This proved to be useful when exploring new
avenues and again is something that with a little instruction, high school students could
do. The idea that we can use our calculators as a tool to help us solve a problem is
something of an issue in schools today, and the ability to use the calculator proved to be a
particularly helpful idea. We also needed the idea of the Tangent function, which may be
introduced at the end of Algebra II or in Trigonometry. Without this concept, we would
not have been able to find the angle measure of each successive triangle and hence we
would not be able to find the total angle.
However, when using the calculator, we had some problems that arose about
error. These issues are another great topic for a high school stats or even algebra class.
As we have noticed in our class just recently, error plays a pretty big part in how well we
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can define the number line. We struggled with the idea of how accurate our calculators
were and even attempted to attack the problem by using the TI-92, which had an exact
value function. This however had many problems since the calculator kept trying to
rationalize the denominator and since was dealing with numbers that were too big to
handle. This could bring up an interesting question for high school students: Why do we
rationalize the denominator? Although this idea is not being stressed as much now as it
was when we went through high school, it is an interesting thing to explore, for if our
calculator was not programmed to rationalize the denominator, it would have been able to
deal with the numbers in exact form.
However, no matter how accurate of answers we use for the hypotenuses, we
would still have error when using the tangent function to find the central angles, and then
consequently when trying to find the total angle. However, we still needed to decide
whether the error would be enough to affect our problem, and so we turned to Calculus to
try and find the error, but the error differentials needed a continuous function, and we
could not find a function to approximate our discrete recursion, so we were stuck. This
would be a great chance for students to talk about why something needs to be continuous
to differentiate, or even a cool problem that could be attacked using calculus.
Now, after trying to attack the error, we came up with the idea that if we could
only come up with an equation, we would be able to use our ideas from Calculus to find
the error. So we began to try to find the rectangular coordinates of each new point, and to
our surprise we found the coordinates to be in polar coordinates (something we hadn’t
really explored). What a great opportunity this would be to talk about spirals and
graphing in polar coordinates. Maybe you could even introduce this problem and ask the
students to find the coordinates of each successive point and they could discover the polar
coordinate system without really knowing it is the polar coordinate system. This could
be a great way to talk about why we bother with polar coordinates at all.
Now, we had a general idea for an equation, but were still not getting an equation
and so explored the limit in order to get some ideas about what kind of equation it could
be like. This is a key concept in early Calculus and even Pre-Calculus to look at what is
commonly called the end behavior of a function. For example, 1/x^3 has the same end
behavior as 1/x. So we began to look at what happens to the length of the hypotenuse as
we move to the infinite triangle. This is where our proof of the bound of the hypotenuse
came in and is very much related to a lot of the calculus proofs seen in high schools,
although they are very rarely asked to reproduce them. Asking students to prove that the
limit as we go to infinity is indeed infinity would be a great question to ask, for students
are often asked this question only when given an equation and they can see it easily or
they already know the end behavior of the function given to them. This question
becomes very interesting when you do not have an explicit equation.
Finally, this problem has some implications of what a limit means. For example, a secant
line goes through two points, but the tangent line is just the secant line with the difference
between the two points being shrunk to zero. We found that our function (for the length
of the hypotenuse) is not bounded, but we did not get a chance to look at what happens as
the area is shrunk to zero, and some weird things could happen when this is done, in the
same way that the tangent line only goes through one point.
So, one can see that this problem starts off quite simple and needs more and more
knowledge of math to continue in the fashion that our group did, but who knows which
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avenues could have been explored had students not known some of the information we
did. One thing is for sure, this problem gets at the heart of what mathematics is supposed
to be, and exploration of ideas given the knowledge we have at our disposal. For
example, when students first begin solving quadratics, they are told that we do not know
how to solve quadratics that are entirely above or below the x-axis, but this does not
mean we stop solving them, and eventually they learn how to deal with these imaginary
numbers. High school students might be given this problem and amaze us at the route
they take, as we may have amazed people in the route we took. What is most important
is that we let students explore problems in mathematics and let them know that some
things are just not proven in one sitting. It took ages for people to deal with concepts that
are looked at as very simple problems now. For example, the use of the concept of zero
was unused by the Greeks and Romans. The concept of nothingness was something that
made no sense to them.
It seems interesting that all of the math done at the high school level is really an
extension of defining the real numbers. All of the things we have talked about and
explored in this project could be used to some extent to define the real number line,
something none of us really had any idea of until now.
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Weekly Project Reports
Fri, 15 Sep 2000
We have answered the basic questions posed in the handout describing the projects. Lan,
Andy, and I had a pretty good idea of what we would get before we even started to draw
the figure. Are suspicions were confirmed. We found the shape to be a nautilus shell. If
the area of the triangle was not held constant, but gets smaller and smaller with
consecutive triangles, then the same basic shape appears. As to how well this new activity
represents the original depends on how much smaller we make the triangles.
One of the sub-problems our group is currently working on is how to find the side lengths
of the Nth triangle, and is there any type of recurrence relation we can use to find it.
Another problem is trying to find the "last triangle." Meaning, is there a triangle that will
use the unit one base from the first triangle as part of its hypotenuse? (I cannot really
describe in words what we mean...I hope this helps.) We tried working on these
questions because of our experience with infinite series and tiling exercises from
geometry. In addition, finding the nth triangle just seemed like a natural question to ask.
Some of the mathematical obstacles we have encountered are dealing with all of the
radicals and trying to find good approximations so that we can draw the figure accurately.
As we saw in class, sq.root of 3 is irrational, and thus does not have terminating decimal.
(Its pretty hard to draw a length of sq.rt. 3 inches) In an attempt to deal with the
radicals, we tried finding a simple recurrence or equation or the like. Concerning what
we could change to make the problem easier, we thought the basic problem was pretty
straightforward and easy. So, we could not think of anything that would make the
problem easier. However, we did experiment with different restrictions such as making
one of the sides a constant length or as suggested in the description, change the area of
the triangle. Again, we ran into trouble with all of the radicals. c. One of the more
general questions arising from our sub-problem of finding the Nth triangle is; Is there a
recurrence relation among the triangles? (We think there probably is) And, is this
recurrence applicable to anything outside of math496? Our overarching theme is
geometry, series, and recurrence relations. A more specific question deals with what our
group calls the last triangle. As we complete the nautilus, we rotate a full 360 degrees
around our original point x. How many triangles will it take to complete a full 360
degrees? (or if you please 2pi radians)
One of the related questions to this problem deals with the golden ratio. For example, if
you start with a rectangle whose sides are related by the golden ratio and then draw and
line making off a square section. The resulting rectangle also has the golden ratio as the
proportion of its sides. This process can be preformed any number of times. This then
relates to the idea of limits, as in what is the limit of the area of the squares that we are
marking off.
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We thought that this whole problem could be presented to a high school geometry class.
The students really need very few skills to explore this problem. They do need to know
about radicals, the Pythagorean theorem, and have a background in algebra. But, as the
high school system is now, the last skill should be a given. We would characterize this
investigation as finding patterns and relations between triangles. We also thought that
this could help the students with their construction skills.
Sat, 23 Sep 2000
The sub-problem that we are currently working on is the calculator or possibly computer
algorithm to calculate the length and angle precisely. We successfully made the program
for the TI-83, but are still trying to work the program for a higher calculator that would
output exact answers.
An obstacle that we are trying to work is in discovering the significance of the error with
estimation from a TI-83. Now we are trying to consider the output of individual numbers
before taking the square root of the answer we want in order to decrease some error.
A more general question is whether this error is very significant for our purposes.
Looking at and comparing the extended lengths of the triangle to give an area of one
would allow us to find a relation of how fast the length is increasing. The problem is the
error in the angle after the variable of n gets large enough. Just having an error of 1%
would affect the result greatly.
Error was dealt with a lot in statistic and the sciences, such as physics. The significance
of the results determines the dependability it.
A high school setting would be able to do this type of analysis with trial and error.
Finding the significance in the error and how that affects results. Though there error is
not great for low number of trials, the error upon error creates a margin where results are
no longer dependable. Showing this type of question to a high school setting would have
to deal with large numbered variables of length compared to the very small angles or
degrees.
Sat, 30 Sept
We have not really progressed much this week, and have been hung up on some
programming problems. We talked to you about how to get the TI-92 to run our
program, and we eventually figured it out. After getting the program to run, we found
that the calculator had a hard time finding the exact values for the side lengths and the
degree measures pat about 5 or six, so we scrapped using the TI. Sub Problem? Now
that we know the Ti does not help us get exact answers, we are concentrating a lot on
error and how much will be ok. Obstacles? We have had a hard time remembering how
to estimate error when doing a series of calculations. We could probably look at how
much error occurs from the last digit being changed, but not really sure.
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General Question? We are looking for away to relate all of our data to see if the triangles
are eventually bounded by a circle, but are being stuck determining how much the error
really matters.
Specific Question? Where and how much does rounding effect our problem, and more
specifically as we move to the infinite triangle? Similar problem? We have all seen how
much rounding can affect answers, especially in problems with many calculations. It also
makes a difference at what point you round, and since we are always using the previous
triangle to find the next, our error keeps compounding.
High school? We could very easily ask a high school class to see how much error can
occur from rounding at different spots in an equation. For example, when using a
monthly payment reduction. If a student round some of the numbers in the equation to
early, then they may get a smaller or larger amount for how long it takes to pay off a debt
or what the monthly payments should be?
Sat. 7, Oct.
We met and decided to write down the angle measure, and total angle measure (measure
of current triangle angle and previous ones) until our total surpassed 360 degrees in order
to find out how long it would take to make a complete circle. We found that it took about
27-28 triangles for this to occur and it was off by about 4 degrees on each side. We are
wondering how much rounding could effect our answer and more importantly if the last
triangle to complete the circle will fall exactly on the first (i.e. they will share
a side) We then met with you and discussed how to relate our findings to an equation,
and how to try to "squeeze our graph between two things we know.
Our sub problem for now is determining what our graph looks like with excel. We have
plotted the points and found a best-fit graph, but now pretty sure we would have to
restrict our domain since our graphs are all discrete (we are graphing angle measure vs.
triangle # and hypotenuse length vs. triangle #.) We are still looking into finding error
using a differential, but we are not sure if it is possible since our graph is discrete and not
continuous.
We are having some trouble remembering the equations for error using calculus, and
have not made a concentrated effort to look it up yet.
Our general question is still whether a circle bound our figure at infinity. A more specific
question is trying to find how the sides and angles relate, and trying to use this equation
to see if we get a circle as we move out to infinity.
A related question comes in high school when working with polar coordinate systems and
the graphs that they produce (i.e. spirals) A good question for high school students is to
ask them what happens to figures as you move to infinity(end behavior) and how that is
effected by certain variable.
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Sat, 21 Oct 2000
As we discussed in class, our group, the conch group, did not really work on the project
due to the midterm and conflicting personal schedules. However, we are still working on
finding an equation to express the relation between the hypotenuse and the triangle
number. We are pretty convinced that this relation has no bound or limit due to the fact
that for any base side length, we can construct a right triangle with area one, and the
hypotenuse will always be longer than the base by Pythagorean theorem. We are still
working on the problem of error in our calculations. We are pretty convinced that the
amount of error we are experiencing is not very significant. We have completed the
calculations for four complete revolutions around our central point using Excel. Excel
will keep track of 15 digits (I think). We have reached the conclusion that we will not
create a triangle in such a way that the "last hypotenuse" after one revolution will
coincide with the base of the first triangle. We are still trying to figure out if any triangle
will have a hypotenuse that will coincide with base of the first triangle.
Mon, 30 Oct 2000
Our group is still trying to prove one way or the other if our recurrence has a limit. We
are pretty sure that it doesn't have a limit, but we are struggling with the idea that the
spiral may converge to a circle. After our meeting, we began to explore the problem
using area one half instead of one. We came to a general consensus that the recurrence
will not have a limit no matter what the area of the triangle is, assuming of course that the
area is held constant. We found that a triangle of area one half would differ from
triangles of area one by a constant. Using the formula 1/2bh=A for the area of a triangle,
to find a triangle of 1/2 is easy. Since we are starting with a base of 1, we have
1/2(1)(h)=1/2. As we began to calculate more triangles of area 1/2, we found the spiral
approximated a circle faster than the triangles of area 1. Again, we saw that the
hypotenuse would keep growing without bound, but it would grow slower.
We are also playing with the idea of trying to plot the vertices of the consecutive triangles
and trying to find an equation that would approximate the equation of a spiral instead of
trying to find an equation that could approximate the relation between the hypotenuse and
the triangle number, or the hypotenuse and angle measure etc... What we found that that
each vertices could be found by (hcos(x), hsin(x)) where h is the length of the hypotenuse
and x is the cumulative angle measure of the triangles. This gave us even more strength
in our decision that the recurrence does not have a limit. Since the points (hcos(x),
hsin(x)) is just a circle if h is constant, but our h will be changing and in fact getting
larger. Now we just have to PROVE our idea.
Sun, 5 Nov 2000
First, our group had a meeting with you and since then have been attempting to prove that
the lengths of our hypotenuse are not bounded for the case of area one. We tried a couple
of proofs, but none have seemed to lead anywhere. We are currently working on the
bounded problem for the area 1 case. Dave and I spoke to you on Friday about how to go
about, and I believe that we will have a concrete proof by middle of next week.
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We are having a hard time remembering what it means something to be an upper bound,
but we have looked back and have started the proof, but only after the talk with you on
Friday did we realize we could make the problem easier by first looking when we set one
of our variables equal to a number. The more general question we are asking is are our
hypotenuse bounded as we shrink area smaller and smaller, and we are starting by
showing that 1 case is not bounded. We are trying to see if our spiral stays a spiral or if it
reduces to a circle as we make area smaller and smaller. This problem comes up when
graphing spirals in high school, or when working with polar coordinates. This question
has many applications to high school math education, the first being able to approximate
what a recursion does with other functions that we know, and being able to talk about
spirals and what they look like (end behavior).
Mon, 13 Nov 2000
This past week was not very productive since we all tried to concentrate on the
homework. We did try to prove the contradiction of a triangle lengths being bounded by a
circle and we believe that we did prove it this time. We will still have to confirm it once
again to make sure there were no errors.
Another question came to mind: what would happen if we decrease the area of the
triangle by 1/n ? Will a circle bound the new lengths. We know this makes the problem
more difficult since there is another variable and each calculation is different from the
other. However, this would show quicker results to whether it is bounded or not.
In trying to graphically prove or show whether the original problem was bounded, we are
still trying to make approximation in the coordinate plane to plot and maybe get an
estimation of what the function should be similar to.
In trying to adapt these ideas in a high school setting, we would use geometry and try to
get the students to look a various similar cases to see whether it helps the original idea.
Students’ work well pictorially to new ideas and doing the algebra to find distances
would also be great practice.
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