Maths Project List:
-Probability Project Overview -Renaissance Era Game: One in Thirty Overview -Probability Analysis -Reflection -Measuring your World: DP Update Part 1 ~~Project Overview~~ -Measuring your World: DP Update Part 2 ~~Volume and Area~~ -Quadratics Overview & Postcard Reflection |
Probability Project Overview:
Probability is the measurement to which an event is likely to occur.
There is two ways to measure the probability of a event occurring, one is a empirical probability and the other is a theoretical probability.
Empirical probability is observing an event and the number of times that event occurred, divided by the total number of events observed. An example would be if it rained 3 days in February, the empirical probability would be 3 / 28 because February rained 3 out of the 28 days possible. It is expected that you'll understand that an empirical probability has an outcome that is certain. The probability is the exact probability of an event happening because it is supported by recorded evidence.
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Theoretical probability is measuring the total number of outcomes for a desired event divided by the total amount of outcomes that are possible. An example of this would be if 10 students with jackets were picked out of 100 students, the probability would be 10/100. A theoretical probability is referring to a probability of an event that could possibly happen. It can be understood that a theoretical probability is measuring how likely an event occurs in a theoretical situation. |
Probability of Multiple Events:
The probability of multiple events is basically asking you what the chances are to either get one event happening or another event happening. To calculate the probability of multiple events, you would need to obtain the probability of at least 2 events. Let's use 2 dice for this example, let's say you roll the dices and you get a one and a 4. The probability of that would be 1 / 36. If you roll again and get a 3 and a 5, that probability would also 1 / 36. However, what would be the probability of either of those events happening is the question here. To solve this, you would add up the probabilities resulting into the answer: 2 / 36. We have learned this concept by rolling dice and recording data on what we rolled. Then we used area diagrams to determine the probability of rolling those numbers on a dice roll.
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Expected Values:
The expected value of something means the predicted value of that event. To calculate this, you multiply the sum of all possible outcomes by the probability of that event. An example of this would be like if you flipped a coin 50 times. If the probability is 1 / 2, the expected value of landing on heads would be 25 times. I actually did this experiment twice where I flipped a coin 50 times and recorded my data. On both of my experiments, I managed to get 25 heads and 25 tails which somehow ended up being the expected value of the event.
Anyways, to calculate the expected value, you would use this equation:
Number of all possible outcomes x Probability of event = the expected value.
Anyways, to calculate the expected value, you would use this equation:
Number of all possible outcomes x Probability of event = the expected value.
Two-Way Tables & Tree Diagrams:
A two-way table is a table that organizes two events and categorizes the probabilities of both variables. This is used to figure out the probability of either multiple possibilities happening, either on of the possibilities happening, or one of the probabilities happening given that a certain event occurred. An example could be that a rock paper scissors game is taking place and one player chooses rock. Then, they won. The probability of that player choosing rock and winning would be a chance of 5 / 20, according to this data. Another example could be if a player picked scissors and lost, the probability of a player losing given that they picked scissors is 8 / 20. Something to learn from this is to look at the variables in question and review the circumstances to formulate the probability of the events occurring.
A tree diagram is used to show off the possibly of two events occurring right after each other. For example, let's say that we have 30 rock paper scissors players. 1 / 3 of them choose scissors and 20 of them chose something else. 12 people total won and 18 people total lost. What is the chance of a player choosing scissors and winning? To solve this, you would have to consider how many players picked scissors. If you multiplied the # of players by the probability of choosing scissors, you would get 10 players picking scissors. Then to find out which of the 10 players have won their game, you would then multiple the probability of winning the game, 2/5. If you multiply 10 by 2/5 you get 4 which 4 / 30 is the probability of choosing scissors and winning. The important thing to learn from this is to keep track of each of the probabilities and then choosing the branch that'll give you the answer to the question. |
Conditional Probability:
Conditional Probability is a probability that is conditional basically. Conditional means there's a condition or a requirement to that probability. That requirement being that the probability is considering an event that already occurred. The probability is written out like this:
Pr=[A l B] ---> the probability of event A given that event B occurred.
For an example, we could say that in a Gamestop store, there is 2 trucks delivering video games to refill their stock. One truck would give out 32 Nintendo Switch games and 20 play station games. The other truck would give out 34 Nintendo Switch games and 25 play station games. How many play station games would Gamestop get from the first truck?
To solve this, you need to consider that the condition is that the games are coming out of the first truck and the question is asking for play station games. So, the equation to use is: Pr(probability)[play station games l the first truck drops off the games = the # of play station games from the first truck. With that knowledge in mind, you can conclude that the answer is: Pr[A l B ] = 20 play station games / 52 possible games.
What to take out of this is that Conditional probabilities always look at certain variables to determine the probability. Always look at the wording of the problem to figure out what to solve for.
What to take out of this is that Conditional probabilities always look at certain variables to determine the probability. Always look at the wording of the problem to figure out what to solve for.
Joint Probability:
A joint probability is a measurement of calculating a probability consisting of an event happening at the same time as another event occurs and the likelihood of both possibilities coming together. An example of this is: if you had a spinner with 6 sections labeled 1-6 on each section, all at equal length. If you spun it twice, what is the probability of the spinner landing on a 2 for the first round and a 5 for the second round? To solve this, you consider that there is only 6 possible sections and they are equally sized so the probability of landing on one of them is 1 / 6. Though it spun twice, so we know that spinning it once to get 2 is 1 / 6, if we spin it again with the same odds, the probability would still be 1 / 6. So to find the probability of getting both a 2 and a 5 by spinning the spinner twice, you would have to multiply the probability of 1/6 to 1/6, which results in the probability of 1 / 36. A good way of thinking about this is to remember that the probability of 2 independent events happening together, you will need to multiply them together. That is just how joint probability works, it is jointed together.
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Marginal Probability:
A Marginal probability is a probability that uses an event and a total from the margins to come up with a probability. For example, if the question asked you, how many people participated in the lottery? The way to solve the problem is to look at the margin of those who are participating which is 81 and there is 100 people recorded total so the probability would be 81 / 100. Another problem could be: how many people got a significant increase in cash the week of the lottery? There is 15 people total recorded with having a increased pay in the week out of 100 people documented so the probability is 15 / 100. (The other 14 who didn't win the lottery must've got a raised at their job or something.) The thing to learn here is that Marginal probability uses the marginals to make a probability using the total margin. (In this case, the 100).
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Renaissance Era Game ~ One in Thirty Overview:
One in thirty is a card game that dates back to at least 1440 AD and it was popular in Ireland and Spain. During that year, this game was mentioned by a priest named Bernardino of Siena at a anti-gaming lecture which most likely was a lecture against gambling. The game stuck around during the 15th to the 17th centuries before it disappeared later in the renaissance era and wouldn't get mentioned again until when it appeared in french casinos around the 1700's under the name "Vingt-et-un" which translates to "twenty one". One in thirty is the ancestor to Blackjack & Twenty one and after One in thirty's fall, Twenty one & Blackjack picked up the pieces and continued it's legacy. The games are pretty similar to each other except they changed One in thirty's 3 cards to 2 and the goal is to get 21 instead of 31.
Now to how it is played:
The game, One in Thirty can be played with 2 or more people. The dealer of the cards always starts with the person on their left. The scoring is: 10 for Jacks, Kings, and Queens. 11 for Aces and the rest of the cards are face value. For the points to add up, the cards must be in the same suit. For example, if you have two heart cards and one spade card. The two heart cards will add up their score while the spade card won't do anything.
That's the basic rules, here is how to play the game:
The game starts when the person on the left gets three cards, you can either choose to pick a card to replace with a new card on top of the deck or grab a card from the previous card to put onto the discard pile. When you put a card in the discard pile, it must be face up and show the card. The game continues with people picking a card to replace a card until a person is confident enough to “knock” which when that happens, everyone reveals the cards. Whoever gets closest to 31 wins.
Now to how it is played:
The game, One in Thirty can be played with 2 or more people. The dealer of the cards always starts with the person on their left. The scoring is: 10 for Jacks, Kings, and Queens. 11 for Aces and the rest of the cards are face value. For the points to add up, the cards must be in the same suit. For example, if you have two heart cards and one spade card. The two heart cards will add up their score while the spade card won't do anything.
That's the basic rules, here is how to play the game:
The game starts when the person on the left gets three cards, you can either choose to pick a card to replace with a new card on top of the deck or grab a card from the previous card to put onto the discard pile. When you put a card in the discard pile, it must be face up and show the card. The game continues with people picking a card to replace a card until a person is confident enough to “knock” which when that happens, everyone reveals the cards. Whoever gets closest to 31 wins.
We did adapt the game a bit. We added one rule that forced a player to take one card from the deck before they can knock. This prevented players from knocking immediately at the start of the game and it can create a longer and more balanced game.
I chose this game to exhibit because I am fond of card game strategy games. It's fun to puzzle over if you should discard a card in hope to get a better score and to figure out which cards to discard so you don't accidentally give your opponents a card they may need. There is a lot of components to it that may not be noticeable at first but it really makes the game a lot more fun.
Below are some of the pictures that were taken of the game, please enjoy!
Below are some of the pictures that were taken of the game, please enjoy!
One / four replicas of the Renaissance cards that we made for exhibition. If I am not mistaken, this is the jack card from the Renaissance Era.
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Here is the table with the game being played out. Each player has 3 cards handed out to them, same with how the picture is showing them here. The discard pile has the cards face up to be seen by our players. This would be the ideal set up for this game. Next to the cards, we would put our replicas we made of the Renaissance cards in the middle of the table, left or right of the playing cards. (Which you can see two of them here, the other two are currently missing.)
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One / four replicas of the Renaissance cards that we made for exhibition. Despite it not looking like so, this is the ace of cards. This is the ace from the Renaissance Era.
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Probability Analysis:
Out of the three cards you draw in the beginning of One in Thirty, what is the probability of getting at least one of either the jack, the queen, the king, or the ace?
The first thing I did with the problem was to add up all the cards that are either a jack, queen, king, or ace. I realized that there is 4 of each card in a standard deck so I knew that I had to add 4 for all of them. I ended up with 16 and I subtracted that by 52 which is the number of cards in a standard deck and I got 36. I understood that drawing a card and for it to not be a jack, queen, king or ace, would be a 36 / 52 probability. Though at first, I though I could multiply 36 / 52 by itself 2 times to get the probability of never getting a jack, queen, king, or ace. Afterwards, I realized that wasn't the case and that every time you draw a card, that is one less card in the deck. Then I started to map out a probability tree to figure out what I needed to multiply to get a probability of getting none of the jacks, queens, kings, or aces. You could say that I used the habit of mathematician of "Describe & Articulate" to draw a diagram of the situation and to help me visualize. Anyways, what I found out was that each time a card is given to a player, that is one card removed from the deck. So I kept subtracting until I finally found the equation I was looking to use. After all that time of "Looking for Patterns", I figured out that I should multiply 36 / 52 by 36 / 51 by 36 / 50 to get the probability of getting 0 of those cards. Pr[0] = 36/52 x 36/51 x 36/50 = 1944/5525 which equals 0.35. Now that we have the probability of having none of them, the only way of figuring out the probability of getting at least one of them is only one step away. You subtract 0.35 by 1 and you should get 0.64, which is a 64% chance of getting at least one of either a jack, queen, king, or ace.
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Reflection:
This project has been pretty fun. Despite me having learned some probability from Dr. Cate's class, a lot of the information I learned in class was fresh and new. It was also refreshing to relearn some of the concepts we learned last year in a new way. There were some challenges on the way such as expected value and some probability work being difficult to understand. I remember that I was confused with having fractions in a probability tree. Numbers would make sense to put on a probability tree because you are pointing out how many people are in that event. Though having fractions in the probability tree threw me off because we just did it with numbers in the previous worksheet and I didn't understand the change at the time. Though now I understand and it seems obvious to me. After that, I think I was able to keep up with the concepts of probability. Things kept clicking in my head on what the problems were asking me and I almost did things flawlessly with a couple hiccups here and there. I feel confident that I understand these concepts now and I am ready to move onto the next unit. Though for the next unit, I hope to be better at answering things quicker. I am still overthinking things and I need to start simple. Or I could start small. Either way, I hope to be better at organizing my time and to think simpler. I feel that is my next path onto becoming a better mathematician.
Measuring your World: ~Project Overview~
This project started when we were doing problems on the Pythagorean's theorem. We proved the Pythagorean Theorem to be true by using a rug diagram that consists of two squares and 4 triangles. If you observe closely, the diagram follows the Pythagorean theorem formula which proves the Pythagorean theorem to be true. After proving the Pythagorean Theorem, we use the formula of the Pythagorean Theorem, "a two, plus b two, equals c 2" into the Distance Formula which secretly is the same thing but phrasing it so that it can be used for coordinate planes. The distance formula is d = √(x2-x1)2 + (y2-y1)2
After using the distance formula, I was able to use the information that I learned then to find the equation of a circle. The middle of the circle would obviously be coordinates 0, 0 because that is the center of the graph which will be the center of our circle. The point that marks the outside crust of the circle will be the point of (X, Y) which if you drop an perpendicular line, you can make a right triangle. The bottom of the triangle (the base) will be x because it is on the x coordinate plane and the opposite side of the hypotenuse is the y coordinate which is on the y coordinate plane. This accounts for two sides of the triangle but the hypotenuse of that triangle should be equal to the radius of the circle so let's call the hypotenuse, the radius! Now that is figured out, now you can times everything by itself because this triangle only accounts for half of the circle and if you want to have an equation of the whole circle, you would need to square it. The x and y added together will equal the radius so if you have x2 + y2, it should equal r2 which is the equation of a circle centered at the origin. So now we arrive to the conclusion that the equation of a circle centered a the origin, (0, 0) is x2 + y2 = r2.
We have also defined the unit circle of having a radius of 1 and having all the necessary angles / coordinate points already mapped out for you. It's a good reference point to go back to for Trigonometry. Using the Unit Circle, you can easily find the coordinates points at 30, 60, and 90 degrees and you can find the coordinate points for any other degree with the unit circle as well. If you know the degrees of a triangle, you can use cosine and sine to find the coordinate points. To find the remaining points on the circle, you can use symmetry to find out what the angles are on each of the triangles in the unit circle. For example, the top side of the circle has 45 degrees on one top side and 135 degrees on the top side and the distance between 45 degrees and 135 degrees is 90 degrees. So the circle goes by, 45 degrees, 90 degrees, then back to 45 degrees, then 45 degrees more, then it goes 90 more degrees, and finally goes 45 degrees more. Add that all up and you should get 360 degrees total which is a circle. I can prove this by starting at 0 degrees and adding my way up to arrive at the destinated degrees which match up with 45 degrees and 135 degrees. 45 + 90 = 135 degrees. Then add 45 to that to get 180, which if you add 45 degrees to that you get the mystery degree that you were looking for. You can also find the other one by following this pattern and you should find that the mystery degrees are 225 and 315 degrees.
After we used the knowledge given to us by using the unit circle, I can use angles as the variable, θ. Which this symbol is also known as "Theta". This can be extremely helpful in math because it allows us to use this angle as a "variable" for a trigonometry function such as cosine, sine, and tangent. This is a handy tool because it can now only allow you find the angle and lengths from one angle but you can use this to find other angles as well. For example, if you know the Cos(θ), to find the sin(θ), you would subtract 90 by theta because those are the other angles around you. With this you can finally find all the angles in a triangle pretty easily.
Now onto the Tangent function, Tangent is basically a single point. Tangent can be used as a coordinate point and it can allow you to find the distance of the radio line. We can use the knowledge already given to us to figure out how to use the tangent(θ). The cos(θ) is x and the sin(θ) is y, so if that is the case, tan(θ) must be y over x. (y / x) Which if you think logically, you can conclude that the Tan(θ) is equal to the Sin(θ) divided by the Cos(θ).
Using the knowledge of similar triangles and hypotenuse, opposite, and adjacent sides, I can figure out the trigonometric functions. The formula of Cos(θ) is Adjacent over Hypotenuse, the formula of Sin(θ) is Opposite over Hypotenuse, and Tan(θ) is of course, opposite over adjacent. Now if you know this, to find a side length that is unknown, you just need to use the correct trigonometric function formula and multiply it by the side length you do know and then you do the math. Presto, you have your side length. After that, we used the knowledge about the trigonometric formulas and applied it the inverse Cosine, Sine, and Tangent functions. It follows the same rules of the original Cosine, Sine, and Tangent functions but now instead of finding the side lengths, we are finding the remaining angles of the triangles. To find the remaining thetas in the triangles, all you need to use is the arcCosine, arcSine, and arcTangent functions and you should be able to find your answer there!
The Mount Everrest problem was a problem that has one side marked as 170 feet and the other side lengths were unknown. We knew two angles, an angles of 72 and 49. However we didn't know the other angle yet, but that was an easy fix because all triangles' angles add up to 180. So all was needed to do was to add 72 by 49, and subtract that by 180 to find the angles of the other side, which was 59. Next is what makes this problem interesting, we drop a perpendicular line down in the triangle. Now you need to find the height of the perpendicular line which uses the known side length multiplied by the Cosine of 41. Which gives you a height of 128.30. Now to find the remaining sides, all you would need is to use the Trigonometric formulas multiplied by the height to get your side lengths. Which unknowingly at the time, brought us to the Law of Sines. The Law of Sines is a law that takes apart the problem and uses the side lengths and the sine of those lengths divided by each other. For example, Sin C / c = Sin B / b = Sin A / a.
And finally we arrive to the Law of Cosine which uses the distance formula to get the Formula of Cosines. Unlike the Law of Sines which uses the lengths of a triangle with Sine, the Law of Cosine relates to the lengths of the sides of a triangle to the cosine of it's angle. So the distance formula for the Law of Cosines is (a-b cos(θ))2 + (-b sin(θ))2 = c2, which if we simplify the formula, we get c2 = a2 + b2 - 2ab Cos θ! Which is the formula used for the Law of Cosines!
After using the distance formula, I was able to use the information that I learned then to find the equation of a circle. The middle of the circle would obviously be coordinates 0, 0 because that is the center of the graph which will be the center of our circle. The point that marks the outside crust of the circle will be the point of (X, Y) which if you drop an perpendicular line, you can make a right triangle. The bottom of the triangle (the base) will be x because it is on the x coordinate plane and the opposite side of the hypotenuse is the y coordinate which is on the y coordinate plane. This accounts for two sides of the triangle but the hypotenuse of that triangle should be equal to the radius of the circle so let's call the hypotenuse, the radius! Now that is figured out, now you can times everything by itself because this triangle only accounts for half of the circle and if you want to have an equation of the whole circle, you would need to square it. The x and y added together will equal the radius so if you have x2 + y2, it should equal r2 which is the equation of a circle centered at the origin. So now we arrive to the conclusion that the equation of a circle centered a the origin, (0, 0) is x2 + y2 = r2.
We have also defined the unit circle of having a radius of 1 and having all the necessary angles / coordinate points already mapped out for you. It's a good reference point to go back to for Trigonometry. Using the Unit Circle, you can easily find the coordinates points at 30, 60, and 90 degrees and you can find the coordinate points for any other degree with the unit circle as well. If you know the degrees of a triangle, you can use cosine and sine to find the coordinate points. To find the remaining points on the circle, you can use symmetry to find out what the angles are on each of the triangles in the unit circle. For example, the top side of the circle has 45 degrees on one top side and 135 degrees on the top side and the distance between 45 degrees and 135 degrees is 90 degrees. So the circle goes by, 45 degrees, 90 degrees, then back to 45 degrees, then 45 degrees more, then it goes 90 more degrees, and finally goes 45 degrees more. Add that all up and you should get 360 degrees total which is a circle. I can prove this by starting at 0 degrees and adding my way up to arrive at the destinated degrees which match up with 45 degrees and 135 degrees. 45 + 90 = 135 degrees. Then add 45 to that to get 180, which if you add 45 degrees to that you get the mystery degree that you were looking for. You can also find the other one by following this pattern and you should find that the mystery degrees are 225 and 315 degrees.
After we used the knowledge given to us by using the unit circle, I can use angles as the variable, θ. Which this symbol is also known as "Theta". This can be extremely helpful in math because it allows us to use this angle as a "variable" for a trigonometry function such as cosine, sine, and tangent. This is a handy tool because it can now only allow you find the angle and lengths from one angle but you can use this to find other angles as well. For example, if you know the Cos(θ), to find the sin(θ), you would subtract 90 by theta because those are the other angles around you. With this you can finally find all the angles in a triangle pretty easily.
Now onto the Tangent function, Tangent is basically a single point. Tangent can be used as a coordinate point and it can allow you to find the distance of the radio line. We can use the knowledge already given to us to figure out how to use the tangent(θ). The cos(θ) is x and the sin(θ) is y, so if that is the case, tan(θ) must be y over x. (y / x) Which if you think logically, you can conclude that the Tan(θ) is equal to the Sin(θ) divided by the Cos(θ).
Using the knowledge of similar triangles and hypotenuse, opposite, and adjacent sides, I can figure out the trigonometric functions. The formula of Cos(θ) is Adjacent over Hypotenuse, the formula of Sin(θ) is Opposite over Hypotenuse, and Tan(θ) is of course, opposite over adjacent. Now if you know this, to find a side length that is unknown, you just need to use the correct trigonometric function formula and multiply it by the side length you do know and then you do the math. Presto, you have your side length. After that, we used the knowledge about the trigonometric formulas and applied it the inverse Cosine, Sine, and Tangent functions. It follows the same rules of the original Cosine, Sine, and Tangent functions but now instead of finding the side lengths, we are finding the remaining angles of the triangles. To find the remaining thetas in the triangles, all you need to use is the arcCosine, arcSine, and arcTangent functions and you should be able to find your answer there!
The Mount Everrest problem was a problem that has one side marked as 170 feet and the other side lengths were unknown. We knew two angles, an angles of 72 and 49. However we didn't know the other angle yet, but that was an easy fix because all triangles' angles add up to 180. So all was needed to do was to add 72 by 49, and subtract that by 180 to find the angles of the other side, which was 59. Next is what makes this problem interesting, we drop a perpendicular line down in the triangle. Now you need to find the height of the perpendicular line which uses the known side length multiplied by the Cosine of 41. Which gives you a height of 128.30. Now to find the remaining sides, all you would need is to use the Trigonometric formulas multiplied by the height to get your side lengths. Which unknowingly at the time, brought us to the Law of Sines. The Law of Sines is a law that takes apart the problem and uses the side lengths and the sine of those lengths divided by each other. For example, Sin C / c = Sin B / b = Sin A / a.
And finally we arrive to the Law of Cosine which uses the distance formula to get the Formula of Cosines. Unlike the Law of Sines which uses the lengths of a triangle with Sine, the Law of Cosine relates to the lengths of the sides of a triangle to the cosine of it's angle. So the distance formula for the Law of Cosines is (a-b cos(θ))2 + (-b sin(θ))2 = c2, which if we simplify the formula, we get c2 = a2 + b2 - 2ab Cos θ! Which is the formula used for the Law of Cosines!
Measuring your World: ~~Volume and Area~~
For my volume and area presentation, I decided to use the Holland IV Submarine to measure because I thought that it was a requirement to include your World's Fair project into this math presentation. Afterwards I realized it wasn't a requirement for whatever reason but I had already started so I continued to set out what I was going to do. My plan was to attempt to find the volume of the submarine and to find the area of the circler part of the submarine. I was able to find the measurements of the submarine online and I went to work. To find the volume of the submarine, I imagined the submarine as a cylinder. So I decided to follow this formula to get the volume:
However, I couldn't find the overall height of the submarine any where in my research and instead I got a measurement of the draft of the submarine. I couldn't figure out how I was suppose to put this in my equation so for a while, I was at a loss. Though, after a while of thinking, I decided to look at the problem from a different perspective.
I turned the problem around and I realized that there is still hope to figuring out the area of that submarine. I was given the length, width, and the diameter of the problem. The diameter is 10ft 3inches, the width is 10ft 4 inches, and the length is 53 feet and 10 inches. Around this time, I decided to skip ahead to work on the area of the circle of that cylinder to give my brain a break but that is what allowed me to figure out the problem. I used the Habit of Mathematician, Start Small, and I found the radius by finding the area of the circler part of the submarine. I realized during that problem that the radius is the diameter divided by 2 so I divided 10 feet by 2 to get 5 feet and I divided 3 inches by 2 to get 1 and 1/2 inches. So the radius was 5 feet and 1 1/2 inch, when applied to the equation of an area of a circle (pi * 5.125^2), you get an area of 82.52 ft. Doing this problem first helped me get the radius for my cylinder problem because the measurements are the same which helped me put things into perspective.
If I rotate the cylinder 90 degrees, the height turns into the length! Which I have the length of the submarine and I already got the radius of the submarine, all I had to do then was just plug it into the equation. So I calculated pi * 5.125^2 * 53 feet and 10 inches (10 / 12 = 0.83 so 53 + 0.83 = 53.83) which the volume equaled to 4,373.34.
So the volume of a submarine is 4,373.34!
Though that wasn't the whole project. We also had to do a presentation of a trigonometry concept which I decided to have a fun story to it relating with facts about the submarine. "The Holland Vi has a range of 1,000 miles when powered by the gasoline engine." So I decided to make a problem about the fact. It was based off a problem we did in class called the "Mount Everrest Problem" which had a triangle with two unknown sides and only two known angles. The problem seems impossible at a first glance but there are some ways of solving it using Trigonometry! I first dropped a perpendicular line to separate the one triangle into two. Then I sought out the height which is the only length known times the split angle from where the perpendicular line came from and then that's your height. After that, you can use the sin function to times the remaining lengths to find the others lengths until you can figure out all the lengths of that triangle. The purpose of the problem was to find the missing sides of the triangle with only one side and two angles which is completely possible to do and I demonstrated it in my presentation.
Despite the difficulties I faced with figuring out how to use my measurements for the problem, I think I did pretty well with coming up with problems and completing them. These problems made me use my brain a lot and I had to conjecture and test solutions that might work. In the end, I think Start Small and thinking from a different perspective was the two skills that helped me the most in making this presentation. Without them, I would've definitely haven't got the answer I was looking for and I might've had to choose another problem to work on. I am glad that I was able to do this problem though because I realized that the height is not always the height and mathematical equations might not always be as easy as it seems. Thinking outside the box can be a helpful skill and it will prove to be useful in future math problems. If I had to do something differently, I wouldn't have searched so hard for the height because if I hadn't, I would've got the presentation done much quicker. Though, it was a learning experience and I am glad I went through it.
If I rotate the cylinder 90 degrees, the height turns into the length! Which I have the length of the submarine and I already got the radius of the submarine, all I had to do then was just plug it into the equation. So I calculated pi * 5.125^2 * 53 feet and 10 inches (10 / 12 = 0.83 so 53 + 0.83 = 53.83) which the volume equaled to 4,373.34.
So the volume of a submarine is 4,373.34!
Though that wasn't the whole project. We also had to do a presentation of a trigonometry concept which I decided to have a fun story to it relating with facts about the submarine. "The Holland Vi has a range of 1,000 miles when powered by the gasoline engine." So I decided to make a problem about the fact. It was based off a problem we did in class called the "Mount Everrest Problem" which had a triangle with two unknown sides and only two known angles. The problem seems impossible at a first glance but there are some ways of solving it using Trigonometry! I first dropped a perpendicular line to separate the one triangle into two. Then I sought out the height which is the only length known times the split angle from where the perpendicular line came from and then that's your height. After that, you can use the sin function to times the remaining lengths to find the others lengths until you can figure out all the lengths of that triangle. The purpose of the problem was to find the missing sides of the triangle with only one side and two angles which is completely possible to do and I demonstrated it in my presentation.
Despite the difficulties I faced with figuring out how to use my measurements for the problem, I think I did pretty well with coming up with problems and completing them. These problems made me use my brain a lot and I had to conjecture and test solutions that might work. In the end, I think Start Small and thinking from a different perspective was the two skills that helped me the most in making this presentation. Without them, I would've definitely haven't got the answer I was looking for and I might've had to choose another problem to work on. I am glad that I was able to do this problem though because I realized that the height is not always the height and mathematical equations might not always be as easy as it seems. Thinking outside the box can be a helpful skill and it will prove to be useful in future math problems. If I had to do something differently, I wouldn't have searched so hard for the height because if I hadn't, I would've got the presentation done much quicker. Though, it was a learning experience and I am glad I went through it.
~Quadratics Overview & Postcard Reflection~
For the Quadratics unit, we observed many concepts to help us understand the process of how to use Quadratic equations.
I'll review them really quick one by one so you can see what we learned this year.
I'll review them really quick one by one so you can see what we learned this year.
Kinematics:
Vertex Form & Standard Form:
Next we learned about Vertex form and how to convert that into Standard Form and how to convert it back to Vertex Form. Both forms can be useful for Quadratic equations and it can help you find certain things you need to find! Like the x intercepts and the vertex of the parabola. These can be extremely helpful for when you need to sketch a graph of a parabola from scratch.
Distributing the Area:
For distributing the area, we make a square and then we add two rectangles on the right and bottom sides. After that, you add a small square on the remaining section to make a huge square. Then you put your x values there and then plus or minus whatever is the normal numbers. For example, (x+3)(x+2). You make a square where you multiply the xs' together to get x^2. Then you would get 3x and 2 x on the two rectangles which gives you 5x! Then all that's left is the small square which you multiply 3 by 2 to get 6! After that is done, you combine like terms to get: x^2 + 5x + 6!
Maxima & Minima:
We learned that when the vertex is at the highest point, that is called the Maxima and if the vertex is at the lowest point, it is called the Minimum. This can determine if the parabola is facing down or upwards.
Factored Form:
Same with Standard Form and Vertex form, we learned how to convert the forms and how to use them in the equations that we need the forms for. This was the final form we had to learn which was a bit hard at first but we got the hang of it in the end.
For this project, we had to make a project about a parabola problem and we had to make the problem ourselves to prove that we understand the Quadratic concepts that we learned from this unit. I did mine on Mario's jump and the information is above!
I had a lot of difficulty with the concepts at first but the test / final we did helped me understand the concepts more and it forced me to study. I felt more confident in myself when I was going into the project and I think I did a lot better than what I would've done beforehand. What I might've done differently next time is to make sure that I absolutely understand the concepts before moving onto the next project. I kinda did worksheets to finish them but I only understood somewhat about the concept and I wish I was able to understand it better when we first learned them. Though, overall I think the unit worked well and I learned a lot from this unit! Everything worked out okay in the end and I am content with how everything went! Thanks for everything and I hope you have a wonderful day / summer / year!