Unit Problem:
Orchard Hideout [Radius 50]
Desmond Tsosie
Problem Statement: A 50 Radius Orchard is becoming a hideout with a tree being planted on every square unit of the orchard. The trees grow 1.5 inches squared per year with the line of sight from the center to the outside of the hideout depleting each year. How many years will it take for each tree to block the line of sight? (Each unit is 10 ft with a 2.5 in circumference.)
Process:
First, I decided my line of sight ran through the two closest trees (0,1) & (1, 49) and decided to use these two trees to determine the amount of time the trees would take to create an Orchard Hideout, I also decided to focus on the (0, 1) unit and completely ignore the second point I previously mentioned because it is irrelevant if the line of sight is blocked off by both and to be blocked off efficiently and in the same amount of time it will be the same exact amount of time to block the line of sight. Then I decided to create a smaller version of this one unit that is a square unit with 100 unit area and a simple circle. What is the initial radius of the tree in this unit though? That was the next thing I had wondered, I used the formula for Circumference to find the initial area of the tree C= 2.pi.r and plugged in 2.5, by dividing it by two then pi x radius, I had gotten 0.5 for the IV of the tree.
Then I attempted to find the final area of the tree for comparison of growth it takes for the tree to grow 1.5 each year to equal the final area. This will be found by comparing the area to where it initially stops next to the line of sight, Throughout the unit I have been able to find the final of area by creating a triangle from the point we are theoretically at to the point under the unit compared to the line of sight and the distance between those two points. Of course as stated above I would be using the formula on (0.1) understanding that the side length is 1 (10) {100) we can use the midpoint formula to find the exact side lengths of this triangle on (0,1). (x1+x2) (y1+y2) / 2 converted into (1+49(x)) divided by two (25, y3) and (0+1(y)) divided by two (25, 0.5), WOO! Both adj and opp sides are found, now we can use the SOH CAH TOA “Tan” To find the last angle while plugging in (25, 0.5) which is 11.45, Wait, 114.50, wait 1.146. Clearly I had some trouble finding it for a minute on a broken calculator. (phone)
[ Opp: .5 Adj: 25 Angle: 1.145. ]
In finding the final radius I had used Sin to find it, Sin(1.145) = x(/1) I used the broken calculator (phone) to find this equation and it turned out to be .019 x 120 (units of 10 ft) and now I had all of my angles to find the final radius! 2.3
Next I used the handy dandy area equation on the board to find the area of our beautiful magnificent circle.~ FA= 3.14 x 2.3^2 = 52.21 [Final Area]
And for the final step I used the equation to figure out the amount of time it takes for the trees to grow, 52.21 = 0.5 + 1.5 (-0.5) 51.71 / 1.5 = 34.47 Years! (I was generally told that this answer was incorrect and that the answer for the final radius was 18.05, though three of my friends have said this I genuinely can not interpret where I could’ve made a mistake, I did use that radius to get the backup amount of time it takes 11.7 years)
Solution:
After 34.37 (possibly 11.71) Years Maddie and Clyde will have built a spectacular Orchard Hideout.
Reflection:
At the time of doing this POW I had heaps of work to complete in separate classes, a majority of the work done in this POW I had slight assistance from a few of my friends (Vincent, ariana, and maddie) one of the main peers helping me complete it being Shea. He called me out for all the wrong calculations I had done which I am so grateful for. I am so intrigued to see if I have gotten it right. It should be 34 years since I've done the equation over and over again. When looking back at this POW I wish I could’ve used my time wisely, I had so much to do during this time though I think I did a pretty good job for the time I used to complete it. I’ve definitely gotten a hang of solving these equations growing from radius of 3 to 6. It was certainly hilarious that we sort of just jumped right to 50 somehow. I feel pretty proud of my work but a majority of it was assisted, I had a tough time in the beginning of this project and didn’t know what I was doing but through perseverance and hard work and help from my beautiful talented peers I can say that I understand how to find the amount of time It’ll take to create my own orchard hideout that someday I will be trapped in with no escape and deadly branches poking my eyes and blocking me from my passions and dreams. (descriptive points +1 credit?)
Desmond Tsosie
Problem Statement: A 50 Radius Orchard is becoming a hideout with a tree being planted on every square unit of the orchard. The trees grow 1.5 inches squared per year with the line of sight from the center to the outside of the hideout depleting each year. How many years will it take for each tree to block the line of sight? (Each unit is 10 ft with a 2.5 in circumference.)
Process:
First, I decided my line of sight ran through the two closest trees (0,1) & (1, 49) and decided to use these two trees to determine the amount of time the trees would take to create an Orchard Hideout, I also decided to focus on the (0, 1) unit and completely ignore the second point I previously mentioned because it is irrelevant if the line of sight is blocked off by both and to be blocked off efficiently and in the same amount of time it will be the same exact amount of time to block the line of sight. Then I decided to create a smaller version of this one unit that is a square unit with 100 unit area and a simple circle. What is the initial radius of the tree in this unit though? That was the next thing I had wondered, I used the formula for Circumference to find the initial area of the tree C= 2.pi.r and plugged in 2.5, by dividing it by two then pi x radius, I had gotten 0.5 for the IV of the tree.
Then I attempted to find the final area of the tree for comparison of growth it takes for the tree to grow 1.5 each year to equal the final area. This will be found by comparing the area to where it initially stops next to the line of sight, Throughout the unit I have been able to find the final of area by creating a triangle from the point we are theoretically at to the point under the unit compared to the line of sight and the distance between those two points. Of course as stated above I would be using the formula on (0.1) understanding that the side length is 1 (10) {100) we can use the midpoint formula to find the exact side lengths of this triangle on (0,1). (x1+x2) (y1+y2) / 2 converted into (1+49(x)) divided by two (25, y3) and (0+1(y)) divided by two (25, 0.5), WOO! Both adj and opp sides are found, now we can use the SOH CAH TOA “Tan” To find the last angle while plugging in (25, 0.5) which is 11.45, Wait, 114.50, wait 1.146. Clearly I had some trouble finding it for a minute on a broken calculator. (phone)
[ Opp: .5 Adj: 25 Angle: 1.145. ]
In finding the final radius I had used Sin to find it, Sin(1.145) = x(/1) I used the broken calculator (phone) to find this equation and it turned out to be .019 x 120 (units of 10 ft) and now I had all of my angles to find the final radius! 2.3
Next I used the handy dandy area equation on the board to find the area of our beautiful magnificent circle.~ FA= 3.14 x 2.3^2 = 52.21 [Final Area]
And for the final step I used the equation to figure out the amount of time it takes for the trees to grow, 52.21 = 0.5 + 1.5 (-0.5) 51.71 / 1.5 = 34.47 Years! (I was generally told that this answer was incorrect and that the answer for the final radius was 18.05, though three of my friends have said this I genuinely can not interpret where I could’ve made a mistake, I did use that radius to get the backup amount of time it takes 11.7 years)
Solution:
After 34.37 (possibly 11.71) Years Maddie and Clyde will have built a spectacular Orchard Hideout.
Reflection:
At the time of doing this POW I had heaps of work to complete in separate classes, a majority of the work done in this POW I had slight assistance from a few of my friends (Vincent, ariana, and maddie) one of the main peers helping me complete it being Shea. He called me out for all the wrong calculations I had done which I am so grateful for. I am so intrigued to see if I have gotten it right. It should be 34 years since I've done the equation over and over again. When looking back at this POW I wish I could’ve used my time wisely, I had so much to do during this time though I think I did a pretty good job for the time I used to complete it. I’ve definitely gotten a hang of solving these equations growing from radius of 3 to 6. It was certainly hilarious that we sort of just jumped right to 50 somehow. I feel pretty proud of my work but a majority of it was assisted, I had a tough time in the beginning of this project and didn’t know what I was doing but through perseverance and hard work and help from my beautiful talented peers I can say that I understand how to find the amount of time It’ll take to create my own orchard hideout that someday I will be trapped in with no escape and deadly branches poking my eyes and blocking me from my passions and dreams. (descriptive points +1 credit?)
POW 2 // Matching Stripes
Problem Statement:
Describe the best amount of special tiles that will have to be used to walk across a square with 63 by 90 units, and find a general formula for the problem.
Process:
- Right away I noticed that there were 16 non special squares and 8 special squares in this problem meaning ⅓ of the squares were special. Was this going to be a general formula already shaping up in front of me? I created a variety of squares such as this one just adding one unit to each side to raise the area of the marching path. 5x7, 6x8, and 7x9. And sure enough from 5x7 the shaded area is 12 and the non shaded area is 24. This gave me an understanding that all possibilities were going to be even and also doubled by the number in between both units. For 6x8 you understand that 7 is in between both numbers and multiplying that by 2 is 14 which is the shaded area and if you multiply that by two you get the side of the line crossing across the square pathway. 28. Pretty interesting. So I used this information to determine the formula that I will use in the future.
- I began to create equations that could be used to find the ⅓ of the tiles and sure enough you can use a simple equation where you multiply the units to find the area and then divide that area by 3 giving you ⅓ of the square which eventually leads to the answer for the special tiles. Though something is different if we begin using this formula on the different units, they begin to not follow the rule previously stated. It was altered and changed into a third -1, -4. Which is very strange and particular. The formula is still present it’s just altered in such a way that there was more to the equation than just stating it as #.a.b/3. (The number between a x b)
- I continued to gather data through area /3 and discovered that the greatest common divisor changed throughout the equations once it rose to 6x8. I then decided to use the amount of special squares discovered when we used axb/3 and instead of multiplying it I then added both side lengths on the tiles and subtract by their greatest common divisor finding the answer for a general formula to find any special tile amount in a given area.
Solution:
I decided to create a simple formula that would be (a.b - d = c) a.b being the side lengths of the tiles the columns given. D being divisor of the tiles, and c being the shaded (special) tiles. And this equation works throughout all tile lengths, 5x6, 7x8, etc. and beyond what is given. Plugging this equation into any given a and b will give the correct answer (i hope) So I plugged this into the tile equation we need to find the solution for and I discovered this:
90+63 - 9 = 144
Then using the original equation I had found which is finding the median of them then multiplying that my two. 76.5 x 2 gives me 153 which is incorrect if we draw it out disproving my first original theory therefore the answer of the question given would show that 144 is the correct answer!
Reflection:
I am very tired and I did all of this work in one night! I think I got the correct answer that you would want. Hopefully I didn’t mess up too badly. I really think I was onto something with finding the median of both row and column and multiplying it by two. I mean 153 is very close to 144. I originally thought that this POW was going to be the actual Unit Problem so I thought it would be due by Thanksgiving Break. I would definitely appreciate it if I didn’t get a late penalty with the fact I emailed beforehand asking for an extension. Though I understand if it has a penalty anyways, This problem is fun though it is definitely challenged by understanding the area and calculating specific areas, finding formulas, and etc.
POW 1
Equally Watered // Pow #1
Desmond Tsosie
Problem Statement: There are two flowers in a garden and you need to find all the possibilities to water both flowers equally with the same amount of water from the one sprinkler, all flowers need to be the same exact distance from the sprinkler.
Process:
Solution: The solution to this problem mainly depends on the design of the problem. If the flowers are able to move then there are a variety of ways, but if you can only move the sprinkler you are very limited in this problem!
Reflection: This POW got in my head, sometimes when I see problems that look this simple it seems like a trap and I´m always pushing to see if there is a secret way of solving a problem that is invisible, but I truly think that the POW was truly just a simple, nice way of getting into pre calculus and getting an understanding of how we´ll be doing things and getting us ready to for the more difficult problems ahead, or there is a real possibility that I totally missed something and the problem isn't really as simple as it seems, maybe there is a catch and something will be revealed when we our next lesson arrives. Though it is fun to finally feel what it is like to do a POW again and It definitely got me ready for more. I´m warmed up and excited to see what's to come.
Desmond Tsosie
Problem Statement: There are two flowers in a garden and you need to find all the possibilities to water both flowers equally with the same amount of water from the one sprinkler, all flowers need to be the same exact distance from the sprinkler.
Process:
- My initial thoughts on this problem are that there are a variety of ways to water 2 flowers from one sprinkler, but it seems you have to keep the sprinkler relatively in the middle of both flowers, but leslie can definitely move the sprinkler up or down if it is directly in the middle of the flowers. Assuming that the sprinkler sprays water to a certain distance that means there is one a select amount of possibilities. And If you don't have a limit to the length that the sprinkler sprays water it is infinite, there are some factors to this problem that aren't explained. The first problem to the question is to assume. If there was a distance given I think there would be a select number of ways.
- In the second question I assumed that if there were 3 flowers you could put it in the middle and then just lower the sprinkler if were thinking of the sprinkler and flowers as a graph we would just move the sprinkler down on the y-axis, but that isn't the case, you can only put the sprinkler in the middle because if you were to move the sprinkler downward it would take away the amount of water for the middle flower that I'm assuming is on the y-axis. Now when you add the placement of the sprinkler and the placement of the flowers you get a variety of solutions, technically speaking we could move the flowers closer, further, move the arrangement clockwise or counterclockwise. But the way of finding the distance and the arrangement of the flowers to the sprinkler it all depends on perspective. So when talking about the problem, it does not make a difference in arrangement and differences, But in real life you have a variety of ways.
- For question three, once you get to 3 you can't change the placement of the sprinkler. It can only be in the middle and this is the only way it can equally water all flowers. Unless you are focusing on changing the flowers and not the sprinklers. Then you can place the flowers in different orders, for example if you were thinking of it as on a graph, you could place the sprinkler on 0 and imagine the flowers are in order as (-5, 0), (5,0), (2, 3), and (-2, 3) then you could move the flowers in variety of ways such as that way.
Solution: The solution to this problem mainly depends on the design of the problem. If the flowers are able to move then there are a variety of ways, but if you can only move the sprinkler you are very limited in this problem!
Reflection: This POW got in my head, sometimes when I see problems that look this simple it seems like a trap and I´m always pushing to see if there is a secret way of solving a problem that is invisible, but I truly think that the POW was truly just a simple, nice way of getting into pre calculus and getting an understanding of how we´ll be doing things and getting us ready to for the more difficult problems ahead, or there is a real possibility that I totally missed something and the problem isn't really as simple as it seems, maybe there is a catch and something will be revealed when we our next lesson arrives. Though it is fun to finally feel what it is like to do a POW again and It definitely got me ready for more. I´m warmed up and excited to see what's to come.
Malls and Meadows
Problem Statement:
For this problem we were given three plots of land to work with, one for 300 acres of farm land given to our city by Mr. Goodfellow, 100 acres for an army base by the U.S Government, and 150 acres of land for mining. And between all three plots of land given we need to use this land between business and recreational land. The business community had won a bid stating that 300 acres of the land would go to development, then through town hall we compromised on using 200 acres of mining land and the army base would go towards recreational uses, and that the amount of army base land and farmland together would have to add up to 100 acres. To complete this project we used the variables Gr, Gb, (for Mr. Goodfellows Business and Recreational Land) Ar, Ab, (For the Army Bases Business and Recreational Land) and Mr. Mb. (For Mining Land Business and Recreational Land) For this problem we need to find the extremes and discover which one leads to the least amount our city would have to use.
Process:
In the beginning of our project we were given a chart on the minimum cost for our recreational and business land for each of the three plots of land. These were the costs that were decided.
Then using a town hall we used the following constraints.. We determined the only constraints we would use are I, II, and III out of all 12, first we discovered we couldn’t use (Gd + Md + Md > 300) because it simply wouldn’t work if Gd was 300 already if Ad and Md were 0 it also wouldn’t work if Gd was 300 and Gr because then VI wouldn’t work.
I Gr + Gd= 300
II Ar +Ad = 100
III Mr+Md = 150
These profits equations then helped us determine the lowest costs that would populate the areas the way we’d want them to.
Solution:
Fr [ 97 ]
Fb [ 203 ]
Mr [ 150 ]
Mb [ 0 ]
Ar [ 97 ]
Ab [ 3 ]
Conclusion:
This unit has taught our class how to persevere through the highs and lows of solving a complex mathematical problem. Overtime, we learned and practiced the skills we needed to approach and then attack the Medows or Malls problem. The most important skills being matrices and constraints. We were taught how to add, subract, multiply, and solve these smaller scale problems to prepare us for our most difficult problem yet! John had us take our time when it came to forming a general understanding of the more foundational math needed to solve the larger problem, and because of this our class grew in a way that we hadn’t before. Not only did this problem encourage us to work together, it also taught us how to slow things down. Since this problem had so many steps, it forced us to use our crictical thinking minds. We had to keep track of so many different aspects of the problem all while staying organized with the information we had already gained. Ending the unit with the Medows or Malls problem helped our class become not only better mathematicians, but also better investigative problem solvers. The longevity of the unit problem forced our class to engage in the mathmatical processes that are crucial towards our development. In the end, this problem which deemed itself so complicated, turned out to be the best work our class has done all year.
For this problem we were given three plots of land to work with, one for 300 acres of farm land given to our city by Mr. Goodfellow, 100 acres for an army base by the U.S Government, and 150 acres of land for mining. And between all three plots of land given we need to use this land between business and recreational land. The business community had won a bid stating that 300 acres of the land would go to development, then through town hall we compromised on using 200 acres of mining land and the army base would go towards recreational uses, and that the amount of army base land and farmland together would have to add up to 100 acres. To complete this project we used the variables Gr, Gb, (for Mr. Goodfellows Business and Recreational Land) Ar, Ab, (For the Army Bases Business and Recreational Land) and Mr. Mb. (For Mining Land Business and Recreational Land) For this problem we need to find the extremes and discover which one leads to the least amount our city would have to use.
Process:
In the beginning of our project we were given a chart on the minimum cost for our recreational and business land for each of the three plots of land. These were the costs that were decided.
Then using a town hall we used the following constraints.. We determined the only constraints we would use are I, II, and III out of all 12, first we discovered we couldn’t use (Gd + Md + Md > 300) because it simply wouldn’t work if Gd was 300 already if Ad and Md were 0 it also wouldn’t work if Gd was 300 and Gr because then VI wouldn’t work.
I Gr + Gd= 300
II Ar +Ad = 100
III Mr+Md = 150
These profits equations then helped us determine the lowest costs that would populate the areas the way we’d want them to.
Solution:
Fr [ 97 ]
Fb [ 203 ]
Mr [ 150 ]
Mb [ 0 ]
Ar [ 97 ]
Ab [ 3 ]
Conclusion:
This unit has taught our class how to persevere through the highs and lows of solving a complex mathematical problem. Overtime, we learned and practiced the skills we needed to approach and then attack the Medows or Malls problem. The most important skills being matrices and constraints. We were taught how to add, subract, multiply, and solve these smaller scale problems to prepare us for our most difficult problem yet! John had us take our time when it came to forming a general understanding of the more foundational math needed to solve the larger problem, and because of this our class grew in a way that we hadn’t before. Not only did this problem encourage us to work together, it also taught us how to slow things down. Since this problem had so many steps, it forced us to use our crictical thinking minds. We had to keep track of so many different aspects of the problem all while staying organized with the information we had already gained. Ending the unit with the Medows or Malls problem helped our class become not only better mathematicians, but also better investigative problem solvers. The longevity of the unit problem forced our class to engage in the mathmatical processes that are crucial towards our development. In the end, this problem which deemed itself so complicated, turned out to be the best work our class has done all year.
Pre-Calc Reflection
This is the first time I've been interested in math, throughout the years I've always felt like I wouldn't understand math.. EVER! That I wouldn't be able to feel a sense of accomplishment through mathematics. Now I finally feel like I've been able to explore math in a fun and understandable way. I felt like the explanations on matrices and finding radius and the math behind shapes finally made sense. I felt as if I could be the top of my class if I tried hard enough. For the first semester I tried my hardest to get the best grade, even though that trend did drop because of events / trips. I have never once done that in a math class! I feel like I accomplished a lot and now I can continue to find success in math if I try hard enough, in the past years I needed help from peers and teachers to truly understand the content, in this class I was able to challenge myself and do things on my own, one of those tasks for example was our exhibition project. I started the project all by myself and even though I had help from my fellow classmate Porter, I was able to do a project that couldve required many people, I stepped up and I never felt that i could have that confidence. This class has given me confidence in my voice and learning ability.