DEV PROJECT

QUESTION 1: Archery

2008-01-12T19:36:00.000-08:00

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Sebastian and his friends decided to start of their super fantastic day by going out and playing archery together. His friends and him were feeling like the game should be more interesting. Who ever got to split his or her arrow first that's trying to send to the bulls eye will not have to pay for their dinner. What are the odds of Sebastian splintering his arrow that he is trying to send into the bullseye if he has arrows with 1 cm diameter shafts if the target is 30 cm in diameter, a bullseye of 3 cm in diameter and he cannot miss the target. (Breaking the lines of a target counts as a hit and moves to a higher point value).

SOLUTION 1: Archery

2008-01-12T16:01:00.000-08:00

- Sebastian has to get the bullseye first. Find the probability of him hitting the target itself which is favorable outcomes over all possible outcomes.

- because we're dealing with circles, we'll be dealing with the area and where we want to hit and where all the possible places are to hit.

- Keep in mind that the favorable outcome is the bullseye but the person doesn't have to hit the bullseye with the whole arrow, only a portion of it has to hit. So, Sebastian's farvorable area to hit would be everything within the blue circle because the farthest the arrow could be would be just hitting the outer cicumference of the bullseye and that would count.

- a closer view of Sebastian's favorable area. The favorable area would be the bullseye area plus the whole area that the arrow can land around it while sitting on it.

- By using the radii of the two circles we have an area that is favored for the arrow to hit. We can than apply the same concepts to the outer edge of the target and we can continue our calculation.

- The radius of the first one would be 2 cm because the bullseye's diametre is 3 cm, the radius of the bullseye would be 1.5 cm. The diameter f the shaft of the arrow is 1 cm, the radius of the shaft must be 0.5 cm. Together the radii make the favorable arrow because that's where Sebastian can hit to get a bullseye. That's divided by all possible outcomes which is π(15.5)² because the diameter of the whole target is 30 cm, which must mean that the radius is 15.5 cm. The pi's reduce.

- Sebastian has to nick his own arrow and this uses the same logic. The radius of the arrow that his trying to hit is .05 cm and his own arrow is 0.5 cm, which gives us a radii of 1 cm together. Sebastian's favorable outcome is π(1)². All possible outcome is still the same because he can hit anywhere else on the target which is π(15.5)².

- Then to find the probability of this happening one arrow after the other, multiply the probabilities together.

- Sebastian's probability of splintering his own arrow that he's trying to send into the bullseye is:

- Which is really really small. However, that's the chance that everyone has as well. In the end no one got it so everyone decided to go out and eat.

QUESTION 2 : Dinner

2008-01-11T17:14:00.000-08:00

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After archery and showing his friends whose really pro, Sebastian wanted to go out and eat with them. He wanted to finish in 45 minutes or less so they have more time to go back to his house and play poker. On the menu there were: 4 appetizers (1 of which have bread), 6 beverages (2 of which are smoothie-like), 10 meals (6 of which have steak), and 5 desserts (3 of which are brownie-like). It usually takes Sebastian 5 minutes to finish a regular appetizer, beverage, dessert each and 25 minutes to finish a regular meal. How many dinner choices does he have when it takes an extra 2 minutes to finish an appetizer with bread, 1 minute to finish a smooth-like beverage, 5 minutes to finish a meal with steak, and an extra 2 minutes to finish a brownie-like dessert in 45 minutes or less if he gets an appetizer, beverage, main course, and dessert?

SOLUTION 2: Dinner

2008-01-11T16:33:00.000-08:00

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We'll start where it says How to Solve.
First make a table of what you know:

4 appetizers, 3 are REGULAR and take 5 minutes to finish and 1 has bread which takes 7 minutes to finish. A regular appetizer takes 5 minutes to finish but with bread it takes 2 MORE minutes to finish.

6 beverages, 4 are REGULAR and take 5 minutes to finish and 2 are smoothie like which takes 6 minutes to finish.

10 main course, 4 are REGULAR and take 25 minutes to finish and 6 have steak which takes 30 minutes to finish.

5 desserts, 2 are REGULAR and take 5 minutes to finish and 3 are brownie like which takes 7 minutes to finish.

Since Sebastian wants to have an appetizer, beverage, main course, dessert and finish in 45 minutes or less there's really only two cases. There are only two cases because:

He can have everything that requires more time to eat or "extra" time except for the main course because that would elapse his time that he wanted to finish. It only takes him 40 minutes to complete dinner if everything was "regular" and since it takes a total of 5 minutes to finish an "extra" appetizer + "extra" beverage + "extra" dessert that works out perfectly to 45 minutes.

He can also have everything remain regular and only change his main course because if he decides to have an "extra" as a main course it would take another 5 minutes to complete his dinner which works out to 45 minutes.

CASE 1

In the first case we would have four slots for four different dishes because each of those dishes have a number of choices to be chosen. There are 4 different dishes to choose for appetizers because he could either choose a regular appetizer or one that requires more time to finish. The second slot would be 6 because he could either choose a regular beverage or one that requires more time to finish. The third slot would be 4 because it HAS to be a regular dish. The reason why it has to be a regular dish is because if he choose the first one to be an extra and the second to be an extra he cannot have a main dish that requires more time or he'll go over 45 minutes. The fourth slot has to be 5 because he can either choose a regular or extra.

That number also includes if he chooses everything to be regular and if he would just have two things such as an appetizer and dessert to be the only ones that require a bit more time to eat.

CASE 2

In the second case the only thing that can be "extra" or can be changed is the main course because if it was regular it only takes 25 minutes and if the main course has steak it takes 5 minutes more to finish which then would work out to 45 minutes. So what we would have is all regular items and the main course would change.

He would only have 3 choices for an appetizer because they are regular and don't require any more time to finish. 4 choices for a beverage. For the main course 6 choices because they all have steak in them which would take 5 minutes more to complete. It's not 10 because in the previous case we already have main courses without steak with the rest being regular, so by making the main course 10 different choices, we would have dinners that were the same. 2 choices for dessert because there are 2 regular desserts.

To find the total number of dinner choices Sebastian can have we just simply add cases 1 and 2 and there's your answer.

QUESTION 3: Poker

2008-01-10T21:08:00.000-08:00

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It was a show down between Sebastian and his friend in the game of Texas Hold 'Em. They were the only two left playing and this last game was going to determine the winner. Since Sebastian went in blind he wanted to see if his chances are good and if he should go all in or not. His friend was nice enough to let him see his cards first and even the burn cards. What are the chances of Sebastian winning if whats on the table is a 7 and 5 of hearts, a king and queen of diamonds, and a jack of spade; the burn cards are a 3 of spade, 6 of club, and a 7 of club; his friends hand is a queen of hearts and a 2 of diamonds; and aces are high and 2's are wild cards?

Not sure how to play? click here to learn how to play

SOLUTION 3: Poker

2008-01-10T14:46:00.000-08:00

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Since Sebastian knows what his opponent has and what cards are used, he knows that the only thing that would beat his friends hand is a straight or higher three of a kind. Theres no flush potential so everything else higher than a flush is eliminated. What Sebastian needs is a:

king and a king, to make a higher three of a kind. There are three kings left and two of them can be his own cards. He doesn't know because he went in blind but it's possible but he has pocket kings.
Two and a Two, to make a three of a kind kings, because two's are wild and can be any card. The highest face on the table is a king and the pocket two's can make a three of a kind.
King and a Two, two's are wild and with the kings it can be a higher three of a kind.
Ace and a Ten, that would make a straight because it'll be Ace, King, Queen, Jack, and Ten.
Ace and a Two, the two can have a face value of a ten.
Two and a Ten, the two can have a face of an Ace since it's a wild card.
Ten and a Nine, that also makes a straight, king, queen, jack, ten, and nine.
Ten and a Two, because the two can have a face value of a nine.

To get a king and a king or a Two and a Two:
Sebastian already knows that a king is used on the table so there are three kings left. Out of those 3 kings he only needs 2 and it doesn't matter the order in which he gets the kings because the outcomes still the same. So its only 3C2, because out of the three kings left he only needs two of them and suit doesn't matter. This also applies to pair of two's, same concept.

3C2 + 3C2 = 6

To get a King and a Two:
There are three kings left and he only needs one of them and since order doesn't matter it'll be 3C1. One of the Two's are used in the opponents hand so there are 3 Two's left and out of those 3 Two's he needs one of them which is also 3C1. Then they have to be multiplied.

(3C1)( 3C1)= 9

To get a Ace and a Ten:
There are four Aces left because they are not used and he only needs one of them so 4C1 and there are four Tens that are not used either and out of those four Tens he needs one of them so 4C1. This also applies to Nine and a Ten because the Nines are not used either so out of the four Nines he needs one.

(4C1)(4C1)(2)= 32

you multiply 16 by two because it's the same thing that happens to getting a Nine and a Ten.

To get an Ace and a Two:
There are four Aces left because they are not used and he only needs one of them so 4C1 out of the four Aces he needs one. For the Two there are three left because one of the Two's is in the opponents hand, so out of the 3 he only needs one, so 3C1. This also applies to getting a Ten and a Two, and a Nine and a Two.

(4C1)(3C1)(3)= 36

To find the favorable outcomes:
Add the numbers up because these are all the hands that he wants to get and that's what favorable outcomes are, the outcomes you want.

3C2 + 3C2 = 6

(3C1)( 3C1)= 9

(4C1)(4C1)(2)= 32

(4C1)(3C1)(3)= 36

total of 83

To find All possible outcomes:
Since ten cards were used (5 for the table, 3 for the burn cards, and 2 are the opponents cards) there are only 42 cards that remain. Out of those 42 cards he'll need two for his own hand, so it would be 42C2, which is 861.

To find the probability it would be favorable outcomes over all possible outcomes. And whatever that number is, in this case .0964, is the probability.

QUESTION 4: Homework

2008-01-09T02:00:00.000-08:00

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After a nice day of just hanging out with friends, Sebastian had homework to finish. Homework was probably what he dreaded the whole day. He had done most of it before the weekend but was stuck on two questions. One of them involved logarithms and trigonometric identities.

Sebastian had to prove that, whatever that thing is. It's like a logarithmetric identity.

SOLUTION 4: Homework

2008-01-09T01:03:00.000-08:00

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Pick a side to work on and then break things down and work on it one at a time. Remember to not bring anything over the equal sign because when you're proving something they should work out the same if they are equal.

Break things down so it's easier and work on simplifying the broken parts of the logarithmetric (merge of the words logarithm and trigonometric) identity.

Break the logarithmetric expression up to make things easier.
Now you have three different trigonometric identities on the left hand side.
Work out the trigonometric identities one by one.
The first expression in the denominator it has csc²θ - cot²θ - sin²θ. If you notice csc²θ - cot²θ is a trigonometric identity which equals to one.
csc²θ - cot²θ = 1
Now in the denominator we have 1-sin²θ. That is also a trigonometric identity:
1-sin²θ = cos²θ
To simplify things we have sin²θ/ cos²θ which is also the same thing as tan²θ.

Moving on to the next trigonometric identity to solve:

In the numerator we have sec²θ - tan²θ - cos ²θ
The trigonometric sec²θ - tan²θ = 1
So now, what we have in the numerator is 1 - cos²θ
That is also a trigonometric identity is 1 - cos²θ = sin²θ
Now we are left with sin²θ/cos²θ and that is the same thing as tan²θ

Now we can re-write the expression with the new simplified trig identity. Now, it is the log of base tan²θ to the power of tan²θ to the exponent of a trigonometric identity. Since the base is the same as the base of argument, that must mean that the whole log is equal to the exponent of argument.

Now pick one side to work on and try to transform that side into looking identical to the other side. The easiest side would be the left hand side because if you look carefully there's a lot more you can do to that side.

In the numerator we have csc²θ + cos(2θ). The trigonometric identity of cos(2θ) is equal to cos²θ - sin²θ.
Now what we have in the numerator is csc²θ + cos²θ - sin²θ.
In the denominator we have cot²θ + sin³θ and cot²θ is the exact same as cos²θ/sin²θ
Now to add the denominator both fractions have to have the same denominators. We multiply the bottom and top by sin²θ. Now what we have in the denominator is cos²θ-sin^5θ all over sin²θ.
The numerator is csc²θ + cos²θ - sin²θ. csc²θ is the same thing as 1/sin²θ.
To add ever thing together we have to get all the others denominators equal to sin²θ.
Now what we will have in the numerator is 1 + cos²θsin²θ - sin^4θ all over sin²θ.
What we have now is an improper fraction. So what we do is take the reciprocal of one fraction and multiply it to the other. The sin²θ reduce. What we have now is 1 + cos²θsin²θ - sin^4θ all over cos²θ-sin^5θ.
On the top cos²θ is the same thing as 1- sin²θ. In the numerator it'll be 1+ sin²θ(1-sin²θ) - sin^4θ.
When the numerator is expanded we have 1+ sin²θ - sin^4θ - sin^4θ, which is 1+ sin²θ - 2sin^4θ.
Q.E.D we have the exact same thing on the right hand side of the equation. The identity is now 1+ sin²θ - 2sin^4θ / cos²θ-sin^5θ.

QUESTION 5: Homework pt. 2

2008-01-08T14:11:00.000-08:00

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Moving on to the last question that Sebastian didn't finish. It dealt with transformations, he hates transformations. The question showed two different graphs and it asked to find a way by just changing two numbers in one equation to look like the other. Somewhat like a trigonometric identity but dealing with graphs. The two graphs were:

f(x) = 3 [sin3(x + 1)] + 1
g(x) = 3cos (x - π/2) + 3

Change two numbers in f(x) so that: f(x) = g(x)

SOLUTION 5: Homework pt. 2

2008-01-07T14:26:00.000-08:00

What he did first was graph both graphs together to see what they both look like.

By looking at the graph he knows that the red graph, f(x), is a lot more compressed horizontally compared to the black graph. The vertical shift is off because f(x) winds around y= 1 while g(x) winds around y= 3.

The standard form for sin and cos are:

Where A is the amplitude, that determines the stretch of the graph vertically. B is not the period but determines the period, the larger the number the graph gets compressed horizontally and when the number is between 1 and 0 the graph stretches horizontally. C is the phase shift, it shifts the graph horizontally and D is the vertical shift.

Since both graphs have the same amplitude there's no need to change an of those numbers.
When you expand f(x):

f(x) = 3 [sin3(x + 1)] + 1
f(x) = 3 [sin3x + 3] + 1
f(x) = 3 sin 3x + 3 + 1
f(x) = 3 sin 3x + 4

Parameter D has to change and be zero for the graph of f(x) because the cos and sin graphs are similiar in many ways the only thing that makes them different is by a phase shift; sin(x) = cos(x - π/2) OR sin(x + π/2) = cos(x).

Since the parameter C of g(x) is -π/2 there is no need to change the parameter of C for the graph of g(x). If we get rid of parmeter D, once you expand it'll be:

f(x)2 = 3 sin 3x + 3

He has to change parameter B in f(x) because they have to have the same periods. To make it easier set parameter B, which is 3 to y, where y is a constant.

g(x) = f(x)2
3cos(x - π/2) + 3 = 3sin[y(x + 1)]

Dividing both sides by three we get:
cos(x - π/2) + 1 = sin[(yx) + 1]
cos(x - π/2) + 1 = sinyx + 1

Moving the one over to one side of the equations, the one's cancel leaving us with:
cos(x - π/2) = sinyx

Because cos(x - π/2) = sin(x), y must be equal to 1, therefore the two numbers that need to be changed are:

3 [ sin 1(x - 1)] + 0 = 3 cos (x - π/2) + 3

REFLECTION

2007-12-29T16:36:00.000-08:00

• Why did you choose the concepts you did to create your problem set?

The units that Ivanna and I chose for our project were the units that we both had difficulty in. The reason why we chose them because it would mean that we would have to practice it enough to actually teach it or explain it to someone else who didn't know a thing about that unit. By going through the units that were difficult for us would mean that we have to understand everything to teach it, so it would help us learn the unit itself.

• How do these problems provide an overview of your best mathematical understanding of what you have learned so far?

Well, since the units we chose were the units where we struggled in probably won't show our best. However we did try our hardest to learn the units as much as possible to generate hard questions and solve them. Our best work would probably be how hard we tried to learn the units and how hard we worked on the project.

• Did you learn anything from this assignment? Was it educationally valuable to you?

I learned some things from the assignment. Not very big but some things that will help in the future. I got a better understanding for the units and it defiantly got me to remember the units. I know how to explain how i arrived at my answers better as well. I learned how to generate my questions easier. By working backwards it will work out, like choose some numbers and make the question work for the answer. When I first saw this I was amazed at how easier it made everything. I tried it myself and it works better because now by working backwards I know how to explain it. This assignment was of much value to me, it got me to think and remember my units.

REFLECTION

2007-12-23T21:40:00.000-08:00

For this project, we chose the concepts that Sam and I thought we had a firm enough grasp on to teach a regular problem, but still push our limits. We did this so we could get the most out of the project, as it is all about our learning through the teaching of others. By choosing problems in which we have great difficulty, we had to break it down to a level which we could understand, so we believe that everyone and anyone else would be able to understand it as well. By choosing problems on the zenith of our knowledge we show the extremes of what we can do while still being able to form coherent sentences explaining it afterwards. A sense of great achievement comes from pushing the brink as for ourselves we have done. "Go for Gold" by comparison, in our opinion, is much easier and we would much have rathered do that (to the uninformed it is an old math exam that you have to ace to get a mark). Even though it was so hard, that is what gives it educational value. Learning is best achieved by teaching because to teach you must know what your doing back and forth, so it therefore forces us to do the job well. One thing that we really found was that making problems is just as hard if not harder than solving them. Since the project is online it also forces you to do it well as you really don't want to seem... well... dim.