BQ7: Unit V

How To Get The Difference Quotient Formula:

To Begin with, you should look at the graph below. It shows you a function with one point being (x,f(x)) and another point being (x+h,f(x+h)). Notice how there is a line drawn going through these points. This line is called a secant line. The difference quotient is really just the slope of this secant line going through point A and B. 
The letter h used in the formula is really just delta x in a shorter way to write it.








The tangent line is found as the letter h, the distance between the first point to the next point, gets smaller. This is because tangent line does not go through the graph but only touches the surface of the function once. As the distance between the points reaches 0, it is closer to being a tangent line which represents the derivative of the function, which you want to ultimately find and be familiar with.

BQ #6: Unit U Concepts 1-8

1. What is continuity? What is discontinuity?

Continuity means that the function will be predictable, meaning it has no undefined points, no breaks, no "holes", and no jumps. To have a continuous function you must make sure that you can draw this function without needing to lift up your pencil from the paper. You also have a clue that you have a continuous function when the limit (INTENDED height) of a function is the same as its value (ACTUAL height).

You know that a function is at discontinuity when the limit is not the same as the value because the function has an undefined point, a break, a "hole", or a jump. At discontinuity, it is impossible to draw your function without needing to lift up your pencil from the paper at some point. 
You must keep in that there are two families of discontinuity: Removable and Non-Removable.

2. What is a limit? When does a limit exist? When does a limit not exist? What is the difference between a limit and a value?

A limit is the INTENDED height of a function. It means that you are getting VERY CLOSE to a certain height as you are getting VERY CLOSE to a certain x-value. A limit differs from a value because as it is only the INTENDED height of a function, the value is the ACTUAL height of a function. For a value you do not just get VERY CLOSE to a certain height, you ACTUALLY GET there. If the function has the same limit and value it is a continuity, there is no ambiguity to it, but if the function has different limit and value then you know it must be a discontinuity. 

Consider that there are two families for discontinuity. The limit exists only in the removable discontinuity, also known as the point discontinuity(it is the only one in that family), because the same height is intended to be reached from both the left and the right sides of a function. You put your fingers on both ends of the function,left and right, and you bring them together to meet at the same place, even if the place isn't there (point/"hole"). If they do not meet at he same place, then there is a non-removable discontinuity and the limit DOES NOT EXIST. The limit does not exist in 3 types of the non-removable discontinuity family. One is called jump discontinuity (the limit DNE because of different left and right); another is called infinite discontinuity (the limit DNE because of unbounded behavior); and finally there is oscillating behavior (the limit DNE because there is not definite point that is reached at a certain x value).

3. How do we evaluate limits numerically, graphically, and algebraically? 

a) Numerically- For instance, the picture below shows the table needed to find limits numerically. If finding the limit as x approaches 3 (#9), we would plug the function of #9 into our calculator at "y=", we would hit "graph" and "trace" for the numbers getting VERY CLOSE to the x-value 3. Our table like shown below will help us organize our information. In the middle box for the x value (on top if the ?) we would put the 3. The 3 boxes on the left should be the x values getting VERY CLOSE to 3 from the left side--2.9, 2.99, 2.999--and the 3 boxes on the right should be the x values getting VERY CLOSE to 3 from the right side--3.1, 3.01, 3.001.



https://drive.google.com/file/d/0B4NSkh2FgPbXR2RSOUU0cFVvWWs/image?pagenumber=6&w=800


b) Graphically-  You just take a look at a graph like the one shown below and write down what you see happening in terms of the limit at each x value you need to take a look at.






https://drive.google.com/file/d/0B4NSkh2FgPbXR2RSOUU0cFVvWWs/image?pagenumber=5&w=800


c) Algebraically- There is direct substitution, in which you plug in the x value directly to the function to get the limit. You can get 4 different types of answers. There is  numerical answer (ex. 3)...YAY we are done! There is a answer with a 0 divided by some number, which equals 0...YAY we are done! There is an answer when a number is divide by 0, which we know is undefined, and therefore the limit DNE...YAY we are done! Finally there is a 0 divided by a 0, which means that it is "not yet determined" and you must keep working on another method to figure it out.        Note: remember 0/0="hole"
Another method is dividing out/factoring. IF you try direct substitution and you get a 0/0, then you may move on to factoring out a polynomial, from whichever numerator or denominator as needed, and get rid of the zeroes. Then you use direct substitution on whatever is left after getting rid of the zeroes.
The last method is called rationalizing/conjugate because if you have a square root in either the numerator or denominator you must multiply by its conjugate to cross out the zero. Once again you must consider this only if you have definitely tried the direct substitution method first and you got 0/0.

Reflection #1: Unit Q

#1. What does it actually mean to verify a trig identity?

You show that the Pythagorean Theorem is correctly used in the triangles to find sine and co-secant, cosine and secant, and tangent and cotangent. For example,  if you are given sine=3/5 you know that cosine must equal 4/5. You get the 4 from the Pythagorean Theorem 3^2+b^2=5^2; b=4. Meanwhile, you show that you get the the 5 as the denominator as well as the c value in the Pythagorean Theorem because we know that sine is S-O-H, opposite/hypotenuse and cosine is C-A-H, adjacent/hypotenuse. Also, tangent is T-O-A, opposite/adjacent. ( Soh cah toa is use for right triangles only).

To better visualize the concept, please take a look at the images below:





http://www.mathwarehouse.com/trigonometry/images/sohcohtoa/sohcahtoa-all.png





http://hyperphysics.phy-astr.gsu.edu/hbase/imgmth/ttrig.gif



#2. What tricks and tips have you found helpful?

I found that knowing the trig identities out of the top of your head was really helpful. However, there is no shame in not being able to memorize them quick enough or at all. I do think is better if you do know them because then you are more familiar with the concept and you have a better idea of how to get where you want to get. I suggest you do Mrs. Kirch's blog posts with thoughtfulness and to practice as much as possible and to ask questions. I also remember having problems in which knowing how to factor pretty good will help you a lot in simplifying or verifying.





#3. Explain your thought process and steps you take in verifying a trig function. Do not speak in specific manners, but speak in general terms of what you would do no matter what they give you.

My thought process would be:

• can I substitute an identity?
• can I use ZPP?
• are the numbers in the problem familiar with unit circle numbers?
• do I have to Square anything?
• does it deal with fractions?

1. - 

do I know how to deal with fractions(ex. when you multiply a fraction you do not have to have like terms!)

BQ #4: Unit T Concept 3: Graphing tangent and cotangent

To understand why "normal" tangent graph is uphill, but "normal" cotangent graph is downhill, you must understand how each is applied in the unit circle. Tangent is sine over cosine. Cosine is 0 at 1/2 radian and at -1/2 radian so this determines where the asymptotes are placed. According to where the asymptotes is how the graph is formed. For instance -1/2 radian comes first then 1/2 radian and quadrant 4 and 1 are in between. In quadrant 4 tangent is negative so it goes below and in quadrant 1 tangent is positive so it goes above.

Cotangent is just above, below because the asymptotes end up in different places since cotangent is cosine over sine and sine is 0 in 0 radian and 1 radian.

BQ #3: Unit T Concept 1-3: Graphing sine, cosine, cosecant, secant, tangent, cotangent

 


The above pictures show the relation between sine and cosine with all the other trig function graphs. Notice that sine and cosine graphs are similar in the way they swingle continuously. In contrast to the other graphs, which obtain asymptotes and do not continue in the way that sine and cosine do. Those rest of the trig functions that have asymptotes go on forever in a vertical way. In Mrs. Kirch's words, those trig function graphs require us to raise our pencil to continue to outline the rest of the graph.




You are recommended to go to desmos.com in order to have a full experience on how different values affect the graph.

BQ #5: Unit T Concept 1-3: Graphing sine, cosine, cosecant, secant, tangent and cotangent

Sine and cosine do not have asymptotes, meanwhile all the other trig functions have asymptotes. But, why? Remember that an asymptote forms when there is an undefined ratio/fraction. This means that there is a number divided by 0.
This is based on the trig function's ratios. Sine and cosine have ratios of y/r and x/r. "r" always equals to one in the unit circle so we can be sure that the ratio will never be undefined.
Meanwhile, the ratios for all the other trig functions have a denominator, not of r=1, but of either y or x values, which could happen to be values of 0.

BQ #2: Unit T Concept 1-3: graphing sine, cosine, secant, cosecant, tangent, and cotangent

Trig Graphs relate to the Unit Circle. We must consider the fact that in trig graphs, the Unit circle is expanded into a line form, instead of a circle form, in order to make sense in a graph.

*A period means the graph goes through one cycle while covering certain radian units on the graph. An amplitude are half the distance between the highest and lowest point of the graph.

*Periods.

For instance, sine, cosine, cosecant, and secant have a period of 2 radians. This is because it contains a 4 part repeating unit all the the way around the unit circle (360 degrees=2 radians). This is shown in the unit circle when we look at the pattern of where the sine and cosine are positive and negative.
Sine is positive on Quadrant I and II and negative on Quadrant III and IV. The pattern is positive positive negative negative. And it continues, but we must realize we had to go all around the unit circle in order to continue the pattern. The same with cosine, whose pattern is pattern is positive negative negative positive. The pattern continues but only have we went around the unit circle. Again, around the circle means it is 2 radians.

In contrast with tangent and cotangent, which have a period of only 1 radian. This is because they only have a 2 part repeating unit which just reaches half way through the unit circle at 180 degrees=1 radian until it repeats again. The pattern is positive negative positive negative all around the unit circle. There was one repetition in pattern so the pattern continues every 1 radian.

*Amplitudes

Sine and cosine have amplitudes because they are the ones with the restriction: -1<sine or cosine<1. This means that their highest point is 1 and its lowest point is -1. Relating to the unit circle, we know that the lowest it goes it to -1 and the highest it goes is to one (remember a unit circle has a unit of 1 all around) The sine an cosine ratios are y/r and x/r. The highest the y and x values could be is 1 and the lowest they can be is -1. ("r" is the ration equals to one).
The other trig functions instead go up forever or down forever because they do not have restrictions.

SP #7: Unit Q Concept 2: Find all trig functions given one trig function and a quadrant

The Picture below will show you a problem as well as how to find all trig functions using IDENTITIES (Reciprocal and/or Ration and/or Pythagorean). 







This problem shows that you can find all trig functions by using the identities that we learned about in Unit Q Concept 1. It shows that it is helpful to know these identities in order to have expanded knowledge that would help us solve problems that could become more difficult. It follows that it is a good practice to get to know the identities of trig functions out of the top of your head.
Please notice that we figure out the quadrant that will be used by checking the answers of the trig functions given, whether they are negative answers or positive answers. Also notice on the right how we figured out the answers of  each trig function, whether they were negative or positive, based on the quadrant that we found out will be used.





And please do not forget to to check out the SECOND WAY to find all the trig functions of this problem. Just click on:


                         Kathy's Awsume Second Way

WPP # 13-14: Unit P Concepts 6-7 Word problems using law of sines and law of cosines

TO SEE THE PICTURES SHOWING THE WORK MADE IN COLLABORATION WITH KATHY C. PLEASE CLICK ON: 
                                                                KATHY'S AH-MAZING X10 BLOGPOST

Hershey has stopped at a stoplight and noticed that his best friend, Marlene, is due west of him at the next stoplight, 30 feet away. Both are going to Hershey's Bakery. Hershey walks N 30* W to get there while Marlene goes N 72* E to get there. What is both of their distances to walk over to Hershey's Bakery?




 After having a nice, long conversation with Hershey, they decided to see each other again the next day. They both leave the bakery at the same time. Marlene, in hurry to get to her cousin's Quinceanera, is headed at a bearing of 315* and is traveling 50 MPH. Hershey on the other hand goes home at 30 MPH at a bearing of 078*. How far apart are they after two hours?

I/D #3: Unit Q Concept 1: Rational Reciprocal and Pythagorean Identities

1. Where does sin^2 x+cos^2 x=1 come from to begin with?

To begin with. you should know that an "identity" is "a proven fact that is always true". Therefore, we assume that a Pythagorean Identity, which is the equation seen above, is called such because that equation can definitely be proven as a fact that is always true. Remember that the Pythagorean Theorem using variables x, y, and r is x^2+y^2=r^2. Now doesn't this look familiar? Yes, we did get it from the a^2+b^2=c^2 equation we've learned before, but notice that just by changing the variables to x, y, and r gives us the exact equation we learned that is of a unit circle.

Now let us play with this equation a little. Who knows what amazing discovery we will come across. If we wanted to set the Pythagorean Theorem (which we just found out is similar to the unit circle equation) equal to 1 as shown in the Pythagorean Identity above, what would we do to it? Oh, of course! Divide everything by r^2 because r^2/r^2 is equal to 1. We are left with x^2/r^2+y^2/r^2=1. You may think, what kind of mess is this? Well, do not fear, Math For Cheese Buckets, like you and me, is here to make it clear! First of all, you should know that the equation can be re-written as (x/r)^2+(y/r)^2=1. 

Moreover, the ratio of cosine is (x/r) and the ration of sine is (y/r). Wait, we've seen these before. Yes, a ton of times in the last 2 units, but wait a sec, we just saw them right now, in the re-re-written equation of the Pythagorean Theorem. We can look at the equation in the previous paragraph and switch the (y/r)^2 with sine^2 x and the (x/r)^2 with cosine^2 x. Guess that we did come across a discovery. We just derived the Pythagorean Identity from the Pythagorean Theorem. 

To show that this identity is true, we will choose one of the "Magic 3" ordered pairs from the unit circle.

• 30 degrees= (radical 3 over 2, 1/2)
• 45 degrees= (radical 2 over 2, radical 2 over 2)
• 60 degrees= (1/2, radical 3 over 2

I will show 30 degrees and 45 degrees (know that 30 degrees is similar to 60 degrees, just switched).

• 30 degrees: (radical 3 over 2)^2 + (1/2)^2 = 1. You cancel out the radical with the squared and end with just the 3 on the top of the first fraction. You square the 2 and end up with a 4 at the bottom of the first fraction, so you have 3/4. For the second fraction you square the 1 and still end with a 1 in the top of that second fraction, and you square the 2 to end up with a 4, so you have 1/4. (3/4)+(1/4)=1.
• 45 degrees: (radical 2 over 2)^2 + (radical 2 over 2)^2 = 1. You can cancel out the radical with the squared for the top of both fractions to end up with a 2 on top of both fractions. Then, square the 2 at the bottom of both fractions and end up with a 4 at the bottom of both fractions: (2/4)+(2/4)=1.

2. Show and explain how to derive the two remaining Pythagorean Identities.

sine^2 x + cosine^2 x = 1. You divide the equation by cosine^2 x first. We know that sine x divided by cosine x equals tangent x by the Ratio Identities. So, we can "power up" through our understanding that it will be the same thing if we squared the sine and cosine we can square the tangent. The first part of the equation becomes tangent^2 x. Obviously, cosine^2 divided by itself is equal to 1. The second part of the equation is 1. On the right side of the equation 1 divided by cosine^2 x will become secant^2 x, because we know by the Ratio Identities that 1 divided by cosine x is secant x, therefore we can again "power up" to show that secant x will become secant^2 x. Our final equation: Tangent^2 x + 1 = secant^2 x.

The picture below can give you a visual. Please ignore the bottom part. Just pay attention up until the equation that is squared.





2.Jhttps://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDHvpTDOkgrgF25cX4WadLd509U46TFysmPmVoq8oK0KN0IxD0MwY4hDnLRII13ZhlSmX0MGhYvoses04nemL83rJuWro7hu1ftw0c0arJGg6nXItpORf5PT0IZCs5Ng1RTSH4az7p8Oyc/s1600/DSCN397PG

Now we divide the equation,  the one at the beginning of the top paragraph, by sine^2 x. Obviously, sine^2 x divided by itself is 1. Through the Ratio Identities we know that cosine x divided by sine x equals cotangent x. Again, we can "power up" to show that cosine^2 x divided by sine^2 x equals cotangent^2 x. On the right side of the equation we show that 1 divided by sine x equals co-secant x through the Ratio Identities, so once again we "power up" to know that 1 divided by sine^2 x will now equal co-secant^2. Our final equation: 1 + Tangent^2 x = Co-secant^2 x.

The picture below will give you the equations I am deriving the above work from.







l_idehttp://a2h-3rdhour.wikispaces.com/file/view/fundamental_identities.gif/140475861/fundamentantities.gif

BQ #1: Unit P Concept 1-5: Law of Sines, Law of Cosines, Area formulas

#3: Law of Sines
The Law of Sines in a triangle that is SSA, or side-side-angle, meaning we know 2 sides and one angle of the triangle, is an ambiguous case because we can have either one, none, or two answers to the triangle. This is because we only know one definite angle. When we knew at least two definite angles, like in the cases of ASA and AAS, we were able to figure out the third definite angle without wondering if there could be another possible way to draw the triangle. With LIMITED information, it is possible that we can create two different triangles by forming a "bridge" with the side that is given to us for the angle that is not given to us. 

After figuring out the value of the angle using the Law of Sines, we use its reference angle to figure out its other possible value. The reference angle is relevant because you can have the angle in the sine quadrant (quadrant II) besides the all quadrant (quadrant I), since it is less than 180 and could be in the boundaries of the Triangle Sum Theorem.  

You know there is either no triangle when you hit a "wall" in the beginning or only one triangle when you hit a "wall" after figuring out the first triangle. The "wall" could be something that is not possible like sine x being greater that 1 or less than negative 1 and the angles adding up to something greater than 180 degrees.

The picture below will show you a problem that will give you two answers.










#5 AREA FORMULAS


The pictures below will help you see how the area formulas of a triangle work and how by using either of them you will still get the same answers. This will show you that the law of sines works with working out a triangle that is not a special right triangle.

WPP #12: Unit O Concept 10: Angles of elevation and depression

http://t0.gstatic.com/images?q=tbn:ANd9GcSKvHvNZVc5F7PgZG2008OF-l9bI4BbhjA7zstukLMqiBMq8DVS:www.firstfoo.com/wp-content/uploads/2013/09/wedding-cake-beautiful-light-blue-quinceanera-cake-with-princess-character-on-top-quinceanera-cakes.jpg

The Problem: 15anera: Sugar, oh Honey Honey.

Marlene's cousin is about to celebrate her traditional quinceanera. She put Marlene in charge of the cake design. Marlene has a friend that works in a pastry and she meets with him to ponder over the design. However, one major issue she has is that they need to figure out how tall the cake should be. For aesthetic reasons, Marlene wants the corner of the table (where the cake will be placed) 25 inches away from the the edge of the top cake. She wants the table edge of the table to be 30 inches from quinceanera doll's eyes at the top of the cake. She wants the angle of the doll's eyes to the corner of the table plus the angle of the corner of the table to the edge of the top cake to be around 60 degrees. If they decide to make the cake 4 feet tall and put a 4 inch tall doll on top, find the angle of depression and elevation to see if 4 feet is close to getting the ideal angle that Marlene is thinking of.

I/D #2: Unit O Concept 7-8: Using the 30-60-90 triangle and using the 45-45-90 angle

30-60-90
To derive the pattern for the 30-60-90 triangles from an equilateral triangle with a side length of 1 we first need to know how an equilateral triangle is labeled. We have the information that it has the side length of 1, so since it is an EQUIlateral(meaning it has EQUAL sides) all three sides of the triangle will be one. Also, since it is equilateral triangle it means that its angles must be equal the same. This is when we absorb past knowledge about triangles. All angles of a triangle must equal to 180 when added. Therefore, in an equilateral triangle the angles will equal 60 each (60 times 3 equals 180). We can create two 30-60-90 triangle from an equilateral triangle by cutting it in half down the middle. We cut it in half down the middle first of all because we cut one of the 60 degrees angles of the equilateral into two 30 degrees, and second of all cutting a line straight down the middle will create two right angles, meaning they are 90 degrees (and now we have two 30-60-90).The lengths that the two 30 degrees angles are reflecting become 1/2 since we cut the equilateral triangle in half and its bottom side length of 1 is split to two halves. We see that the lengths reflected by the 90 degrees angles, the hypotenuse, are still.We do the Pythagorean Theorem, in relation to the special RIGHT triangles we have just formed, to get the side length of the line cutting straight down the middle. When we complete the Pythagorean Theorem we see that length is equal to radical 3 divided by 2. To get whole numbers for all our sides we multiply all the side values by 2, and the side reflected by the 60 degrees becomes radical 3, the side reflected by the 30 degrees becomes 1 and the side reflected by the 90 degrees becomes 2. We place the variable n in front of all these values to show that the pattern could be expanded. In other words, the length of 1 could be any other length, example 7, and the formula will still work.






http://t3.gstatic.com/images?q=tbn:ANd9GcSWUjC1X0IBBlOnbhexm2OgkwZW8m1w6N4IkOT6hzig-hn9nuvF:hs.doversherborn.org/hs/baroodyj/HonorsGeom/Class%2520Notes/Chapter%252011/Lesson11-5/EquilateralTriangle2.gif




45-45-90
We have a square with side lengths of 1. Since it is a square, all the four side lengths are equal to 1 and all the four angles will be right angles, meaning equal to 90 degrees. We cut the square diagonally to make two special right triangles of 45-45-90. (Two 90 degrees angles in the square are split to make four 45 degrees angles).Both sides reflected by both 45 degrees angles are still  length of 1 since we did not cut the square's side lengths and instead cut it through the middle diagonally.Now, this requires us to instead find the one missing length of the diagonal line, which represents the hypotenuse in the two special right triangles we formed. Again, we utilize the Pythagorean Theorem to find this missing value. When we complete the Pythagorean Theorem we find that the value is equal to radical 2. We label all the side lengths with the variable n because we show that the value could be expanded and the lengths do not have to equal 1 in order for the rules to work.







http://blog.powerscore.com/Portals/156640/images/satblog-5.jpg

1. “Something I never noticed before about special right triangles is…that the 30-60-90 was derived from an equilateral triangle. Now all its measurements for its rule makes sense.
2. “Being able to derive these patterns myself aids in my learning because…I am able to apply my knowledge so that when I do not remember exactly how the rules are for special right triangles, at least now I am able to know how to find them.

I/D#1: Unit N Concept after 6 and before 7: How do SRT and UC relate?

Special Right Triangles and the Unit Circle!

Heading for this Section Inquiry Activity Summary:

For the activity, we labeled the special right triangles.
For the 30 degrees triangle (which is the short one because it has the longer side down and the shorter side as its height) the hypotenuse/radius = r = 2x = 1 (because we are focusing on the Unit Circle which we know has a radius of 1). The short side (height) = y = x = 1/2, and this is because we divided r by 2x to make it equal 1 so we must also divide x by 2x to give it a value which is 1/2. the bottom side is x = x times radical 3 = radical 3 divided by 2. Given this information, we can also figure out the ordered pairs of this triangle assuming it is drawn out in the first quadrant of a graph. The point where the 30 degrees angles lies will of course be (0,0). Going to the next point lying on the x-axis, the ordered pair would be (radical 3 divided by 2 (x), 0). And the point lying above would be (radical 3 over 2(x), 1/2(y)). 

http://00.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_30.gif

For the 45 degrees angle the values change, except the r remains = to 1. However, you must know that we got the value of 1 by dividing x radical 2 by x radical 2 because 45 degrees has different rule,according to the Special Right Triangle rules, the hypotenuse is x radical 2. not 2x like in the 30 degrees angles. The value for x and y are the same because they are the same length according to the rules of SRT. This value is radical 2 over 2 (let us set the y and x values as both x and they both become radical 2 over 2 when we divided the x by the hypotenuse value of x radical 2). The point where the 45 degrees lies will of course also be (0,0). Going to the next point lying on the x-axis, the ordered pair would be (radical 2 over 2 (x),0). And the point lying above would be (radical 2 over 2 (x), radical 2 over 2 (y)).

http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image036.gif

For the 60 degrees angle it is very similar to the 30 degrees angle. Just like the angles switch from where there were lying (as in the 30 degrees is no longer on the vertex point of the graph, instead now it is the 60 degrees angle opening), the values will switch too. This just means that x = 1/2 now and y = radical 3 over 2. r will continue to = 1. Meanwhile, the point where the 60 degrees angle lies now will of course be (0,0). The point lying on the x axis will be (1/2 (x), 0). And the point lying above would be (1/2 (x),radical 3 over 2 (y)).

http://00.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_13.gif

This activity helped me to derive from the Unit Circle because I got a better understanding of where all the important values in the Unit Circle came from instead of just accepting them without expanded knowledge. Now I get a better sense of why and how we must learn many certain things in order for us to learn and understand others. This activity helped me have a better view of the 30, 45, and 60 degrees angles and fully comprehend the patterns that go along with them.

The triangles in this activity were all in the first quadrant but they may also lie on the second, third, and fourth quadrant, and refer to the degrees that they represent as reference angles. The number values of x and y do not necessarily change. Nevertheless, when we go from the first quadrant to the second one we will make sure that all the x values of each ordered pair of each angle are negative since the triangle is on the quadrant where the x-axis is negative while the y-axis is still going positive. On the third quadrant both the x and y values become negative and on the fourth quadrant only the y value becomes negative.

The below picture will give an example of how the 30, 60, and 45 degrees angles would look like in the quadrants II, III, and IV.

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBOzfx3dUOBGPrCKd12aLaE85rhswxJHDtn_zO-5usBtImcuEhiPG66Uo7w_FibbJWWa0Du17esK5BaD-Cw14-FEgOpdfDdDNrc6k1yZJr27s_88FRNv7t55RSrkjIK12HwCRyrb4OfUc/s1600/012.



The coolest thing I learned from this activity was that you can find the ordered pairs of those points of a unit circle. I never thought you can actually look at a point in a curved circle and know its ordered pair (like actual x and y values!)
This activity will help me in the unit because I will have a complete understanding of where and how to get the values of certain points in the unit circle. I will therefore, understand the sine, cosine, and tangent better since they have to do a lot with these angles and ordered pairs.
Something I never realized before about special right triangles and the unit circle is that the sides of the triangle are measured x and y for a purpose of showing us the ordered pair, I literally thought they were labeled like that just because. I also did not realize the connection between the SOH CAH TOA and the unit circle but now the abbreviations make actual sense.

 The Unit Circle at It's fullest!












http://t3.gstatic.com/images?q=tbn:ANd9GcR_fq7L0KBbyBoR514RtTD7Gf4NwC2FxBCPY-8S2EgBpovRuFj_:alvalxy.wikispaces.com/file/view/%25E6%259C%25AA%25E5%2591%25BD%25E5%2590%258D.jpg/306597904/%25E6%259C%25AA%25E5%2591%25BD%25E5%2590%258D.jpg

RWA: Unit M Concept 6- Hyperbolas

                                                                   Definition:
A hyperbola is a set of all points such that the difference in the distance from two points is a constant.

In a hyperbolas it matters which term (x or y) comes first in the equation because the equation of the hyperbola contains a minus instead of a plus like the ellipse. Therefore, the graph will look different based on which term (x or y) comes first.
The equation of the hyperbola can either be (x-h)^2/a^2 all minus (-!) (y-k)^2/b^2=1 or (y-k)^2/a^2 all minus (x-h)^2/b^2. Notice that the bottom terms (a and b) are not switched around and instead the a stays with the first term always and the b with the second term always. Also notice that when the x and y terms are switched the center values (h,k) also switch, meaning that the h value always matches the x value in the point and the k value always matches the y value in the point of the graph. DO NOT mix those two values when writing out the center for the graph.
http://t0.gstatic.com/images?q=tbn:ANd9GcTAP1wXzUwZiM2d8qshPZqrjVmj3GTa2FeHAOLsXfTbpKlsz5nr:mysite.du.edu/~jcalvert/math/hype1.gif
When x comes first in the equation you should already know that the graph will have the hyperbola going left and right. Also, this means that the transverse axis (main axis) is horizontal and for the vertices the x changes value and the y stays the same ((x,y) from the point of the center, which you get from (h,k)). Meanwhile, when y comes first in the equation the graph will have the hyperbola going up and down and the transverse axis will be vertical, meaning the vertices x will stay the same but the y value will change.
After figuring out these simple connections between the algebraic and graphic hyperbolas, it will be simple to discover other hyperbola elements such as the conjugate axis, the co-vertices and the foci. Remember that the foci of hyperbolas are outside the box drawn in the graph.
                    To have a better visual with the information given above please do not be a cheesebucket and check out video below:







                                                   REAL LIFE APPLICATIONS:
5. Cooling Towers of Nuclear Reactors - The hyperboloid is the design standard for all nuclear cooling towers. It is structurally sound and can be built with straight steel beams. When designing these cooling towers, engineers are faced with two problems: (1) the structutre must be able to withstand high winds and (2) they should be built with as little material as possible. The hyperbolic form solves both of these problems. For a given diameter and height of a tower and a given strength, this shape requires less material than any other form. A 500 foot tower can be made of a reinforced concrete shell only six or eight inches wide. See the pictures below (this nuclear power plant is located in Indiana). 

6. Stones in a Lake - When two stones are thrown simultaneously into a pool of still water, ripples move outward in concentric circles. These circles intersect in points which form a curve known as the hyperbola. 

Source: 
http://www-prod.pen.k12.va.us/Div/Winche...

http://t0.gstatic.com/images?q=tbn:ANd9GcQTbwo-QERHjE7sHXVRwPD7K_YE5lnNO3FSv2Nj0zuqhowGicuBSQ:education.ti.com/en/timath/~/media/Images/Activities/US/Math/Algebra%2520II/Exploring%2520Hyperbolas/ExploringHyperbolas3.jpg

To understand the picture above of the real world application for hyperbolas go ahead and check out the link below:

press here: http://www.pleacher.com/mp/mlessons/calculus/apphyper.html

Work Cited:

http://t0.gstatic.com/images?q=tbn:ANd9GcTAP1wXzUwZiM2d8qshPZqrjVmj3GTa2FeHAOLsXfTbpKlsz5nr:mysite.du.edu/~jcalvert/math/hype1.gif

http://t0.gstatic.com/images?q=tbn:ANd9GcQTbwo-QERHjE7sHXVRwPD7K_YE5lnNO3FSv2Nj0zuqhowGicuBSQ:education.ti.com/en/timath/~/media/Images/Activities/US/Math/Algebra%2520II/Exploring%2520Hyperbolas/ExploringHyperbolas3.jpg

http://www.pleacher.com/mp/mlessons/calculus/apphyper.html

<iframe width="500" height="375" src="http://www.schooltube.com/embed_force/2eec3f755fdc4a84a617/" frameborder="0" allowfullscreen="allowfullscreen" mozallowfullscreen="mozallowfullscreen" webkitallowfullscreen="webkitallowfullscreen"></iframe>

http://kirchmathanalysis.blogspot.com/p/unit-m.html

WPP #10: Unit L Concept 9-14

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