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

INQUIRY ACTIVITY SUMMARY
1.) 30º TRIANGLE

Well to start we have to label all the sides. The hypotenuse has to equal to 1, and due to the rules of the 30º,60º,90º right triangle the side that is adjacent to 30º (x) has a value of x radical 3, the side opposite (y) has a value of x and the hypotenuse (r) as 2x. Then we divided by 2x in the hypotenuse,  since we did that we must do the same and then simplify the other sides by 2x . So x-radical 3/2x turns to radical 3/2 which is your x value. Then we do the same for our y value and x/2x turns to 1/2 and that is your y value. Then we draw the coordinate plane on the triangle.The origin is (0,0), while  going across the x axis is (radical 3/2,0) and then going directly up from the previous coordinate is (radical 3/2,1/2).

2.)45º TRIANGLE

For the 45º triangle you have to label it accordingly by the special rule for the 45º.  (x) is the side adjacent to the given degree with value of x, the (y) is opposite to the angle but similar to the adjacent side too so it has the same value of x and the hypotenuse, (r) is equaled to 1 still and is x radical 2. Then you are suppose to divide the other two sides by x radical 2 so x/x radical 2 is 1/radical 2 but we could NEVER leave a radical in the denominator so we multiply it by radical 2 and it should be radical 2/ 2. Since the x and y have the same value y would have to be also radical 2/2. Then draw the coordinate plane of the triangle. You start at the origin (0,0) then go to the right and that is (radical 2/2,0) then go up from the previous coordinate and it would be (radical 2/2,radical2/2).

3.) 60º TRIANGLE

The 60º triangle is exactly the same as the 30º triangle. (x) is adjacent but has a value of x, (y) would be the opposite angle with a value of  x radical 3 and the hypotenuse (r) still equal to 1 and is 2x. Then divide the other two sides by 2x like the 30º triangle and x/2x would be 1/2 the value of x, for the value of y it is x radical 3/2x and that would be radical 3/2. When drawing it in the coordinate plane you start at the origin (0,0) then move alongside to the x and it is (1/2,0) then go up with the previous coordinate plane and that is (1/2, radical 3/2).

4.) HOW DOES THIS ACTIVITY HELP YOU TO DERIVE THE UNIT CIRCLE?

This activity helps me derive the unit circle by figuring out the certain points and if you know these three you can know the rest of the other points but you need to know there signs if its going to be negative or positive.

5.)

Depending on which quadrant the SRT is found, the value will change by its sign positive or negative. If the triangle is found on Quadrant II, then x from the coordinate pair will be negative. In Quadrant III, both x and y will be negative. Finally in Quadrant IV, only y will be negative. In Quadrant I, all trig functions are positive, while in Quadrant II only sine and cosecant are positive and the rest negative. Meanwhile in Quadrant III, tangent and cotangent are the only positive. Finally in the Quadrant IV, only cosine and secant are positive. you can also memorize it by All Students Take Calculus.

6.) INQUIRY ACTIVITY REFLECTION

"THE COOLEST THING I LEARNED FROM THIS ACTIVITY WAS..." well i took cp Algebra ll so when i saw this i didnt know what this was at all as in the Unit Circle but the activity was Geometry so i knew that and learned that this activity was actually the first three in the unit circle in quadrant l.

" THIS ACTIVITY WILL HELP ME IN THIS UNIT BECAUSE..." I noticed that if we know the first quadrant then we can fill out the whole unit circle that is amazing and to help you more you can sing the song to get the ordered pairs and other methods as well.

"SOMETHING I NEVER REALIZED BEFORE ABOUT SPECIAL RIGHT TRIANGLES AND THE UNIT CIRCLE IS..." That i never knew that they were so related to one another. i didnt know that the coordinates can be different shapes as well.

RWA#1: Unit M Concpet 4-6- Conic Sections in Real LIfe

1. Parabola- Set of all points equal distance from a focus and the directrix
2.Algebraically

Graphically










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The parabola has FOUR key features: vertex, directrix, focus, and the axis of symmetry. The vertex is the point at which the parabola starts. The directrix is a line outside the parabola that is perpendicular to the axis of symmetry. The axis of symmetry is a line that breaks the graph into two equal sections. The focus is the point inside the parabola. Lastly the distance from the focus to the vertex and the vertex to the directrix is the same and known as "p". 
For graphing the Parabola you need to make sure the formula is in standard form. After you have done that you can find your parts you know what the vertex is it is (h,k). Then plot the focus which you should know is p by now and you plot it away from the vertex meaning it should be inside the parabola. To know if p is going up, down, left or right you need to know that when "p" is positive and the equation is x^2 it goes up, if "p" is positive and equation is y^2 it goes right, if "p" is negative and equation is x^2 it opens down, if "p" is negative and the equation is y^2 it opens left. the effect the focus does to the shape is the distance from the vertex to the Focus which determines how skinny or fat the parabola is going to be. The closer the focus is to the vertex the fatter the graph will be, the further away the focus is from the vertex the skinnier the graph will be.The eccentricity of the parabola is equal to 1 all the time.You then put in the directrix which is a line p units away from the vertex in the opposite direction of the focus. Then put in the axis of symmetry which is perpendicular to the directrix. Notice that the vertex and the focus are both going to be ordered pairs while the directrix and the axis of symmetry are going to be lines.
3. RWA

http://upload.wikimedia.org/wikipedia/commons/4/48/Headlight_lens_optics_schematic.png The cars headlights can be a great example of having a parabolic shape."The light is placed in the focus of a parabolic mirror, as the light travels and meets the mirror, it is reflected in lines parallel to the axis . This is why the light beam from the headlights of cars are so strong."( http://www3.ul.ie/~rynnet/swconics/UP.htm) source should be the vertex that bounces all of the dispersed lights from the reflector in a straight line straight forward.

  "Parabolic reflector has the property that if a light source is placed at the focus of the reflector then the light rays will reflect from the reflector as rays that is parallel to the axis. To have a safe night drive you don't want all the light rays to be parallel to the axis. Some of the light must be aimed far down the road to the side upward for signs or bridges."(http://www.pleacher.com/mp/mlessons/calculus/appparab.html)

Here it talks more about the headlight and how you have to find the right vertex so that when it bounces all the rays stay straight.

4. Works Cited Page
 http://upload.wikimedia.org/wikipedia/commons/4/48/Headlight_lens_optics_schematic.png
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Car Headlights

Headlights