I/D #1: Unit N Concept 7: How do SRTs and the UC relate?

Inquiry Activity Summary: 
The activity we did in class had us recalling our knowledge of special right triangles from geometry. We discovered that these same right triangles are seen within our unit circle. Thus, by setting our radius to 1 (because a unit circle ALWAYS has a radius of 1), we can determine the lengths of each side of the triangle. As a result, we can then find the actual ordered pairs of the points of the triangles that create the angles of 30, 45, and 60 degrees.

(http://upload.wikimedia.org/wikipedia/commons/1/15/Triangle_30-60-90_rotated.png)

1. This is a 30 special right triangle. The side across from the 30 degrees (the shortest side) is equal to x. The side across from the 60 degrees (adjacent to the 30 degrees) is equal to  x√3. The side across from the 90 degree angle is equal to 2x. 

(https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiFw_yhWi1CV1w3RqOUrjPmoZ4KTj8Enp8JqJlpqLkJvTeYwXGRKzfpj0MRUIvUT9pBNS55uVLVffyPa7jigCyFXy9PoHL1nH7i_7B9MP4ROYjwGhXuZc4jKDB0XhyIThUL59BJtzcNFAk/s1600/special+right+triangle.jpg)

2. This is a 45 special right triangle. The side across from the 45 degrees is equal to x. Because we have two equal angles (another 45 degree), the side opposite of the other 45 degree (or adjacent to the 45 degree angle to the very left) is also x. The hypotenuse has the length of x√2. 

(https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiFw_yhWi1CV1w3RqOUrjPmoZ4KTj8Enp8JqJlpqLkJvTeYwXGRKzfpj0MRUIvUT9pBNS55uVLVffyPa7jigCyFXy9PoHL1nH7i_7B9MP4ROYjwGhXuZc4jKDB0XhyIThUL59BJtzcNFAk/s1600/special+right+triangle.jpg)

3. This is a 60 special right triangle. This triangle is basically like the 30 special right triangle,  but the 30 and 60 degrees have switched places. The side across the 60 degree is x√3. The side across the 30 degree is x. The side across the 90 degree is 2x. 

4. The use of these triangles enables us to understand why ordered pairs resemble things like (1/2, rad3/2). We now understand the meaning of these numbers: they are the values of the sides created by the angles. In order to find the values of all the sides of these triangles WITHIN a UC, we take all the values of the sides and divide by the hypotenuse. We divide the hypotenuse by itself in order to have it equal to 1. Because we divided the hypotenuse by itself, we must do the same to each of the other sides, and divide by the hypotenuse as well. 

(http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/
image022.gif)

This triangle is on a coordinate plane, a graph. Take into consideration of what is your x-axis and your y-axis and remember that your triangle begins at the CENTER (0,0). 

By now we know the values of each of the sides for a typical 30 degree special right triangle. Now, we want to find the length within a UC. As stated before, we need to divide each side by the hypotenuse. The hypotenuse is equal to 2x so 2x/2x=1. Our radius/hypotenuse is 1, which we want. Move on to the 30 degree side: that equals x, so x/2x (cancel out the x's) equals 1/2. Our vertical side (our "y" side) is equal to 1/2. Move on to the 60 degree side: x √3, so x√3/2x (x's cancel out) equals √3/2. Our horizontal side (our "x" side) is equal to √3/2. Now, looking at the blue point in the picture above, we now know that on the x-axis (because this is in a graph), the distance is √3/2. We also know that the distance, looking at the y-axis, to the point is 1/2. This is how we get the ordered pair (√3/2, 1/2).

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

Again, remember this is on a graph. Your angle begins at (0,0). Your horizontal distance is essentially your x value. Your vertical distance is your y value.

Divide each side by the hypotenuse once more. Your hypotenuse in a 45 degree angle equals x√2. The hypotenuse divided by itself is 1. Each side across the 45 degree (they are the same length) is x. Divide x by x√2. The x's cancel leaving 1/√2. Rationalize by multiplying the top and bottom by √2. Thus, each leg of the triangle in a UC is equal to √2/2. Now, with this knowledge, we can find the ordered pair of the point, shown above. The horizontal side/x value is equal to √2/2. The vertical side/y value is equal to √2/2. Taking these two to find our ordered pair, we are given the coordinate of (√2/2,√2/2). 

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

This is the same exact idea we have discussed in the two previous pictures (basically the 30 degree angle values but switched). We divide all sides by the hypotenuse (2x) and we get the values of 1/2 for the horizontal/x value side and √3/2 for the vertical/y value side. Again, we know this is on a graph, with the triangle beginning at the origin. We use these values to find the coordinate of the 60 degree triangle which is (1/2, √3/2). 

The values of the ordered pairs are explained through these special right triangles, basically the distance of one leg and another. It is also worth noting that for the quadrant angles (0, 90, 180, 270, & 360 degrees) have much simpler ordered pairs. The 0 degree does not create any angle and lies on the positive x-axis. Because we know that the radius is 1, we know that the ordered pair for the 0 degree is (1,0) because 1 is radius (on the x axis) and it lies on the x-axis (explaining the 0). Similar situations are found for the other quadrant angles: 90 degree has the ordered pair of (0,1), 180 degree has the ordered pair of (-1,0), 270 has the ordered pair of (0,-1), and 360 shares the same ordered pair as 0 degrees (they share the same coterminal side). 

5. These triangles lie within the first quadrant. However, we can redraw these triangles in the other quadrants but still find the same patterns seen in the first quadrant. 

(http://01.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_34.gif)

This picture displays the 30 degree triangle just in different quadrants. As you can see, they share the same values as we saw in the first quadrant. HOWEVER, looking closely, you should notice the negatives on certain values. In quadrant 2, your x value turns negative, as it should, considering that is the negative x-axis. In quadrant 3, both valeus are negative since you have the negative x-axis and the negative y-axis. Finally, quadrant 4 shows the ordered pair to have a negative y value because you have the negative y-axis. 

(http://01.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_45.gif)

This picture has a 45 degree triangle in quadrant 2. Again, we see the same values that we talked of in quadrant 1. However, in this quadrant, our x-value is negative because of our negative x-axis. 

(http://02.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_21.gif)

This picture displays the 60 degree triangle in quadrants 3 and 4. In quadrant 3, the values are the same as in the first quadrant, but both values are negative because it falls in the negative x and y axes. In quadrant 4, the values are the same as well, except only the y-value is negative because it falls in the negative y-axis. 

Inquiry Activity Reflection:

1. “The coolest thing I learned from this activity was…” how we were able to find the values of the ordered pairs through special right triangles.

2. “This activity will help me in this unit because…” it reasoned why we had the numbers that we had. I struggled with understanding why we had such numbers. I used to memorize but never completely understood the meaning of these numbers. Now, if I forget one, I can recall special right triangles and find it easily.

3. “Something I never realized before about special right triangles and the unit circle is…” that just by knowing the basics of these special right triangles, allows you to achieve a better understanding of the unit circle.

Work Cited:
1.http://upload.wikimedia.org/wikipedia/commons/1/15/Triangle_30-60-90_rotated.png
2. https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiFw_yhWi1CV1w3RqOUrjPmoZ4KTj8Enp8JqJlpqLkJvTeYwXGRKzfpj0MRUIvUT9pBNS55uVLVffyPa7jigCyFXy9PoHL1nH7i_7B9MP4ROYjwGhXuZc4jKDB0XhyIThUL59BJtzcNFAk/s1600/special+right+triangle.jpg
3. http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_
RESOURCE/U19_L1_T3_text_final_3_files/image022.gif
4. http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE
/U19_L1_T3_text_final_3_files/image036.gif
5. http://00.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_13.gif
6. http://01.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_34.gif
7. http://01.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_45.gif
8. http://02.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_21.gif

RWA# 1: Unit M Concept 6: Graphing Hyperbolas Given Equations and Defining All Parts

1. "The set of all points such that the difference in the distance from two points is a constant." (Kirch)
2. Algebraically, a hyperbola's equation can either look like this: . Or it can look something like this: . (I will go into further detail of what all of these variables mean later on.) If you were to graph these equations, a hyperbola would look like two parabolas opposite of one another, facing in different directions. Because there are two ways the equation can be written, the hyperbola may appear differently as well, depending on which of the two terms ("x" or "y") is negative. If your "y" term is subtracted, your hyperbola will branch to the left and right. If your "x" term is subtracted/negative, your hyperbola will branch up and down. 

"y" term is negative
(http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx)

"x" term is negative
(http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx

      When discussing hyperbolas, we need to remember these features: center, transverse axis, conjugate axis, vertices, co-vertices, foci, eccentricity, and asymptotes. To begin, to find the center, we need to first remember something very crucial: "h" goes with "x" and "k" goes with "y". Knowing this, we can look at our standard form and find our "h" and our "k", basically allowing us to find our center (h,k). You can plot the center, and go on from there. 
      Moving on, looking at your standard form, whichever term ("x" or "y") comes first will determine whether your transverse axis is horizontal or vertical. If "y" is the first term, that tells us that our transverse axis will be vertical. If "x" is the first term, that tells us that our transverse axis will be horizontal. Knowing this, we will know that our conjugate axis must be the opposite. So if our transverse axis is vertical, our conjugate axis will be horizontal and vice versa. (Side note: By knowing if the axes are horizontal or vertical, we will know if our lines will be "x=" or "y=". And because we know our axes go through our center, we know those "x=h" and "y=k". Less work for us!)The distance of the transverse axis is 2a ("a" is the square root of the denominator beneath the first term) and the distance of the conjugate is 2b ("b" is the square root of the denominator beneath the second term). If you want to know which lines are your transverse and conjugate axes just by looking at a graph, take a look at the branches. If your branches go left and right, you know your transverse axis is horizontal. If your branches go up and down, you know your transverse axis is vertical. If you want to know the length of either of the axes, simply count the units (to find "a" or "b", divide the entire length of the axis by 2). 
       Going on to vertices, these are fairly easy to find. Graphically, the vertices are basically the ends of the transverse axis. Look at the end points and you have the vertices. Algebraically, all you need to know is what your "a" is. If your transverse axis is "x=" then you know that the "x's" in your vertices will remain the same as your center. From there, all you need to do is add and subtract "a" to "k". {ex: center: (0,0); trans: x=0; vertices: (0,0+a) & (0,0-a)} The same can be said for the co-vertices. Look for the end points of the conjugate axis on your graph. Algebraically, find "b" and add that to the value that is changing ("x" or "y")(HINT: If your "y" values changed for your vertices, your "x" values will change for your co-vertices). Your foci is the same idea, with the adding to either the "x" or "y" values of your center (because your foci are "c" units away from your center, correct?). You find "c" by using this formula: a² + b² = c^2. After finding "c", you add that to the SAME VALUE ("x" or "y") that you added for your vertices because your foci lay on the same line (transverse axis) as your vertices. So if your "x" values did not change for your vertices, then your "x" values will not change for your foci. You will take the "k" value of the center and add and subtract "c" from it to find both foci. To plot, your foci will be going the same direction as your vertices, and further than your vertices in fact (I will explain soon). If your transverse axis is horizontal, your foci will be going horizontal. 
        After all of this, now you need to calculate the eccentricity, how much the conic deviates from the shape of a circle. To find eccentricity, you divide "c" by "a" (c/a). The foci has a great effect on the eccentricity. The greater the foci is (as in the farther it is from the center), the more it deviates from being a circle, the greater the eccentricity is. The greater your eccentricity is, the wider it is; the less your eccentricity is, the "skinnier" it is. 
        Finally, let's talk about asymptotes (they are the diagonals of the box we create from the conjugate and transverse). We have given formulas to use to find them: (used when transverse is horizontal) and (when transverse is vertical). You take your variables and plug them in. However, you can also use our knowledge from algebra 1 and use our slope-intercept formula: y=mx+b. Our slope (m) would be either a/b (if our transverse is vertical because our "a" is our "y"/ our "rise", and our "b" is our "x"/ our "run") or b/a (if our transverse is horizontal). From there, you plug in our center (x,y) (the center goes through our lines) and find 'b". REMEMBER, because we have two asymptotes, one slope should be POSITIVE a/b or b/a (it depends on the graph) while the other should be the NEGATIVE of that. You can also count and find the slope on the graph. 

hyperbola and its features
(https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjToBqnI6fP8DDSxy9Ae6udUE1V6aukjRmQAEfbJTcyZaIN1HRvNd8ukmy6sFXLG0RpCuH7hxXR_JEpfTzaU5dr1on3yfIg9bbifyNswsOSSr2Yk3HnLanhiGEEwpKooLBBA1k80elZBQg/s1600/kirch.png)

3. Real world applications of hyperbolas can be seen in cooling towers for nuclear reactors. These towers were built into the shape of hyperboloids (hyperbolas in 3D!) on purpose. When building these, there were problems in creating a building that would be able to withstand high winds and intense conditions without using that much material. Finally, engineers discovered the benefits of creating these towers into the shape of a hyperbola. 

    By building these towers into hyperboloids, they are able to stand up to high winds and use less material. "A 500 foot tower can be made of a reinforced concrete shell only six to eight inches wide." (http://www.pleacher.com/mp/mlessons/calculus/apphyper.html) Also, they help the upward air flow move faster, cooling things faster and better. Before, these towers were built in a cylindrical form but they had encountered problems with the shape. 

Works Cited:

1. http://www.purplemath.com/modules/hyperbola.htm

2. https://people.richland.edu/james/lecture/m116/conics/hypdef.html

3. http://www.structuremag.org/article.aspx?articleID=326

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

5. http://www.youtube.com/watch?v=Z6cwpsDC_5A

6. http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx