Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Multilateration wikipedia , lookup
Line (geometry) wikipedia , lookup
Reuleaux triangle wikipedia , lookup
Rational trigonometry wikipedia , lookup
History of trigonometry wikipedia , lookup
Euclidean geometry wikipedia , lookup
Trigonometric functions wikipedia , lookup
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 5 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 5 and 941 . We quickly abandoned the idea of trying to calculate the side lengths 58 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 1 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? 2 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. 3 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 4 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- 5 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 6 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 7 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. 8 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. 9 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. 10 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. 11 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. 12 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. 13