Mathematics 10+: Logarithms

Logarithms before calculators

Mathematics 10+: Graphing Periodic Functions

The value of b in the graph, y=sin(bx) affects the period of the sine curve.

The graph below represents the function y=sin(x). As you can see, the period of this graph is 2π. The period of an graph is the distance it takes for the sine or cosine curve to begin repeating again.

This function will appear as a red line on my upcoming graphs.

Lets now take a look at the graph with the function, y=sin(2x)

Compared to y=si(x) (red line), the sine curve is much more narrow. Because of the curve being narrow, the period has also become smaller to 2.

While comparing the two graphs and the periodic function  y=sin(bx),  I came up with an equation that defies an graphs period.

When an trigonometric function is formatted in a matter of y=sin(bx), its graphs period can be calculated using 2π/b.

EXPLANATION:

The equation of an periodic function is *f(x) = f(x+p) (p=period). Y=sin(x) has an period of 2π, so when it is applied under the equation, the function will look like sin(bx)=sin(bx+2π). *proved using Fourier series

sin(bx) = sin(bx+2π) can be factored to sinb(x) = sinb(x+2π/b).

sinb(x) = sinb(x+2π/b): Now that this function falls under the periodic function equation, f(x) = f(x+p), it can be used to defy the period of the equation, y=sin(bx)

sinb(x) = sinb(x+2π/b) is equivalent to f(x) = f(x+2π/b), meaning that p(period) = 2π/b.

Therefor the period of an function(graph) = 2π/b.

To prove my findings I created graphs with various values of b.

y=sin(1/2x)

Period = 2π/0.5 = 4π

Input 4π  and y=sin(1/2x) into an periodic function format = sin(1/2x) = sin(1/2x+4π)

Factorize = sin1/2(x) = sin1/2(x+2π)

Simplify = f(x) = (x+2π)

The periodic function using y=sin(1/2x) and 4π turned out to be f(x) = (x+2π), meaning that they are proper to each other.

y=sin(3x)

Period = 2π/3 = 2/3π

Input 2/3π  and y=sin(3x) into an periodic function format = sin(3x) = sin(3x+2/3π)

Factorize = sin3(x) = sin3(x+2)

Simplify = f(x) = (x+2π)

The periodic function using y=sin(3x) and 2/3π turned out to be f(x) = (x+2π), they are proper to each other.

Using my findings, I made a table for the value of b and the resulting period of the function.

B	-5	-4	-3	-2	-1	0	1	2	3	4	5
Period	2pi/-5	pi/-2	2pi/-3	-pi	-2pi	n/a	2pi	pi	2pi/3	pi/2	2pi/5

The equation period = 2π/b can also be used even if b has a negative value.

Axes have four quadrants. First and second quadrant, in which sin is positive, and the third and fourth quadrant, which sin is negative. Meaning that the top half of the axis is considered positive for the sine curve while the bottom half is considered negative. If the function has a negative period, the curve will start of with a negative slope going downwards, thus making the sine curve look horizontally flipped from its usual stance.

Graphing may become very complicated when brackets are included. For instance y=sinb(x) and y= sin(bx). These two functions look very similar and indeed yes they are the same when graphed. However there are some functions that look similar, but may be completely different when graphed.

For example y=sin(bx-c) and y=sinb(x-c). These two look very similar but in fact have very far apart features. This  traces back to the fundamentals of factorization. Lets un-factorize these two functions.

y=sinbx-sinc and y=sinbx-sinbc

As you can already see here, the second equation has an additional b variable. Inorder to find out the difference when graphed, I used geogebra to model both functions.

The black graph represents y=sin(bx-c) while the blue graph represents y=sinb(x-c). When I vary the value of b, the period for both functions change. When I vary the value of c, the blue graph has an vertical shift while the black graph has an vertical shift.

If the value of b affects the period of an sine curve, how does the value of b affect the period of an tan curve?

First off, I modelled the function, y=tan(x) inorder to see the period of an tan graph when the value of b is 1.

The period of this graph is π. Ending at π on the x-axis, the graph completes one cycle, thus being a period. Now lets take a look when b=2.

y=tan(2x)

When b=2, the period = π/2.

As you can see when the value of b increases, the tan curve becomes more narrow.

Inorder to find the period’s equation of the function y=tan(bx) I used the periodic function equation.

Substitute: tan(bx) = tan(bx+π)

Factorize: tanb(x) = tanb(x+π/b)

Simplify: f(x) = f(x+π/b)

Since f(x) = f(x+p), p = π/b)

There for the period of an tan graph can be found using the the equation period = π/b.

Inorder to justify that my formula is correct, I used the function y=tan(3x) as an example.

Period = π/3 = π/3

Input π/3  and y=tan(3x) into an periodic function format = tan(3bx) = tan(3bx+π/3)

Factorize = tan3b(x) = tan3b[x+(π/3)/b]

Simplify = f(x) = (x+π/b)

The periodic function using y=tan(3x) and π/3 turned out to be f(x) = (x+π/b), meaning that they are proper to each other.

While investigating the tan graph, I noticed that the asymptote is always directly middle of the period across the x axis. Using the period equation i can interpret that the asymptote can be found using the equation asymptote = π/2b.

Mathematics 10+: What are the Features of a Cycle?

What are the features of a cycle?

Cyclic behaviors can be seen everywhere in our daily lives, such as wheels that are in motion and the sun’s movement. These types of cycles have physical presences and our easy to observe however, there are cycles in our lives that are intangible.

An very common cycle in Japan will be the cycle of seasons. Japan has four seasons that proceed in the order of, spring, summer, fall, and winter. Even though there are signs and resemblers of each season, the season itself does not have a physical presence, there for it is intangible.

Cycle keep repeating, there for it must have an start, that is equivalent to an end.  For example lets take a look at a clock. The long hand starts from 12 o clock and ends at 12 o clock, therefor 12 ocklock can be considered both the start and end. In broad, an cycle only has two parts, the start/end and the process. The cycle it self begins at the start/end, continues on to the process, then finishes at the start/end so that a new cycle can be conducted.

Cycles vary in many ways. From how they start, end, repeat, behave, and progress. In terms of progress and behavior lets take a look at the clock once again. Clocks conduct the same cycle everyday. Both the long and short hand moves at a certain speed, taking a certain amount of time inorder to complete one cycle. Now lets take a look at the four seasons in Japan. Every year the four seasons have the same cycle, how ever in different behaviors and progression speed. For some years, winter may be shorter than other years. Because of difference in behavior, the cycle may not complete in one year.  However it begins at the start/end (spring), therefor it is a cycle. So the clock has an consistent frequency/period of 24 hours, however on the other side the four season’s frequency/period is more broad. Therefor every cycle has an frequency/period of its own, that can either be consistent or inconsistent.

Since cycles have different frequency/periods how do we measure them? Two main types of measuring cycles are time and progression. Just like i’ve stated above, one day cycle on earth is measured with 24 hours. Since it has an consistent time frequency, it is more convenient to measure using time. Now lets take a look at our life cycle. In my case, I consider one day cycle when I go to sleep. Since the time I go to sleep differ day to day, it is more convenient using numbers instead of time inorder to measure my day cycle.

Cycles can be found in mathematics also. For example the cycle of 24 hours will have to be mathematically modelled when designing a clock. Depending on the cycle, the speed of the minute hand and hour hand will have to be adjusted. This has to be modelled very carefully. If there is a small malfunction in the speed of the hour hand, the cycle will slowly start to go inaccurate day by day. Now accuracy of modeling is even pursued when adjusting the minute hand’s speed. If the minute hand completes a cycle in just 59 seconds, by a day has passed it will cause an inaccuracy of 43200 seconds from the actual time.

After investigating on the realm of cycles and trig behaviors, I have a strong feeling that it is very important to our world. Cycles our everywhere in our daily lives. An example will be time. We are always thinking about time,  during our lives. Your age, todays date, history dates, timers, alarm clock, turnitin deadline, us humans depend on time as a very important measure of life. Without time we would be completely lost of the past, present, and future.

Mathematics 10+: Ambiguous Case

Do you know, or do you wonder why Mrs.Nayak told you there is no such congruence theorem as SSA (or ASS) last year? If there is no such congruence theorem, why can we use the sine rule to find an unknown angle, and the information given is SSA, as in the examples below?

To find out, i’m going to construct a situation in Geogebra where a triangle is made up of one horizontal line, a fixed angle of 30^o, and a fixed adjacent length of 5 cm. I will then use a circle to adjust the length of the side opposite 30^o, and see how many triangles can be formed with 30^o, 5 cm, and a 3^rd fixed length.

Adjacent = 5 Opposite = d A= 30 degrees

I set up my Geogebra so that it looks like the following. From now on I will be addressing the adjacent side (length of 5) as adjacent , the side(s) equivalent to the radius of the triangle as opposite (d), the bottom side (the second side adjacent to A) as bottom, and the 30 degree intersect as A.

The radius of the circle (d), defines the length of the opposite side of 30 degrees. Inorder to adjust the radius smoothly, I used sliders with the equation, d (radius of circle) = x (length of radius).

Opposite (d) < 2.5

When the radius of the circle is smaller than 2.5, the opposite wouldn’t reach inorder to complete the polygon, thus not allowing a triangle to form. There for when d < 2.5 there are no possible shapes. Inorder to verify my findings, I used Geogebra and created a situation in which the opposite’s value is 2.

The circle doesn’t intersect the bottom, meaning that when the opposite has the value of 2, a triangle won’t form. The dotted red line shows the length required to form a triangle.

Opposite (d) = 2.5

When d = 2.5, there is one intersecting point between the circle and bottom, meaning that there is one solution to a triangle.

Why is there only one solution when d = 2.5? The shortest distance from point D to bottom is 2.5. When the opposite is 2.5 degrees, it is perpendicular to bottom, meaning that the formed angle is 90 degrees. There can’t be two solutions when the opposite line is 90 degrees (vertical). When d = 2.5 there is one possible shape.

2.5 < Opposite (d) < 5

When the opposite is larger than 2.5, this are two possible shapes. HOWEVER, this is only when the opposite side is smaller than the adjacent. When 2.5 < d < 5 there are two possible shapes of a triangle. When the length of the opposite is 2.5 < d < 5, the opposite could be positioned to fit in two different angles. I set up a Geogebra situation in which d’s length has the value of 3.5

The red dotted lines represent the two ways the opposite (value of 3.5) , meaning that there are two possible shapes of a triangle.

5 < Opposite (d) 

What happens when the opposite is larger than the adjacent? I set up a geogebra situation in which the opposite has an value of 6.

The circle intersects the bottom once, meaning that there is one possible shape of an triangle. When d >5 there is one possible shape of triangle.

Adjacent = 5 Opposite = d A= 45 degrees

Now that i’ve finished investigating what happens when I change the value of opposite, I came across a question. What happens when I change the value of A? Will it affect the recent observations i’ve done?

To find out, i’m going to construct a situation in Geogebra where a triangle is made up of one horizontal line, a fixed angle of 45^o, and a fixed adjacent length of 5 cm. I will then use a circle to adjust the length of the side opposite 45^o, and see how many triangles can be formed with 45^o, 5 cm, and a 3^rd fixed length.

I set up my Geogebra so that it looks like the following. I will be addressing the adjacent side (length of 5) as adjacent , the side(s) equivalent to the radius of the triangle as opposite (d), the bottom side (the second side adjacent to A) as bottom, and the 45 degree intersect as A.

My observations for this situation weren’t that far apart compared to the recent situation. All the equations apply except instead of 2.5, the triangle forms starting from 3.54. Inorder to verify my findings, I used geogebra inorder to make a situation in which the opposite has a value of 3.54.

As you see when the opposite (d) is 3.54, the circle intersects the bottom. This means that all the equations i’ve observed in the recent situation will be overwritten with 3.54 instead of 2.5. So the equation/possible number of triangles will be like,

Equation	Possible Number of Triangles
Opposite (d) < 3.54	0
Opposite (d) = 3.54	1
3.54 < Opposite (d) < 5	2
5 < Opposite (d)	1

While comparing my results for both situations, I came across a pattern. Since were given the values for an angle and its hypotenuse side, we should be able to use sine inorder to calculate the opposite’s length.

xSine = Opposite/hypotenuse (in this case, the side which i referred as adjacent)

For situation one, the equation and product will look like

30Sin = x/5 I got the answer, 2.5. The answer is equivalent to the shortest value of the opposite, in needed to form one triangle.

For situation two, the equation and product will look like

45Sin = x/5 I got the answer, 3.54 (3.535 rounded up). The answer is equivalent to the shortest value of the opposite, in needed to form one triangle.

The smallest value for opposite (d) so that it is able to form one possible triangle can be defined using the equation aSin = opposite/adjacent.

Adjacent = 4 or 6 Opposite = d A= 30 degrees

Now that we’ve finished exploring different angles and found an equation to find the opposite, I decide to step onto situations with different adjacent length.

What happens when the length of the adjacent changes? How will my observations differ when the adjacent is 4 or 6?

Well first off I used to equation I found earlier.

30Sin = opposite/4. Opposite = 2

30Sin = opposite/6. Opposite = 3

I plotted out both situations on geogebra and got the same result. Other than that the observations were quite similar. In conclusion, the equation/number of possible triangles will look like

Equation	Possible Number of Triangles
Opposite (d) < (aSin =    Opposite/Hypotenuse)	0
Opposite (d) = (aSin = Opposite/Hypotenuse)	1
3.54 < (aSin = Opposite/Hypotenuse) < 5	2
5 < (aSin = Opposite/Hypotenuse)	1

3.54 < (aSin = Opposite/Hypotenuse) < 5

When the value of opposite is somewhere between 3.54 and 5, there will be two possible triangles that can be formed.

As you see the b(opposite) can be set in two different angles inorder to form a triangle. The value for both b is equal. Now, how will I be able to find the degree of angle for A? From my recent observation during this investigation, I found out that aSin = Opposite/Hypotenuse. This means that we have values for B(Angle), a(Side), and b(side). Since we know these three values, we can use the sine rule inorder to find A

First off, set up your sine rule!

Since we are finding for A, we will have to isolate it! We multiply both sides by a so that the denominator for sin(A) will get cancelled out. Next we take away the sin which gives us..

Now since A can be set up in two different angles,we will subtract the answer of A from 180 degrees, inorder to solve the second answer for A.

Submarine and Ship

 

A ship and a submarine are doing a naval exercise together.  The two vessels leave the port at the same time.  The submarine sets off in a direction due east of the port, while the ship sails in a direction, ?? North of the submarine.

 

After the ship has travelled k kilometers, it receives a distress call from the submarine.  The radar is not working well and the ship captain can only determine that the submarine is b kilometers away from the ship.  Where could the submarine be, if the ship knows it is in a line due east from the port, and b kilometers away from the ship?

At this point of the question, it isn’t impossible to solve (unless I use a protractor). However, lets make an assumption that the angle from the ship to submarine is X degrees. Using the sine rule, we can set up a equation for the submarines distance (x) from the port.

Just like the i’ve done in the recent equation, you isolate the x inorder to solve the equation. HOWEVER, we solved this equation using the values A(angle) S(Big Side) s(Small Side). This means that there can be two triangles solved SO, once we solve b we have to subtract it from 180 degrees inorder to achieve the second answer to b.

 

 

 

 

Mathematics 10+: Midyear Reflection

My progress during first semester wasn’t as good as I thought. I was interested in some sections, and not so much for others. I always have fun solving puzzle questions like the one you can find on the AMC competition. I’ve noticed that im not much interested in graphing and equations.During class I like how we are given questions to solve within tables. This gives us opportunities to communicate between one another, and to help each other out. During second semester, I would like to improve my investigations more. Even though I dont struggle on solving problems, I dont have a clear understanding on why things happen. I will achieve this by asking my teacher more questions!

Mathematics 10+: Modeling Inverse Variation

Investigation

There are many real life situations in which inverse variations can be implied. The situation investigated is ” The apparent height of an object varies when viewed at difference distances”.

First off, we started with gathering data. We used a 20 meter tape measure to determine our data from the object (Daiki). The data recorder, started from 20 meters and closed up four meters every time we measured the object’s height. Our data became the following,

If the data recorder comes closer to the object within four meters, the object cannot be measured due to its size, therefor our data ends at four meters. I input the data onto GeoGebra, which graphed the points onto the grid. Inorder to find the best fit line, I input the formula

This formula alone, looks like this

However by using sliders and adjusting the value of a,h, and k, I was able to make the graph fit the coordinates of the data.

Since height cannot be negative, I only captured the positive half of the slope. As you can see, the graph almost completely goes through each point. The values of my variables were, A= 85.8, H= 0, and K= 0.

Conclusion

While creating this graph, I came across a question, what will the domain and range of this graph be? As you can see as the line is closer to 0, it becomes steep and runs horizontal. I looked back to my graph and noticed that the data had stopped at x (distance) at 4 and y (height) at 19. I realized that measurements smaller than these cannot be taken, which means that the domain of this graph will be x>4 and the range will be 0<y<19. The vertical and horizontal asymptote will be 0. The horizontal asymptote is 0, because the data recorder and object cannot be at 0 distance. The vertical asymptote is 0, because no matter how far you go the size of the object will not be 0. As the graph closes to the vertical asymptote, it indicates that the height of the object becomes larger. This is because of the human eye structure, the closer you are to an object, it looks larger. However this wouldn’t fit to the context of the problem because the height of an object wouldn’t be larger than its initial height. On the other side, as the value of x variable gets infinitely large, the value of y variable becomes infinitely small. This doesn’t fit to the context of the problem. This is because we use our eyes to gather data, therefor it would be impossible to measure the objects height if we go too far to the extent of not being able to see the object clearly. Inverse variation also appears in daily life situations. For example,

Use of electricity during winter: as the temperature goes down, the use of heaters increase, thus causing the use of electricity to go up.

Force of Gravity: the further you are from the surface of earth, the weaker the pull of gravity.

Multitasking/procrastinating while working on homework assigned by Mrs.Durkin: the more time spent multitasking/procrastinating on needless things (i.e. Facebook, Youtube, ETC), the less work you finish!

Mathematics 10+: Transformation of Functions

In this investigation, I will use my prior knowledge of the graph of y=x² and y=a(x-h)²+k and my calculator, to investigate the position and shape of variations on functions such as y=|x|   and y=√x.

y=a(x-h)²+k

The equation, y=a(x-h)²+k is the equation of a parabola expressed in vertex formula. This formula is often used to find the vertex coordinates of a parabola. For instance, the vertex coordinates in the formula, y=a(x-h)²+k will be (h,k). So what happens if I change the variables a, k, and h? I used geogebra inorder to show my findings.

The first slider, represents the change in a (0:00 – 0:14). As you can see in the video, the value of a affects the shape of the parabola.

– Increase in a = narrower parabola

– Decrease in a = wider parabola

– a equals zero = no parabola. This is because when you insert a=0 into the equation y=a(x-h)²+k, (x-h)² will be cancelled out, thus causing the equation to be y=k. Which will cause the graphed function to become a linear line.

– When a has a positive value = Upwards parabola. The vertex will be the minimum point.

– When a has a negative value = Downwards parabola. The vertex will be the maximum point.

I illustrated three different value for a, given that the values of h and k are 0.

a=1  a=2 a=3

(Click on the image to enlarge)

I noticed that when a=1, the parabola goes through the coordinate (1,1). When a=2, the parabola goes through the coordinate (1,2). When a=3, the parabola goes through the coordinate (1,3). Because h and k are 0, the equation y=a(x-h)²+k can be simplified down to y=a (Since we are looking at the coordinate (1,y), we insert 1 for x and 0 for h, allowing us to simple it down to 1). There for, the y coordinate will be equal to a. 

The second slider (0:15 – 0:21), represents h. h is equivalent to the x coordinate of the vertex. So the decrease/increase in h will cause a horizontal shift of the graph.

The third slider (0:16 – 0:28), represents k. k is equivalent to the y coordinate of the vertex. So the decrease/increase in k will cause a horizontal shift of the graph.

y=a|x-h|+k

The equation, y=a|x-h|+k is an function with an absolute value. An absolute value is the value of an real number, regardless of its sign (positive or negative).  So, for example, the absolute value of 7 is 7, and the absolute value of -7 is also 7. I illustrated graphs of the function, y=a|x-h|+k when the values of a, h and k are varied. I used geogebra inorder to show my findings.

As you might have noticed, a acts the exact same way like ive stated before while investigating the function y=a(x-h)²+k (increase in a= narrower parabola, decrease in a= wider parabola, a≠0).

h and k also acts as the coordinates of the vertex, just like ive stated before while investigating the function y=a(x-h)²+k. 

Inorder to verify if my analysis on a, h, and k are correct, I illustrated graphs that have the value of all three variables change.

As you can see from the video, yes change in a does affect the shape of the parabola, regardless of the value of h and k. Also, h and k does resemble the coordinates of the vertex, regardless of the value of a.

y= a√(x-h)+k

The equation, y= a√(x-h)+k is an function with a rooted value. I illustrated graphs of the function, y= a√(x-h)+k when the values of a, h and k are varied. I used geogebra inorder to show my findings.

This function looks very unfamiliar compared to those that we have investigated before. The graphed version looks like a side ways parabola with half a side missing. Both a, h, and k performs like the other functions ive investigated before.

Inorder to verify if my analysis on a, h, and k are correct, I illustrated graphs that have the value of all three variables change.

Just like the other functions before, a affects the shape of the parabola, h causes an horizontal shift, and k causes an vertical shift.

Conclusion:

Functions are largely varied with the variables, a, h, and k. For all functions, a affects the shape, h and k affects the vertex. The closer a is to zero, the wider and parallel to the x axis the function gets.

Unusual Function:

The unusual function I made is f(x) = (a (x – h))³ + k. What happens to my graph if I vary the variables a, k, and h? I used geogebra inorder to show my findings.

Just like the functions i’ve investigated before, the rules of a,h, and, k are the same. a affects the shape of the parabola, h and k affects the vertex.

Mathematics 10+: Scratch and Win

Scratch and Win

The fast food chain, MacDuffs’ is running a competition.  With every order, you obtain a card which has 12 circles covered up, and you can scratch off up to four circles. You win if 3 or more PALM TREES are revealed, but lose if 2 or more CRABS are revealed.  If you win, you can scratch off any one of the three squares to show what you have won. The card above is shown with all of the circles and all of the squares revealed.  Macduff’s want the game to be both fun and relatively easy to win.

As a devoted and loyal fan of Macduffs, I investigated the chances of winning if the PALM TREES are in the ratio 2:1 (i.e 2/3 of the circles are PALM TREES); and also explored on what happens if the ratio is changed.

First of all, is the ration really 2:1? 8 palm trees and 4 crabs. The ratio would be 8 palm trees:4 crabs, which can be reduced to 2:1. This verifies the fact that the ratio of palm trees to crabs is 2:1.

There are 12 options when scratching a brand new card. The ration of palm trees to crabs is 2:1, which means that 8 of the options will always be a palm tree, and the other 4 will always be a crab. Since all the covered options look the same and there is no way telling which is a palm tree and which is a crab, the 12 options are equally likely to be scratched. This verifies the fact that the probability of my first choice being a palm tree when scratching a brand new card will be 2/3. Which also defies the probability of my first choice being a crab when scratching a brand new card as 1/3.

Inorder to calculate the chance of winning, I will have to use conditional probability. Conditional probability is the chances of an event happening, given that another is already known to have occurred. An question that requires the use of conditional probability will be, calculating the probability of getting a palm tree on my second go, given that I have gotten a palm tree on my first go. The formula will be 2/3 (probability of palm tree) times 7/11 (probability of palm tree on second go) which will be 14/33. On the second go, the denominator (the number of options) and the numerator (the number of palm trees) decreases by one. I also calculated the probability of getting a crab on my second go, given that I have gotten a palm tree on my first go. The formula will be 2/3 (probability of palm tree) times 4/11 (probability of crab on second go) which will be 8/33.

By applying conditional probability continuosly, I can calculate the probabilities for all the winning combinations. I created a tree diagram to show all combinations for the scratch card.

The blue bubbles with fractions represent palm trees. The fraction defies the probability of scratching a palm tree in the certain round. The white bubbles with fractions represent crabs. The fraction within the bubble defies the probability of scratching a crab in the certain round. As you can see from my tree diagram, their are five possible combinations to win this game. I calculated the probability of each combination.

(Palm Tree, Palm Tree, Palm Tree, Palm Tree) = 1680/11880

(Palm Tree, Palm Tree, Palm Tree, Crab) = 1344/11880

(Palm Tree, Palm Tree, Crab, Palm Tree) = 1344/11880

(Palm Tree, Crab, Palm Tree, Palm Tree) = 1344/11880

(Crab, Palm Tree, Palm Tree, Palm Tree) = 1344/11880

If I know all the combinations of winning, and the probability of each combination, I can add them up to defy the probability of winning this game. The probability of winning this game will be

(1680+1344+1344+1344+1344)/11880 = 7056/11880, which can be reduced to 98/165

Since I know the probability of winning the card, I can go even further, and calculate the probability of winning after the first and second scratch.

If my first scratch is a palm tree, the probability of winning will be, 8/12 (probability of palm tree) times 98/165 (probability of winning) which will be 784/1980

If my first scratch is a crab, the probability of winning will be, 4/12 (probability of crab) times 98/165 (probability of winning) which will be 392/1980

If I get a palm tree on my first two scratches, the probability of winning will be, 784/1980 times 7/11 (probability of palm tree on second go) which will be 5488/21780

If I get one of each (palm tree and crab) on my first two scratches, the probability of winning will be, 8/12 (probability of palm tree on first go) times 4/11 (probability of crab on second go) times 98/164 (probability of winning) which will be 3136/21648

So what happens if we change the ratio of palm tree:crabs? How does it effect the probability of winning? While thinking of ways to tackle this question, I noticed that my tree diagram will look the exactly same, except the fractions will differ slightly. If the ratio of palmtree:crab is 9:3, the probability of winning will be,

[Palm Tree (9/12), Palm Tree (8/11), Palm Tree (7/10), Palm Tree (6/9)] = 3024/11880

[Palm Tree (9/12), Palm Tree (8/11), Palm Tree (7/10), Crab (3/9)] = 1512/11880

[Palm Tree (9/12), Palm Tree (8/11), Crab (3/10), Palm Tree (7/9)] = 1512/11880

[Palm Tree (9/12), Crab (3/11), Palm Tree (8/10), Palm Tree (7/9)] = 1512/11880

[Crab (3/12), Palm Tree (9/11), Palm Tree (8/10), Palm Tree (7/9)] = 1512/11880

When the ratio of palm tree:crab is 9:3, the probability of winning the scratch card will be 9072/11880, which can be reduced to 42/55

From this I observe that the more palm trees there are in the scratch card, the higher the percentage of winning.

At this point, I came across another question. What happens if the ratio is 1:1? Again the tree diagram will look the same, except with different fractions. The probability winning will be,

[Palm Tree (6/12), Palm Tree (5/11), Palm Tree (4/10), Palm Tree (3/9)] = 360/11880

[Palm Tree (6/12), Palm Tree (5/11), Palm Tree (4/10), Crab (6/9)] = 720/11880

[Palm Tree (6/12), Palm Tree (5/11), Crab (6/10), Palm Tree (4/9)] = 720/11880

[Palm Tree (6/12), Crab (6/11), Palm Tree (5/10), Palm Tree (4/9)] = 720/11880

[Crab (6/12), Palm Tree (6/11), Palm Tree (5/10), Palm Tree (4/9)] = 720/11880

When the ratio of palm tree:crab is 6:6, the probability of winning the scratch card will be 3240/11880, which can be reduced to 3/11.

As there were less palm trees compared to 9:3 scratch card, the probability of winning was lower.

Another question I came across was, what if there is only one crab? The ratio will be 11:1. If there is only one crab, the probability of winning is 100%. You can win from any type of combination on the tree diagram. This is because the conditions of losing this game is when you get more than two crabs. But since there is only one crab when the ratio is 11:1, there is no combination that can cause a loss.

To get a better idea of the probability of winning compared to the ratio, I made a graph with the coordinates, Number of Palm Trees for x and Probability of Winning (percent) for y.

I started the x axis from two because any number below, the probability of winning is 0%

Expected winnings,  known as E(X),  is found by the formula:  P(win) x prize money – Cost of Game. By calculating the expected winnings, I can see whether the scratch card is profitable for the card makers or not.

If the scratch card costs 2 dollars, and the prize is 4 dollars, the scratch card is profitable for the card makers.  98/165 (probability of winning) x 4 (prize money) – 2 (cost of game) = 0.37. So the card makers will make a profit of about 37 cents for every scratch card they sell. However, expected winnings arent always correct. If the customers keep winning, the card makers may have a loss. The more scratch cards are sold, the closer the profit for one card will become.

If the scratch card costs 3 dollars, and the prize is 4 dollars, the scratch card isn’t profitable for the card makers. 98/165 (probability of winning) x 4 (prize money) – 3 (cost of game) = -0.63. The card makers will make a loss of about 63 cents for every scratch card they sell.

In a scratch card with the ratio of palm tree:crab is 6:6, if the card cost is 2 dollars and the prize is 5 dollars, the scratch card isn’t profitable for the card makers. 3/11 (probability of winning) x 5 (prize money) – 2 (cost of game) = -0.64. The card makers will make a loss of about 64 cents for every scratch card they sell.

10 Lotto Scratch Card

The scratch card I made consists of 8 options to scratch. This scratch card is intended for children, and will be sold at sweet shops. It will be sold at two dollars, and the prize will be ten dollars. The conditions of winning will be scratching a $$ mark. The rest will be a lollipop picture. The ratio of $$:Lollipop will be 7:1. If you scratch the $$, you get 10 dollars. If you scratch the lollipop, you get a dumdum pop of your choice. This scratch card will be both profitable and enticing. 10 dollars is a large amount of money for children. Two dollars is a very affordable price for children, and given that they’ll still win a lollipop even though they lose, will act as an incentive for them to buy the scratch card.

Reflection:

My scratch card is both profitable and enticing. I’ve managed to do this by creating if from the perspective of a children. 10 dollars, that may be a lot of money for a children. Being able to try for such prize just by paying 2 dollars may act as an incentive for the consumer to buy the scratch card. Even if they dont win 10$, they still get a lollipop. Thus it is still profitable for the card maker. They’ll win one per every eight scratch cards. Gives the card maker 60 cent profit for every card they sell. Its a win-win situation for both the card maker and consumer.

The design of gambling games takes a large role in society now a days. Companies prize massive amount of money inorder to get as many customers as possible. However, only a handful amount of people win the lottery, giving the company a large profit.

Just as I stated above, if I were to make a more profitable and enticing game, I will create a lottery. However this will cost a lot of money to start with, and may not be possible for myself at the moment to make happen. Customers will buy the lottery ticket, dreaming for the massive amount of cash.

Mathematics 10+: Complex Numbers

Imaginary Numbers:

A real number can ALWAYS be placed on a number line.

For example, if x=3, the number line will look like the following (Circles represents points),

Still, some problems can’t be placed on a number line as x has no solution. An example will be the formula,

X²=-2

This formula has no solution since -2 cannot be square rooted. The answer, no solution means that their is no number on the number line that answers the equation X²=-2. This means that the solution to X²=-2 may be somewhere off the number line like the blue circles shown below.

These numbers that cant be placed on the number line, are called imaginary numbers or complex numbers. They are represented with the numeral, i. Inorder to define i, we use the equation,

By using i, we are able to solve problems like X²=-2 as

 instead of using no solution which is a much more inaccurate answer.

Complex numbers is used in the quadratic formula when the discriminant is a negative number. If the discriminant is a negative number, the equation has no solution. But instead, by inserting i, you can get a more accurate answer to the equation.

Resoures:

Owens, D. (2008, December 18). Complex numbers, part 1 [Video file]. Retrieved
from http://www.youtube.com/watch?v=9Fm8aUyf1Yo

Mathematics 10+: About Me

About Me

     I’m Daiki Yamamoto, a tenth grade student learning at Canadian Academy. I’ve lived in Japan for my whole life. It will now be my eighth year studying at CA. I have been here since I was seven years old. I’m full Japanese and Japan has always been my home country. I am a native speaker of Japanese. I haven’t studied at a Japanese school but I speak Japanese at home and am able to read and write almost at my age level. I have played in the varsity soccer team since freshman year. This year, I am hoping to become a starter on the team so that I will be involved with the team more often. I am also joining the tea ceremony course this year. Even though I am Japanese, I’ve studied at an international school for my whole life so I haven’t had many chances to be involved with Japanese culture. Through out the tea ceremony course, I am looking forward on learning more about the Japanese culture.

I am confident with mathematics. I tend to find it easier than other subjects I take. I don’t enjoy studying math but I enjoy solving puzzles that involve math. I believe math in general is strength for me. I still lack some knowledge on solving indices and roots so I plan on perfecting those subjects during this year. All math classes have been successful for me. Learning math is whether you are able to memorize and understand the formula to solve the problem. I haven’t had hard time understanding subjects so I say all my math classes have been successful.